$int_0^{pi/2}log^2(cos^2x)mathrm{d}x=frac{pi^3}6+2pilog^2(2)$???












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I saw in a paper by @Jack D'aurizio the following integral
$$I=int_0^{pi/2}log^2(cos^2x)mathrm{d}x=frac{pi^3}6+2pilog^2(2)$$
Below is my attempt.



$$I=4int_0^{pi/2}log^2(cos x)mathrm{d}x$$
Then we define
$$F(a)=int_0^{pi/2}log^2(acos x)mathrm{d}x$$
So we have
$$F'(a)=frac2aint_0^{pi/2}log(acos x)mathrm{d}x$$
Which I do not know how to compute. How do I proceed? Thanks.






integration







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    $begingroup$
    "The trap is set, now we wait (for Jack)."
    $endgroup$
    – Clement C.
    Dec 30 '18 at 8:47






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    $begingroup$
    Substitute $cos x=t$ and use the derivatives of Beta function.
    $endgroup$
    – Kemono Chen
    Dec 30 '18 at 8:50
















4












$begingroup$


I saw in a paper by @Jack D'aurizio the following integral
$$I=int_0^{pi/2}log^2(cos^2x)mathrm{d}x=frac{pi^3}6+2pilog^2(2)$$
Below is my attempt.



$$I=4int_0^{pi/2}log^2(cos x)mathrm{d}x$$
Then we define
$$F(a)=int_0^{pi/2}log^2(acos x)mathrm{d}x$$
So we have
$$F'(a)=frac2aint_0^{pi/2}log(acos x)mathrm{d}x$$
Which I do not know how to compute. How do I proceed? Thanks.






integration







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    "The trap is set, now we wait (for Jack)."
    $endgroup$
    – Clement C.
    Dec 30 '18 at 8:47






  • 2




    $begingroup$
    Substitute $cos x=t$ and use the derivatives of Beta function.
    $endgroup$
    – Kemono Chen
    Dec 30 '18 at 8:50














4












4








4





$begingroup$


I saw in a paper by @Jack D'aurizio the following integral
$$I=int_0^{pi/2}log^2(cos^2x)mathrm{d}x=frac{pi^3}6+2pilog^2(2)$$
Below is my attempt.



$$I=4int_0^{pi/2}log^2(cos x)mathrm{d}x$$
Then we define
$$F(a)=int_0^{pi/2}log^2(acos x)mathrm{d}x$$
So we have
$$F'(a)=frac2aint_0^{pi/2}log(acos x)mathrm{d}x$$
Which I do not know how to compute. How do I proceed? Thanks.






integration







share|cite|improve this question









$endgroup$




I saw in a paper by @Jack D'aurizio the following integral
$$I=int_0^{pi/2}log^2(cos^2x)mathrm{d}x=frac{pi^3}6+2pilog^2(2)$$
Below is my attempt.



$$I=4int_0^{pi/2}log^2(cos x)mathrm{d}x$$
Then we define
$$F(a)=int_0^{pi/2}log^2(acos x)mathrm{d}x$$
So we have
$$F'(a)=frac2aint_0^{pi/2}log(acos x)mathrm{d}x$$
Which I do not know how to compute. How do I proceed? Thanks.






integration




integration






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asked Dec 30 '18 at 8:44









clathratusclathratus

4,715337




4,715337








  • 3




    $begingroup$
    "The trap is set, now we wait (for Jack)."
    $endgroup$
    – Clement C.
    Dec 30 '18 at 8:47






  • 2




    $begingroup$
    Substitute $cos x=t$ and use the derivatives of Beta function.
    $endgroup$
    – Kemono Chen
    Dec 30 '18 at 8:50














  • 3




    $begingroup$
    "The trap is set, now we wait (for Jack)."
    $endgroup$
    – Clement C.
    Dec 30 '18 at 8:47






  • 2




    $begingroup$
    Substitute $cos x=t$ and use the derivatives of Beta function.
    $endgroup$
    – Kemono Chen
    Dec 30 '18 at 8:50








3




3




$begingroup$
"The trap is set, now we wait (for Jack)."
$endgroup$
– Clement C.
Dec 30 '18 at 8:47




$begingroup$
"The trap is set, now we wait (for Jack)."
$endgroup$
– Clement C.
Dec 30 '18 at 8:47




2




2




$begingroup$
Substitute $cos x=t$ and use the derivatives of Beta function.
$endgroup$
– Kemono Chen
Dec 30 '18 at 8:50




$begingroup$
Substitute $cos x=t$ and use the derivatives of Beta function.
$endgroup$
– Kemono Chen
Dec 30 '18 at 8:50










5 Answers
5






active

oldest

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$begingroup$

Let $$I(a)=int_0^{frac {pi}{2}} (cos^2 x)^a dx$$



Hence we need $I''(0)$.



Now recalling the definition of Beta function we get $$I(a)=frac 12 Bleft(a+frac 12 ,frac 12right)=frac {sqrt {pi}}{2}frac {Gammaleft(a+frac 12right)}{Gamma(a+1)}$$



Hence we have $$I''(a) =frac {sqrt {pi}}{2}frac {Gammaleft(a+frac 12right)}{Gamma(a+1)}left(left[psi^{(0)}left(a+frac 12 right)-psi^{(0)}(a+1)right]^2 +psi^{(1)}left(a+frac 12 right)-psi^{(1)}(a+1)right) $$



Substituting $a=0$ in above formula yields the answer.






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    $begingroup$

    begin{align}J=int_0^{frac{pi}{2}} ln^2left(cos xright),dxend{align}



    Observe that,



    begin{align}I&=4J\
    J&=int_0^{frac{pi}{2}} ln^2left(sin xright),dx\
    int_0^{frac{pi}{2}} lnleft(sin xright),dx&=int_0^{frac{pi}{2}} lnleft(cos xright),dx
    end{align}



    (change of variable $y=dfrac{pi}{2}-x$ )



    begin{align}
    K&=int_0^{frac{pi}{2}} ln^2 left(2sin xcos xright),dx\
    &=int_0^{frac{pi}{2}} ln^2 left(sinleft(2xright)right),dx\
    end{align}



    Perform the change of variable $y=2x$,



    begin{align}
    K&=frac{1}{2}int_0^{pi} ln^2 left(sin xright),dx\
    &=frac{1}{2}int_0^{frac{pi}{2}} ln^2 left(sin xright),dx+frac{1}{2}int_{frac{pi}{2}}^pi ln^2 left(sin xright),dx\
    end{align}



    In the latter integral perform the change of variable $y=dfrac{pi}{2}-x$ and recall $sinleft(pi-xright)=sin x$ for $x$ real,



    begin{align}
    K&=int_0^{frac{pi}{2}} ln^2 left(sin xright),dx\
    &=int_0^{frac{pi}{2}} ln^2 left(cos xright),dx\
    &=J
    end{align}



    On the other hand,



    begin{align}
    K&=int_0^{frac{pi}{2}}left(ln 2+ln(sin x)+ln(cos x)right)^2 ,dx\
    &=frac{pi}{2}ln^2 2+int_0^{frac{pi}{2}}ln^2(sin x),dx+int_0^{frac{pi}{2}}ln^2(cos x),dx+2ln 2int_0^{frac{pi}{2}}ln(cos x),dx+\
    &2ln 2int_0^{frac{pi}{2}}ln(sin x),dx+2int_0^{frac{pi}{2}}ln(sin x)ln(cos x),dx\
    &=frac{pi}{2}ln^2 2+2J+4ln 2int_0^{frac{pi}{2}}ln(cos x),dx+2int_0^{frac{pi}{2}}ln(sin x)ln(cos x),dx\
    end{align}



    begin{align}L&=int_0^infty frac{ln^2 x}{1+x^2},dxend{align}



    Perform the change of variable $x=tan y$,



    begin{align}L&=int_0^{frac{pi}{2}} ln^2left(tan xright),dx\
    &=int_0^{frac{pi}{2}}left(lnleft(sin xright)-lnleft(cos xright)right)^2,dx\
    &=2J-2int_0^{frac{pi}{2}}ln(sin x)ln(cos x),dx\
    end{align}



    Therefore,



    begin{align}K+L&=frac{pi}{2}ln^2 2+4J+4ln 2int_0^{frac{pi}{2}}ln(cos x),dx
    end{align}



    Therefore (recall $K=J$),



    begin{align}J&=frac{1}{3}L-frac{pi}{6}ln^2 2-frac{4}{3}ln 2int_0^{frac{pi}{2}}ln(cos x),dxend{align}



    On the other hand,



    begin{align}L&=int_0^1 frac{ln^2 x}{1+x},dx+int_1^infty frac{ln^2 x}{1+x},dxend{align}



    In the latter integral perform the change of variable $y=dfrac{1}{x}$,



    begin{align}L&=2int_0^1 frac{ln^2 x}{1+x},dxend{align}



    But it is well known that,



    begin{align}int_0^1 frac{ln^2 x}{1+x^2},dx&=frac{pi^3}{16}\
    int_0^{frac{pi}{2}}ln(cos x),dx&=-frac{1}{2}piln 2
    end{align}



    Therefore,



    begin{align}J&=frac{pi^3}{24}-frac{pi}{6}ln^2 2-frac{4}{3}ln 2times -frac{1}{2}piln 2\
    &=boxed{frac{pi^3}{24}+frac{1}{2}piln^2 2}\
    end{align}



    PS:
    See: https://math.stackexchange.com/a/2942594/186817



    (in this post i assume only the value of $zeta(4)$ )






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      2












      $begingroup$

      Differentiation of the Beta function is a viable approach, but since we are dealing with a squared logarithm and not higher powers, it is faster to just exploit the Fourier (cosine) series of $log(cos^2theta)$ and Parseval's theorem. Given



      $$-log(cos^2theta)=2log 2+2sum_{kgeq 1}frac{(-1)^k}{k}cos(2ktheta),tag{1} $$
      since $int_{0}^{pi/2}cos(2jx)cos(2kx),dx =frac{pi}{4}delta(j,k)$, we immediately have
      $$ int_{0}^{pi/2}log^2(cos^2theta),dtheta = 2pilog^2 2+pizeta(2).tag{2}$$



      Since the LHS equals $int_{0}^{1}frac{4log^2 x}{sqrt{1-x^2}},dx$, we have just found the value of the hypergeometric series
      $$ 8sum_{ngeq 0}frac{binom{2n}{n}}{4^n(2n+1)^3}=8cdotphantom{}_4 F_3left(tfrac{1}{2},tfrac{1}{2},tfrac{1}{2},tfrac{1}{2};tfrac{3}{2},tfrac{3}{2},tfrac{3}{2};1right)tag{3} $$
      which also equals $int_{0}^{+infty}frac{log^2(1+t^2)}{1+t^2},dt$ or $frac{1}{2}int_{0}^{1}frac{log^2(x)}{sqrt{x(1-x)}},dx$.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Yeah you're gonna have to explain this a bit more
        $endgroup$
        – clathratus
        Dec 30 '18 at 9:51










      • $begingroup$
        @ClementC.: of course, now fixed.
        $endgroup$
        – Jack D'Aurizio
        Dec 30 '18 at 10:44










      • $begingroup$
        OK, I have a dumb question. You get $frac{pi}{2}sum_{kgeq 1} left(frac{1}{k}right)^2=2pizeta(2)$ for the second term by Parseval, don't you? (or did I mess up with one factor there?)
        $endgroup$
        – Clement C.
        Dec 30 '18 at 10:57








      • 1




        $begingroup$
        @ClementC. the factor $2$ in $(1)$ gives that $$int_{0}^{pi/2}left(2sum_{kgeq 1}frac{(-1)^k}{k}cos(2ktheta)right)^2,dtheta = 4cdotfrac{pi}{4}cdotzeta(2)=pizeta(2).$$
        $endgroup$
        – Jack D'Aurizio
        Dec 30 '18 at 14:11










      • $begingroup$
        @JackDAurizio Ah. Thanks.
        $endgroup$
        – Clement C.
        Dec 30 '18 at 16:10



















      1












      $begingroup$

      Getting rid of the first roadblock:



      Use the fact that $$log(acos x) = log a + logcos x$$
      along with the result that
      $$
      int_0^{pi/2} logcos x,dx = -frac{pi}{2}log 2
      $$
      to get
      $$
      F'(a) = frac{pilog a}{a} -frac{pi}{2a}log 2
      $$






      share|cite|improve this answer









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      • $begingroup$
        Of course, one of the proofs of the intermediary result is due to Jack D'Aurizio.
        $endgroup$
        – Clement C.
        Dec 30 '18 at 8:58



















      1












      $begingroup$

      $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
      newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
      newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
      newcommand{dd}{mathrm{d}}
      newcommand{ds}[1]{displaystyle{#1}}
      newcommand{expo}[1]{,mathrm{e}^{#1},}
      newcommand{ic}{mathrm{i}}
      newcommand{mc}[1]{mathcal{#1}}
      newcommand{mrm}[1]{mathrm{#1}}
      newcommand{pars}[1]{left(,{#1},right)}
      newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
      newcommand{root}[2]{,sqrt[#1]{,{#2},},}
      newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
      newcommand{verts}[1]{leftvert,{#1},rightvert}$




      $ds{I equiv
      int_{0}^{pi/2}ln^{2}pars{cos^{2}pars{x}},dd x =
      {pi^{3} over 6} + 2piln^{2}pars{2}: {LARGE ?}}$      .




      begin{align}
      I & equiv
      bbox[10px,#ffd]{int_{0}^{pi/2}ln^{2}pars{cos^{2}pars{x}},dd x}
      ,,,stackrel{x mapsto pi/2 - x}{=},,,
      4int_{0}^{pi/2}ln^{2}pars{sinpars{x}},dd x
      \[5mm] & =
      left.4,Reint_{x = 0}^{x = pi/2}ln^{2}pars{{1 - z^{2} over 2z},ic},{dd z over ic z}
      ,rightvert_{ z = exppars{ic x}}
      \[5mm] & =
      left.4,Imint_{x = 0}^{x = pi/2}ln^{2}pars{{1 - z^{2} over 2z},ic},{dd z over z}
      ,rightvert_{ z = exppars{ic x}}
      \[1cm] & stackrel{mrm{as} epsilon to 0^{+}}{sim},,,
      -, 4,
      overbrace{Imint_{1}^{epsilon}ln^{2}
      pars{1 + y^{2} over 2y},{dd y over y}}^{ds{= large 0}}
      \[2mm] & phantom{stackrel{mrm{as} epsilon to 0^{+}}{sim},,,,,,,} -
      4,Imint_{pi/2}^{0}bracks{lnpars{1 over 2epsilon} + pars{{pi over 2} - theta}ic}^{2},ic,ddtheta
      \[2mm] & phantom{stackrel{mrm{as} epsilon to 0^{+}}{sim},,,,,,,,}
      -4,Imint_{epsilon}^{1}bracks{lnpars{1 - x^{2} over 2x} + {pi over 2},ic}^{2}{dd x over x}
      \[1cm] & =
      4int_{0}^{pi/2}bracks{ln^{2}pars{2epsilon} -
      pars{{pi over 2} - theta}^{2}},ddtheta
      -4piint_{epsilon}^{1}lnpars{1 - x^{2} over 2x}
      {dd x over x}
      \[5mm] & stackrel{mrm{as} epsilon to 0^{+}}{sim}
      bracks{2piln^{2}pars{2epsilon} - {pi^{3} over 6}} -
      bracks{4piint_{0}^{1}{lnpars{1 - x^{2}} over x},dd x -
      4piint_{epsilon}^{1}{lnpars{2x} over x},dd x}
      \[5mm] & =
      bracks{2piln^{2}pars{2epsilon} - {pi^{3} over 6}} -
      bracks{2piint_{0}^{1}{lnpars{1 - x} over x},dd x +
      2pilnpars{epsilon}lnpars{4epsilon}}
      \[5mm] & stackrel{mrm{as} epsilon to 0}{to}
      bbx{{pi^{3} over 6} + 2piln^{2}pars{2}} approx
      8.1865
      end{align}




      Note that
      $ds{int_{0}^{1}{lnpars{1 - x} over x},dd x =
      -,mrm{Li}pars{1} = -,{pi^{2} over 6}}$      .







      share|cite|improve this answer









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        5 Answers
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        5 Answers
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        active

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        3












        $begingroup$

        Let $$I(a)=int_0^{frac {pi}{2}} (cos^2 x)^a dx$$



        Hence we need $I''(0)$.



        Now recalling the definition of Beta function we get $$I(a)=frac 12 Bleft(a+frac 12 ,frac 12right)=frac {sqrt {pi}}{2}frac {Gammaleft(a+frac 12right)}{Gamma(a+1)}$$



        Hence we have $$I''(a) =frac {sqrt {pi}}{2}frac {Gammaleft(a+frac 12right)}{Gamma(a+1)}left(left[psi^{(0)}left(a+frac 12 right)-psi^{(0)}(a+1)right]^2 +psi^{(1)}left(a+frac 12 right)-psi^{(1)}(a+1)right) $$



        Substituting $a=0$ in above formula yields the answer.






        share|cite|improve this answer









        $endgroup$


















          3












          $begingroup$

          Let $$I(a)=int_0^{frac {pi}{2}} (cos^2 x)^a dx$$



          Hence we need $I''(0)$.



          Now recalling the definition of Beta function we get $$I(a)=frac 12 Bleft(a+frac 12 ,frac 12right)=frac {sqrt {pi}}{2}frac {Gammaleft(a+frac 12right)}{Gamma(a+1)}$$



          Hence we have $$I''(a) =frac {sqrt {pi}}{2}frac {Gammaleft(a+frac 12right)}{Gamma(a+1)}left(left[psi^{(0)}left(a+frac 12 right)-psi^{(0)}(a+1)right]^2 +psi^{(1)}left(a+frac 12 right)-psi^{(1)}(a+1)right) $$



          Substituting $a=0$ in above formula yields the answer.






          share|cite|improve this answer









          $endgroup$
















            3












            3








            3





            $begingroup$

            Let $$I(a)=int_0^{frac {pi}{2}} (cos^2 x)^a dx$$



            Hence we need $I''(0)$.



            Now recalling the definition of Beta function we get $$I(a)=frac 12 Bleft(a+frac 12 ,frac 12right)=frac {sqrt {pi}}{2}frac {Gammaleft(a+frac 12right)}{Gamma(a+1)}$$



            Hence we have $$I''(a) =frac {sqrt {pi}}{2}frac {Gammaleft(a+frac 12right)}{Gamma(a+1)}left(left[psi^{(0)}left(a+frac 12 right)-psi^{(0)}(a+1)right]^2 +psi^{(1)}left(a+frac 12 right)-psi^{(1)}(a+1)right) $$



            Substituting $a=0$ in above formula yields the answer.






            share|cite|improve this answer









            $endgroup$



            Let $$I(a)=int_0^{frac {pi}{2}} (cos^2 x)^a dx$$



            Hence we need $I''(0)$.



            Now recalling the definition of Beta function we get $$I(a)=frac 12 Bleft(a+frac 12 ,frac 12right)=frac {sqrt {pi}}{2}frac {Gammaleft(a+frac 12right)}{Gamma(a+1)}$$



            Hence we have $$I''(a) =frac {sqrt {pi}}{2}frac {Gammaleft(a+frac 12right)}{Gamma(a+1)}left(left[psi^{(0)}left(a+frac 12 right)-psi^{(0)}(a+1)right]^2 +psi^{(1)}left(a+frac 12 right)-psi^{(1)}(a+1)right) $$



            Substituting $a=0$ in above formula yields the answer.







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            share|cite|improve this answer










            answered Dec 30 '18 at 12:04









            DarkraiDarkrai

            6,2971441




            6,2971441























                2












                $begingroup$

                begin{align}J=int_0^{frac{pi}{2}} ln^2left(cos xright),dxend{align}



                Observe that,



                begin{align}I&=4J\
                J&=int_0^{frac{pi}{2}} ln^2left(sin xright),dx\
                int_0^{frac{pi}{2}} lnleft(sin xright),dx&=int_0^{frac{pi}{2}} lnleft(cos xright),dx
                end{align}



                (change of variable $y=dfrac{pi}{2}-x$ )



                begin{align}
                K&=int_0^{frac{pi}{2}} ln^2 left(2sin xcos xright),dx\
                &=int_0^{frac{pi}{2}} ln^2 left(sinleft(2xright)right),dx\
                end{align}



                Perform the change of variable $y=2x$,



                begin{align}
                K&=frac{1}{2}int_0^{pi} ln^2 left(sin xright),dx\
                &=frac{1}{2}int_0^{frac{pi}{2}} ln^2 left(sin xright),dx+frac{1}{2}int_{frac{pi}{2}}^pi ln^2 left(sin xright),dx\
                end{align}



                In the latter integral perform the change of variable $y=dfrac{pi}{2}-x$ and recall $sinleft(pi-xright)=sin x$ for $x$ real,



                begin{align}
                K&=int_0^{frac{pi}{2}} ln^2 left(sin xright),dx\
                &=int_0^{frac{pi}{2}} ln^2 left(cos xright),dx\
                &=J
                end{align}



                On the other hand,



                begin{align}
                K&=int_0^{frac{pi}{2}}left(ln 2+ln(sin x)+ln(cos x)right)^2 ,dx\
                &=frac{pi}{2}ln^2 2+int_0^{frac{pi}{2}}ln^2(sin x),dx+int_0^{frac{pi}{2}}ln^2(cos x),dx+2ln 2int_0^{frac{pi}{2}}ln(cos x),dx+\
                &2ln 2int_0^{frac{pi}{2}}ln(sin x),dx+2int_0^{frac{pi}{2}}ln(sin x)ln(cos x),dx\
                &=frac{pi}{2}ln^2 2+2J+4ln 2int_0^{frac{pi}{2}}ln(cos x),dx+2int_0^{frac{pi}{2}}ln(sin x)ln(cos x),dx\
                end{align}



                begin{align}L&=int_0^infty frac{ln^2 x}{1+x^2},dxend{align}



                Perform the change of variable $x=tan y$,



                begin{align}L&=int_0^{frac{pi}{2}} ln^2left(tan xright),dx\
                &=int_0^{frac{pi}{2}}left(lnleft(sin xright)-lnleft(cos xright)right)^2,dx\
                &=2J-2int_0^{frac{pi}{2}}ln(sin x)ln(cos x),dx\
                end{align}



                Therefore,



                begin{align}K+L&=frac{pi}{2}ln^2 2+4J+4ln 2int_0^{frac{pi}{2}}ln(cos x),dx
                end{align}



                Therefore (recall $K=J$),



                begin{align}J&=frac{1}{3}L-frac{pi}{6}ln^2 2-frac{4}{3}ln 2int_0^{frac{pi}{2}}ln(cos x),dxend{align}



                On the other hand,



                begin{align}L&=int_0^1 frac{ln^2 x}{1+x},dx+int_1^infty frac{ln^2 x}{1+x},dxend{align}



                In the latter integral perform the change of variable $y=dfrac{1}{x}$,



                begin{align}L&=2int_0^1 frac{ln^2 x}{1+x},dxend{align}



                But it is well known that,



                begin{align}int_0^1 frac{ln^2 x}{1+x^2},dx&=frac{pi^3}{16}\
                int_0^{frac{pi}{2}}ln(cos x),dx&=-frac{1}{2}piln 2
                end{align}



                Therefore,



                begin{align}J&=frac{pi^3}{24}-frac{pi}{6}ln^2 2-frac{4}{3}ln 2times -frac{1}{2}piln 2\
                &=boxed{frac{pi^3}{24}+frac{1}{2}piln^2 2}\
                end{align}



                PS:
                See: https://math.stackexchange.com/a/2942594/186817



                (in this post i assume only the value of $zeta(4)$ )






                share|cite|improve this answer











                $endgroup$


















                  2












                  $begingroup$

                  begin{align}J=int_0^{frac{pi}{2}} ln^2left(cos xright),dxend{align}



                  Observe that,



                  begin{align}I&=4J\
                  J&=int_0^{frac{pi}{2}} ln^2left(sin xright),dx\
                  int_0^{frac{pi}{2}} lnleft(sin xright),dx&=int_0^{frac{pi}{2}} lnleft(cos xright),dx
                  end{align}



                  (change of variable $y=dfrac{pi}{2}-x$ )



                  begin{align}
                  K&=int_0^{frac{pi}{2}} ln^2 left(2sin xcos xright),dx\
                  &=int_0^{frac{pi}{2}} ln^2 left(sinleft(2xright)right),dx\
                  end{align}



                  Perform the change of variable $y=2x$,



                  begin{align}
                  K&=frac{1}{2}int_0^{pi} ln^2 left(sin xright),dx\
                  &=frac{1}{2}int_0^{frac{pi}{2}} ln^2 left(sin xright),dx+frac{1}{2}int_{frac{pi}{2}}^pi ln^2 left(sin xright),dx\
                  end{align}



                  In the latter integral perform the change of variable $y=dfrac{pi}{2}-x$ and recall $sinleft(pi-xright)=sin x$ for $x$ real,



                  begin{align}
                  K&=int_0^{frac{pi}{2}} ln^2 left(sin xright),dx\
                  &=int_0^{frac{pi}{2}} ln^2 left(cos xright),dx\
                  &=J
                  end{align}



                  On the other hand,



                  begin{align}
                  K&=int_0^{frac{pi}{2}}left(ln 2+ln(sin x)+ln(cos x)right)^2 ,dx\
                  &=frac{pi}{2}ln^2 2+int_0^{frac{pi}{2}}ln^2(sin x),dx+int_0^{frac{pi}{2}}ln^2(cos x),dx+2ln 2int_0^{frac{pi}{2}}ln(cos x),dx+\
                  &2ln 2int_0^{frac{pi}{2}}ln(sin x),dx+2int_0^{frac{pi}{2}}ln(sin x)ln(cos x),dx\
                  &=frac{pi}{2}ln^2 2+2J+4ln 2int_0^{frac{pi}{2}}ln(cos x),dx+2int_0^{frac{pi}{2}}ln(sin x)ln(cos x),dx\
                  end{align}



                  begin{align}L&=int_0^infty frac{ln^2 x}{1+x^2},dxend{align}



                  Perform the change of variable $x=tan y$,



                  begin{align}L&=int_0^{frac{pi}{2}} ln^2left(tan xright),dx\
                  &=int_0^{frac{pi}{2}}left(lnleft(sin xright)-lnleft(cos xright)right)^2,dx\
                  &=2J-2int_0^{frac{pi}{2}}ln(sin x)ln(cos x),dx\
                  end{align}



                  Therefore,



                  begin{align}K+L&=frac{pi}{2}ln^2 2+4J+4ln 2int_0^{frac{pi}{2}}ln(cos x),dx
                  end{align}



                  Therefore (recall $K=J$),



                  begin{align}J&=frac{1}{3}L-frac{pi}{6}ln^2 2-frac{4}{3}ln 2int_0^{frac{pi}{2}}ln(cos x),dxend{align}



                  On the other hand,



                  begin{align}L&=int_0^1 frac{ln^2 x}{1+x},dx+int_1^infty frac{ln^2 x}{1+x},dxend{align}



                  In the latter integral perform the change of variable $y=dfrac{1}{x}$,



                  begin{align}L&=2int_0^1 frac{ln^2 x}{1+x},dxend{align}



                  But it is well known that,



                  begin{align}int_0^1 frac{ln^2 x}{1+x^2},dx&=frac{pi^3}{16}\
                  int_0^{frac{pi}{2}}ln(cos x),dx&=-frac{1}{2}piln 2
                  end{align}



                  Therefore,



                  begin{align}J&=frac{pi^3}{24}-frac{pi}{6}ln^2 2-frac{4}{3}ln 2times -frac{1}{2}piln 2\
                  &=boxed{frac{pi^3}{24}+frac{1}{2}piln^2 2}\
                  end{align}



                  PS:
                  See: https://math.stackexchange.com/a/2942594/186817



                  (in this post i assume only the value of $zeta(4)$ )






                  share|cite|improve this answer











                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    begin{align}J=int_0^{frac{pi}{2}} ln^2left(cos xright),dxend{align}



                    Observe that,



                    begin{align}I&=4J\
                    J&=int_0^{frac{pi}{2}} ln^2left(sin xright),dx\
                    int_0^{frac{pi}{2}} lnleft(sin xright),dx&=int_0^{frac{pi}{2}} lnleft(cos xright),dx
                    end{align}



                    (change of variable $y=dfrac{pi}{2}-x$ )



                    begin{align}
                    K&=int_0^{frac{pi}{2}} ln^2 left(2sin xcos xright),dx\
                    &=int_0^{frac{pi}{2}} ln^2 left(sinleft(2xright)right),dx\
                    end{align}



                    Perform the change of variable $y=2x$,



                    begin{align}
                    K&=frac{1}{2}int_0^{pi} ln^2 left(sin xright),dx\
                    &=frac{1}{2}int_0^{frac{pi}{2}} ln^2 left(sin xright),dx+frac{1}{2}int_{frac{pi}{2}}^pi ln^2 left(sin xright),dx\
                    end{align}



                    In the latter integral perform the change of variable $y=dfrac{pi}{2}-x$ and recall $sinleft(pi-xright)=sin x$ for $x$ real,



                    begin{align}
                    K&=int_0^{frac{pi}{2}} ln^2 left(sin xright),dx\
                    &=int_0^{frac{pi}{2}} ln^2 left(cos xright),dx\
                    &=J
                    end{align}



                    On the other hand,



                    begin{align}
                    K&=int_0^{frac{pi}{2}}left(ln 2+ln(sin x)+ln(cos x)right)^2 ,dx\
                    &=frac{pi}{2}ln^2 2+int_0^{frac{pi}{2}}ln^2(sin x),dx+int_0^{frac{pi}{2}}ln^2(cos x),dx+2ln 2int_0^{frac{pi}{2}}ln(cos x),dx+\
                    &2ln 2int_0^{frac{pi}{2}}ln(sin x),dx+2int_0^{frac{pi}{2}}ln(sin x)ln(cos x),dx\
                    &=frac{pi}{2}ln^2 2+2J+4ln 2int_0^{frac{pi}{2}}ln(cos x),dx+2int_0^{frac{pi}{2}}ln(sin x)ln(cos x),dx\
                    end{align}



                    begin{align}L&=int_0^infty frac{ln^2 x}{1+x^2},dxend{align}



                    Perform the change of variable $x=tan y$,



                    begin{align}L&=int_0^{frac{pi}{2}} ln^2left(tan xright),dx\
                    &=int_0^{frac{pi}{2}}left(lnleft(sin xright)-lnleft(cos xright)right)^2,dx\
                    &=2J-2int_0^{frac{pi}{2}}ln(sin x)ln(cos x),dx\
                    end{align}



                    Therefore,



                    begin{align}K+L&=frac{pi}{2}ln^2 2+4J+4ln 2int_0^{frac{pi}{2}}ln(cos x),dx
                    end{align}



                    Therefore (recall $K=J$),



                    begin{align}J&=frac{1}{3}L-frac{pi}{6}ln^2 2-frac{4}{3}ln 2int_0^{frac{pi}{2}}ln(cos x),dxend{align}



                    On the other hand,



                    begin{align}L&=int_0^1 frac{ln^2 x}{1+x},dx+int_1^infty frac{ln^2 x}{1+x},dxend{align}



                    In the latter integral perform the change of variable $y=dfrac{1}{x}$,



                    begin{align}L&=2int_0^1 frac{ln^2 x}{1+x},dxend{align}



                    But it is well known that,



                    begin{align}int_0^1 frac{ln^2 x}{1+x^2},dx&=frac{pi^3}{16}\
                    int_0^{frac{pi}{2}}ln(cos x),dx&=-frac{1}{2}piln 2
                    end{align}



                    Therefore,



                    begin{align}J&=frac{pi^3}{24}-frac{pi}{6}ln^2 2-frac{4}{3}ln 2times -frac{1}{2}piln 2\
                    &=boxed{frac{pi^3}{24}+frac{1}{2}piln^2 2}\
                    end{align}



                    PS:
                    See: https://math.stackexchange.com/a/2942594/186817



                    (in this post i assume only the value of $zeta(4)$ )






                    share|cite|improve this answer











                    $endgroup$



                    begin{align}J=int_0^{frac{pi}{2}} ln^2left(cos xright),dxend{align}



                    Observe that,



                    begin{align}I&=4J\
                    J&=int_0^{frac{pi}{2}} ln^2left(sin xright),dx\
                    int_0^{frac{pi}{2}} lnleft(sin xright),dx&=int_0^{frac{pi}{2}} lnleft(cos xright),dx
                    end{align}



                    (change of variable $y=dfrac{pi}{2}-x$ )



                    begin{align}
                    K&=int_0^{frac{pi}{2}} ln^2 left(2sin xcos xright),dx\
                    &=int_0^{frac{pi}{2}} ln^2 left(sinleft(2xright)right),dx\
                    end{align}



                    Perform the change of variable $y=2x$,



                    begin{align}
                    K&=frac{1}{2}int_0^{pi} ln^2 left(sin xright),dx\
                    &=frac{1}{2}int_0^{frac{pi}{2}} ln^2 left(sin xright),dx+frac{1}{2}int_{frac{pi}{2}}^pi ln^2 left(sin xright),dx\
                    end{align}



                    In the latter integral perform the change of variable $y=dfrac{pi}{2}-x$ and recall $sinleft(pi-xright)=sin x$ for $x$ real,



                    begin{align}
                    K&=int_0^{frac{pi}{2}} ln^2 left(sin xright),dx\
                    &=int_0^{frac{pi}{2}} ln^2 left(cos xright),dx\
                    &=J
                    end{align}



                    On the other hand,



                    begin{align}
                    K&=int_0^{frac{pi}{2}}left(ln 2+ln(sin x)+ln(cos x)right)^2 ,dx\
                    &=frac{pi}{2}ln^2 2+int_0^{frac{pi}{2}}ln^2(sin x),dx+int_0^{frac{pi}{2}}ln^2(cos x),dx+2ln 2int_0^{frac{pi}{2}}ln(cos x),dx+\
                    &2ln 2int_0^{frac{pi}{2}}ln(sin x),dx+2int_0^{frac{pi}{2}}ln(sin x)ln(cos x),dx\
                    &=frac{pi}{2}ln^2 2+2J+4ln 2int_0^{frac{pi}{2}}ln(cos x),dx+2int_0^{frac{pi}{2}}ln(sin x)ln(cos x),dx\
                    end{align}



                    begin{align}L&=int_0^infty frac{ln^2 x}{1+x^2},dxend{align}



                    Perform the change of variable $x=tan y$,



                    begin{align}L&=int_0^{frac{pi}{2}} ln^2left(tan xright),dx\
                    &=int_0^{frac{pi}{2}}left(lnleft(sin xright)-lnleft(cos xright)right)^2,dx\
                    &=2J-2int_0^{frac{pi}{2}}ln(sin x)ln(cos x),dx\
                    end{align}



                    Therefore,



                    begin{align}K+L&=frac{pi}{2}ln^2 2+4J+4ln 2int_0^{frac{pi}{2}}ln(cos x),dx
                    end{align}



                    Therefore (recall $K=J$),



                    begin{align}J&=frac{1}{3}L-frac{pi}{6}ln^2 2-frac{4}{3}ln 2int_0^{frac{pi}{2}}ln(cos x),dxend{align}



                    On the other hand,



                    begin{align}L&=int_0^1 frac{ln^2 x}{1+x},dx+int_1^infty frac{ln^2 x}{1+x},dxend{align}



                    In the latter integral perform the change of variable $y=dfrac{1}{x}$,



                    begin{align}L&=2int_0^1 frac{ln^2 x}{1+x},dxend{align}



                    But it is well known that,



                    begin{align}int_0^1 frac{ln^2 x}{1+x^2},dx&=frac{pi^3}{16}\
                    int_0^{frac{pi}{2}}ln(cos x),dx&=-frac{1}{2}piln 2
                    end{align}



                    Therefore,



                    begin{align}J&=frac{pi^3}{24}-frac{pi}{6}ln^2 2-frac{4}{3}ln 2times -frac{1}{2}piln 2\
                    &=boxed{frac{pi^3}{24}+frac{1}{2}piln^2 2}\
                    end{align}



                    PS:
                    See: https://math.stackexchange.com/a/2942594/186817



                    (in this post i assume only the value of $zeta(4)$ )







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Dec 31 '18 at 10:44

























                    answered Dec 31 '18 at 10:31









                    FDPFDP

                    5,63211526




                    5,63211526























                        2












                        $begingroup$

                        Differentiation of the Beta function is a viable approach, but since we are dealing with a squared logarithm and not higher powers, it is faster to just exploit the Fourier (cosine) series of $log(cos^2theta)$ and Parseval's theorem. Given



                        $$-log(cos^2theta)=2log 2+2sum_{kgeq 1}frac{(-1)^k}{k}cos(2ktheta),tag{1} $$
                        since $int_{0}^{pi/2}cos(2jx)cos(2kx),dx =frac{pi}{4}delta(j,k)$, we immediately have
                        $$ int_{0}^{pi/2}log^2(cos^2theta),dtheta = 2pilog^2 2+pizeta(2).tag{2}$$



                        Since the LHS equals $int_{0}^{1}frac{4log^2 x}{sqrt{1-x^2}},dx$, we have just found the value of the hypergeometric series
                        $$ 8sum_{ngeq 0}frac{binom{2n}{n}}{4^n(2n+1)^3}=8cdotphantom{}_4 F_3left(tfrac{1}{2},tfrac{1}{2},tfrac{1}{2},tfrac{1}{2};tfrac{3}{2},tfrac{3}{2},tfrac{3}{2};1right)tag{3} $$
                        which also equals $int_{0}^{+infty}frac{log^2(1+t^2)}{1+t^2},dt$ or $frac{1}{2}int_{0}^{1}frac{log^2(x)}{sqrt{x(1-x)}},dx$.






                        share|cite|improve this answer











                        $endgroup$













                        • $begingroup$
                          Yeah you're gonna have to explain this a bit more
                          $endgroup$
                          – clathratus
                          Dec 30 '18 at 9:51










                        • $begingroup$
                          @ClementC.: of course, now fixed.
                          $endgroup$
                          – Jack D'Aurizio
                          Dec 30 '18 at 10:44










                        • $begingroup$
                          OK, I have a dumb question. You get $frac{pi}{2}sum_{kgeq 1} left(frac{1}{k}right)^2=2pizeta(2)$ for the second term by Parseval, don't you? (or did I mess up with one factor there?)
                          $endgroup$
                          – Clement C.
                          Dec 30 '18 at 10:57








                        • 1




                          $begingroup$
                          @ClementC. the factor $2$ in $(1)$ gives that $$int_{0}^{pi/2}left(2sum_{kgeq 1}frac{(-1)^k}{k}cos(2ktheta)right)^2,dtheta = 4cdotfrac{pi}{4}cdotzeta(2)=pizeta(2).$$
                          $endgroup$
                          – Jack D'Aurizio
                          Dec 30 '18 at 14:11










                        • $begingroup$
                          @JackDAurizio Ah. Thanks.
                          $endgroup$
                          – Clement C.
                          Dec 30 '18 at 16:10
















                        2












                        $begingroup$

                        Differentiation of the Beta function is a viable approach, but since we are dealing with a squared logarithm and not higher powers, it is faster to just exploit the Fourier (cosine) series of $log(cos^2theta)$ and Parseval's theorem. Given



                        $$-log(cos^2theta)=2log 2+2sum_{kgeq 1}frac{(-1)^k}{k}cos(2ktheta),tag{1} $$
                        since $int_{0}^{pi/2}cos(2jx)cos(2kx),dx =frac{pi}{4}delta(j,k)$, we immediately have
                        $$ int_{0}^{pi/2}log^2(cos^2theta),dtheta = 2pilog^2 2+pizeta(2).tag{2}$$



                        Since the LHS equals $int_{0}^{1}frac{4log^2 x}{sqrt{1-x^2}},dx$, we have just found the value of the hypergeometric series
                        $$ 8sum_{ngeq 0}frac{binom{2n}{n}}{4^n(2n+1)^3}=8cdotphantom{}_4 F_3left(tfrac{1}{2},tfrac{1}{2},tfrac{1}{2},tfrac{1}{2};tfrac{3}{2},tfrac{3}{2},tfrac{3}{2};1right)tag{3} $$
                        which also equals $int_{0}^{+infty}frac{log^2(1+t^2)}{1+t^2},dt$ or $frac{1}{2}int_{0}^{1}frac{log^2(x)}{sqrt{x(1-x)}},dx$.






                        share|cite|improve this answer











                        $endgroup$













                        • $begingroup$
                          Yeah you're gonna have to explain this a bit more
                          $endgroup$
                          – clathratus
                          Dec 30 '18 at 9:51










                        • $begingroup$
                          @ClementC.: of course, now fixed.
                          $endgroup$
                          – Jack D'Aurizio
                          Dec 30 '18 at 10:44










                        • $begingroup$
                          OK, I have a dumb question. You get $frac{pi}{2}sum_{kgeq 1} left(frac{1}{k}right)^2=2pizeta(2)$ for the second term by Parseval, don't you? (or did I mess up with one factor there?)
                          $endgroup$
                          – Clement C.
                          Dec 30 '18 at 10:57








                        • 1




                          $begingroup$
                          @ClementC. the factor $2$ in $(1)$ gives that $$int_{0}^{pi/2}left(2sum_{kgeq 1}frac{(-1)^k}{k}cos(2ktheta)right)^2,dtheta = 4cdotfrac{pi}{4}cdotzeta(2)=pizeta(2).$$
                          $endgroup$
                          – Jack D'Aurizio
                          Dec 30 '18 at 14:11










                        • $begingroup$
                          @JackDAurizio Ah. Thanks.
                          $endgroup$
                          – Clement C.
                          Dec 30 '18 at 16:10














                        2












                        2








                        2





                        $begingroup$

                        Differentiation of the Beta function is a viable approach, but since we are dealing with a squared logarithm and not higher powers, it is faster to just exploit the Fourier (cosine) series of $log(cos^2theta)$ and Parseval's theorem. Given



                        $$-log(cos^2theta)=2log 2+2sum_{kgeq 1}frac{(-1)^k}{k}cos(2ktheta),tag{1} $$
                        since $int_{0}^{pi/2}cos(2jx)cos(2kx),dx =frac{pi}{4}delta(j,k)$, we immediately have
                        $$ int_{0}^{pi/2}log^2(cos^2theta),dtheta = 2pilog^2 2+pizeta(2).tag{2}$$



                        Since the LHS equals $int_{0}^{1}frac{4log^2 x}{sqrt{1-x^2}},dx$, we have just found the value of the hypergeometric series
                        $$ 8sum_{ngeq 0}frac{binom{2n}{n}}{4^n(2n+1)^3}=8cdotphantom{}_4 F_3left(tfrac{1}{2},tfrac{1}{2},tfrac{1}{2},tfrac{1}{2};tfrac{3}{2},tfrac{3}{2},tfrac{3}{2};1right)tag{3} $$
                        which also equals $int_{0}^{+infty}frac{log^2(1+t^2)}{1+t^2},dt$ or $frac{1}{2}int_{0}^{1}frac{log^2(x)}{sqrt{x(1-x)}},dx$.






                        share|cite|improve this answer











                        $endgroup$



                        Differentiation of the Beta function is a viable approach, but since we are dealing with a squared logarithm and not higher powers, it is faster to just exploit the Fourier (cosine) series of $log(cos^2theta)$ and Parseval's theorem. Given



                        $$-log(cos^2theta)=2log 2+2sum_{kgeq 1}frac{(-1)^k}{k}cos(2ktheta),tag{1} $$
                        since $int_{0}^{pi/2}cos(2jx)cos(2kx),dx =frac{pi}{4}delta(j,k)$, we immediately have
                        $$ int_{0}^{pi/2}log^2(cos^2theta),dtheta = 2pilog^2 2+pizeta(2).tag{2}$$



                        Since the LHS equals $int_{0}^{1}frac{4log^2 x}{sqrt{1-x^2}},dx$, we have just found the value of the hypergeometric series
                        $$ 8sum_{ngeq 0}frac{binom{2n}{n}}{4^n(2n+1)^3}=8cdotphantom{}_4 F_3left(tfrac{1}{2},tfrac{1}{2},tfrac{1}{2},tfrac{1}{2};tfrac{3}{2},tfrac{3}{2},tfrac{3}{2};1right)tag{3} $$
                        which also equals $int_{0}^{+infty}frac{log^2(1+t^2)}{1+t^2},dt$ or $frac{1}{2}int_{0}^{1}frac{log^2(x)}{sqrt{x(1-x)}},dx$.







                        share|cite|improve this answer














                        share|cite|improve this answer



                        share|cite|improve this answer








                        edited Dec 31 '18 at 17:32

























                        answered Dec 30 '18 at 9:48









                        Jack D'AurizioJack D'Aurizio

                        290k33282664




                        290k33282664












                        • $begingroup$
                          Yeah you're gonna have to explain this a bit more
                          $endgroup$
                          – clathratus
                          Dec 30 '18 at 9:51










                        • $begingroup$
                          @ClementC.: of course, now fixed.
                          $endgroup$
                          – Jack D'Aurizio
                          Dec 30 '18 at 10:44










                        • $begingroup$
                          OK, I have a dumb question. You get $frac{pi}{2}sum_{kgeq 1} left(frac{1}{k}right)^2=2pizeta(2)$ for the second term by Parseval, don't you? (or did I mess up with one factor there?)
                          $endgroup$
                          – Clement C.
                          Dec 30 '18 at 10:57








                        • 1




                          $begingroup$
                          @ClementC. the factor $2$ in $(1)$ gives that $$int_{0}^{pi/2}left(2sum_{kgeq 1}frac{(-1)^k}{k}cos(2ktheta)right)^2,dtheta = 4cdotfrac{pi}{4}cdotzeta(2)=pizeta(2).$$
                          $endgroup$
                          – Jack D'Aurizio
                          Dec 30 '18 at 14:11










                        • $begingroup$
                          @JackDAurizio Ah. Thanks.
                          $endgroup$
                          – Clement C.
                          Dec 30 '18 at 16:10


















                        • $begingroup$
                          Yeah you're gonna have to explain this a bit more
                          $endgroup$
                          – clathratus
                          Dec 30 '18 at 9:51










                        • $begingroup$
                          @ClementC.: of course, now fixed.
                          $endgroup$
                          – Jack D'Aurizio
                          Dec 30 '18 at 10:44










                        • $begingroup$
                          OK, I have a dumb question. You get $frac{pi}{2}sum_{kgeq 1} left(frac{1}{k}right)^2=2pizeta(2)$ for the second term by Parseval, don't you? (or did I mess up with one factor there?)
                          $endgroup$
                          – Clement C.
                          Dec 30 '18 at 10:57








                        • 1




                          $begingroup$
                          @ClementC. the factor $2$ in $(1)$ gives that $$int_{0}^{pi/2}left(2sum_{kgeq 1}frac{(-1)^k}{k}cos(2ktheta)right)^2,dtheta = 4cdotfrac{pi}{4}cdotzeta(2)=pizeta(2).$$
                          $endgroup$
                          – Jack D'Aurizio
                          Dec 30 '18 at 14:11










                        • $begingroup$
                          @JackDAurizio Ah. Thanks.
                          $endgroup$
                          – Clement C.
                          Dec 30 '18 at 16:10
















                        $begingroup$
                        Yeah you're gonna have to explain this a bit more
                        $endgroup$
                        – clathratus
                        Dec 30 '18 at 9:51




                        $begingroup$
                        Yeah you're gonna have to explain this a bit more
                        $endgroup$
                        – clathratus
                        Dec 30 '18 at 9:51












                        $begingroup$
                        @ClementC.: of course, now fixed.
                        $endgroup$
                        – Jack D'Aurizio
                        Dec 30 '18 at 10:44




                        $begingroup$
                        @ClementC.: of course, now fixed.
                        $endgroup$
                        – Jack D'Aurizio
                        Dec 30 '18 at 10:44












                        $begingroup$
                        OK, I have a dumb question. You get $frac{pi}{2}sum_{kgeq 1} left(frac{1}{k}right)^2=2pizeta(2)$ for the second term by Parseval, don't you? (or did I mess up with one factor there?)
                        $endgroup$
                        – Clement C.
                        Dec 30 '18 at 10:57






                        $begingroup$
                        OK, I have a dumb question. You get $frac{pi}{2}sum_{kgeq 1} left(frac{1}{k}right)^2=2pizeta(2)$ for the second term by Parseval, don't you? (or did I mess up with one factor there?)
                        $endgroup$
                        – Clement C.
                        Dec 30 '18 at 10:57






                        1




                        1




                        $begingroup$
                        @ClementC. the factor $2$ in $(1)$ gives that $$int_{0}^{pi/2}left(2sum_{kgeq 1}frac{(-1)^k}{k}cos(2ktheta)right)^2,dtheta = 4cdotfrac{pi}{4}cdotzeta(2)=pizeta(2).$$
                        $endgroup$
                        – Jack D'Aurizio
                        Dec 30 '18 at 14:11




                        $begingroup$
                        @ClementC. the factor $2$ in $(1)$ gives that $$int_{0}^{pi/2}left(2sum_{kgeq 1}frac{(-1)^k}{k}cos(2ktheta)right)^2,dtheta = 4cdotfrac{pi}{4}cdotzeta(2)=pizeta(2).$$
                        $endgroup$
                        – Jack D'Aurizio
                        Dec 30 '18 at 14:11












                        $begingroup$
                        @JackDAurizio Ah. Thanks.
                        $endgroup$
                        – Clement C.
                        Dec 30 '18 at 16:10




                        $begingroup$
                        @JackDAurizio Ah. Thanks.
                        $endgroup$
                        – Clement C.
                        Dec 30 '18 at 16:10











                        1












                        $begingroup$

                        Getting rid of the first roadblock:



                        Use the fact that $$log(acos x) = log a + logcos x$$
                        along with the result that
                        $$
                        int_0^{pi/2} logcos x,dx = -frac{pi}{2}log 2
                        $$
                        to get
                        $$
                        F'(a) = frac{pilog a}{a} -frac{pi}{2a}log 2
                        $$






                        share|cite|improve this answer









                        $endgroup$













                        • $begingroup$
                          Of course, one of the proofs of the intermediary result is due to Jack D'Aurizio.
                          $endgroup$
                          – Clement C.
                          Dec 30 '18 at 8:58
















                        1












                        $begingroup$

                        Getting rid of the first roadblock:



                        Use the fact that $$log(acos x) = log a + logcos x$$
                        along with the result that
                        $$
                        int_0^{pi/2} logcos x,dx = -frac{pi}{2}log 2
                        $$
                        to get
                        $$
                        F'(a) = frac{pilog a}{a} -frac{pi}{2a}log 2
                        $$






                        share|cite|improve this answer









                        $endgroup$













                        • $begingroup$
                          Of course, one of the proofs of the intermediary result is due to Jack D'Aurizio.
                          $endgroup$
                          – Clement C.
                          Dec 30 '18 at 8:58














                        1












                        1








                        1





                        $begingroup$

                        Getting rid of the first roadblock:



                        Use the fact that $$log(acos x) = log a + logcos x$$
                        along with the result that
                        $$
                        int_0^{pi/2} logcos x,dx = -frac{pi}{2}log 2
                        $$
                        to get
                        $$
                        F'(a) = frac{pilog a}{a} -frac{pi}{2a}log 2
                        $$






                        share|cite|improve this answer









                        $endgroup$



                        Getting rid of the first roadblock:



                        Use the fact that $$log(acos x) = log a + logcos x$$
                        along with the result that
                        $$
                        int_0^{pi/2} logcos x,dx = -frac{pi}{2}log 2
                        $$
                        to get
                        $$
                        F'(a) = frac{pilog a}{a} -frac{pi}{2a}log 2
                        $$







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered Dec 30 '18 at 8:53









                        Clement C.Clement C.

                        50.6k33892




                        50.6k33892












                        • $begingroup$
                          Of course, one of the proofs of the intermediary result is due to Jack D'Aurizio.
                          $endgroup$
                          – Clement C.
                          Dec 30 '18 at 8:58


















                        • $begingroup$
                          Of course, one of the proofs of the intermediary result is due to Jack D'Aurizio.
                          $endgroup$
                          – Clement C.
                          Dec 30 '18 at 8:58
















                        $begingroup$
                        Of course, one of the proofs of the intermediary result is due to Jack D'Aurizio.
                        $endgroup$
                        – Clement C.
                        Dec 30 '18 at 8:58




                        $begingroup$
                        Of course, one of the proofs of the intermediary result is due to Jack D'Aurizio.
                        $endgroup$
                        – Clement C.
                        Dec 30 '18 at 8:58











                        1












                        $begingroup$

                        $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
                        newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
                        newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
                        newcommand{dd}{mathrm{d}}
                        newcommand{ds}[1]{displaystyle{#1}}
                        newcommand{expo}[1]{,mathrm{e}^{#1},}
                        newcommand{ic}{mathrm{i}}
                        newcommand{mc}[1]{mathcal{#1}}
                        newcommand{mrm}[1]{mathrm{#1}}
                        newcommand{pars}[1]{left(,{#1},right)}
                        newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
                        newcommand{root}[2]{,sqrt[#1]{,{#2},},}
                        newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
                        newcommand{verts}[1]{leftvert,{#1},rightvert}$




                        $ds{I equiv
                        int_{0}^{pi/2}ln^{2}pars{cos^{2}pars{x}},dd x =
                        {pi^{3} over 6} + 2piln^{2}pars{2}: {LARGE ?}}$                        .




                        begin{align}
                        I & equiv
                        bbox[10px,#ffd]{int_{0}^{pi/2}ln^{2}pars{cos^{2}pars{x}},dd x}
                        ,,,stackrel{x mapsto pi/2 - x}{=},,,
                        4int_{0}^{pi/2}ln^{2}pars{sinpars{x}},dd x
                        \[5mm] & =
                        left.4,Reint_{x = 0}^{x = pi/2}ln^{2}pars{{1 - z^{2} over 2z},ic},{dd z over ic z}
                        ,rightvert_{ z = exppars{ic x}}
                        \[5mm] & =
                        left.4,Imint_{x = 0}^{x = pi/2}ln^{2}pars{{1 - z^{2} over 2z},ic},{dd z over z}
                        ,rightvert_{ z = exppars{ic x}}
                        \[1cm] & stackrel{mrm{as} epsilon to 0^{+}}{sim},,,
                        -, 4,
                        overbrace{Imint_{1}^{epsilon}ln^{2}
                        pars{1 + y^{2} over 2y},{dd y over y}}^{ds{= large 0}}
                        \[2mm] & phantom{stackrel{mrm{as} epsilon to 0^{+}}{sim},,,,,,,} -
                        4,Imint_{pi/2}^{0}bracks{lnpars{1 over 2epsilon} + pars{{pi over 2} - theta}ic}^{2},ic,ddtheta
                        \[2mm] & phantom{stackrel{mrm{as} epsilon to 0^{+}}{sim},,,,,,,,}
                        -4,Imint_{epsilon}^{1}bracks{lnpars{1 - x^{2} over 2x} + {pi over 2},ic}^{2}{dd x over x}
                        \[1cm] & =
                        4int_{0}^{pi/2}bracks{ln^{2}pars{2epsilon} -
                        pars{{pi over 2} - theta}^{2}},ddtheta
                        -4piint_{epsilon}^{1}lnpars{1 - x^{2} over 2x}
                        {dd x over x}
                        \[5mm] & stackrel{mrm{as} epsilon to 0^{+}}{sim}
                        bracks{2piln^{2}pars{2epsilon} - {pi^{3} over 6}} -
                        bracks{4piint_{0}^{1}{lnpars{1 - x^{2}} over x},dd x -
                        4piint_{epsilon}^{1}{lnpars{2x} over x},dd x}
                        \[5mm] & =
                        bracks{2piln^{2}pars{2epsilon} - {pi^{3} over 6}} -
                        bracks{2piint_{0}^{1}{lnpars{1 - x} over x},dd x +
                        2pilnpars{epsilon}lnpars{4epsilon}}
                        \[5mm] & stackrel{mrm{as} epsilon to 0}{to}
                        bbx{{pi^{3} over 6} + 2piln^{2}pars{2}} approx
                        8.1865
                        end{align}




                        Note that
                        $ds{int_{0}^{1}{lnpars{1 - x} over x},dd x =
                        -,mrm{Li}pars{1} = -,{pi^{2} over 6}}$                        .







                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
                          newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
                          newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
                          newcommand{dd}{mathrm{d}}
                          newcommand{ds}[1]{displaystyle{#1}}
                          newcommand{expo}[1]{,mathrm{e}^{#1},}
                          newcommand{ic}{mathrm{i}}
                          newcommand{mc}[1]{mathcal{#1}}
                          newcommand{mrm}[1]{mathrm{#1}}
                          newcommand{pars}[1]{left(,{#1},right)}
                          newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
                          newcommand{root}[2]{,sqrt[#1]{,{#2},},}
                          newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
                          newcommand{verts}[1]{leftvert,{#1},rightvert}$




                          $ds{I equiv
                          int_{0}^{pi/2}ln^{2}pars{cos^{2}pars{x}},dd x =
                          {pi^{3} over 6} + 2piln^{2}pars{2}: {LARGE ?}}$                          .




                          begin{align}
                          I & equiv
                          bbox[10px,#ffd]{int_{0}^{pi/2}ln^{2}pars{cos^{2}pars{x}},dd x}
                          ,,,stackrel{x mapsto pi/2 - x}{=},,,
                          4int_{0}^{pi/2}ln^{2}pars{sinpars{x}},dd x
                          \[5mm] & =
                          left.4,Reint_{x = 0}^{x = pi/2}ln^{2}pars{{1 - z^{2} over 2z},ic},{dd z over ic z}
                          ,rightvert_{ z = exppars{ic x}}
                          \[5mm] & =
                          left.4,Imint_{x = 0}^{x = pi/2}ln^{2}pars{{1 - z^{2} over 2z},ic},{dd z over z}
                          ,rightvert_{ z = exppars{ic x}}
                          \[1cm] & stackrel{mrm{as} epsilon to 0^{+}}{sim},,,
                          -, 4,
                          overbrace{Imint_{1}^{epsilon}ln^{2}
                          pars{1 + y^{2} over 2y},{dd y over y}}^{ds{= large 0}}
                          \[2mm] & phantom{stackrel{mrm{as} epsilon to 0^{+}}{sim},,,,,,,} -
                          4,Imint_{pi/2}^{0}bracks{lnpars{1 over 2epsilon} + pars{{pi over 2} - theta}ic}^{2},ic,ddtheta
                          \[2mm] & phantom{stackrel{mrm{as} epsilon to 0^{+}}{sim},,,,,,,,}
                          -4,Imint_{epsilon}^{1}bracks{lnpars{1 - x^{2} over 2x} + {pi over 2},ic}^{2}{dd x over x}
                          \[1cm] & =
                          4int_{0}^{pi/2}bracks{ln^{2}pars{2epsilon} -
                          pars{{pi over 2} - theta}^{2}},ddtheta
                          -4piint_{epsilon}^{1}lnpars{1 - x^{2} over 2x}
                          {dd x over x}
                          \[5mm] & stackrel{mrm{as} epsilon to 0^{+}}{sim}
                          bracks{2piln^{2}pars{2epsilon} - {pi^{3} over 6}} -
                          bracks{4piint_{0}^{1}{lnpars{1 - x^{2}} over x},dd x -
                          4piint_{epsilon}^{1}{lnpars{2x} over x},dd x}
                          \[5mm] & =
                          bracks{2piln^{2}pars{2epsilon} - {pi^{3} over 6}} -
                          bracks{2piint_{0}^{1}{lnpars{1 - x} over x},dd x +
                          2pilnpars{epsilon}lnpars{4epsilon}}
                          \[5mm] & stackrel{mrm{as} epsilon to 0}{to}
                          bbx{{pi^{3} over 6} + 2piln^{2}pars{2}} approx
                          8.1865
                          end{align}




                          Note that
                          $ds{int_{0}^{1}{lnpars{1 - x} over x},dd x =
                          -,mrm{Li}pars{1} = -,{pi^{2} over 6}}$                          .







                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
                            newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
                            newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
                            newcommand{dd}{mathrm{d}}
                            newcommand{ds}[1]{displaystyle{#1}}
                            newcommand{expo}[1]{,mathrm{e}^{#1},}
                            newcommand{ic}{mathrm{i}}
                            newcommand{mc}[1]{mathcal{#1}}
                            newcommand{mrm}[1]{mathrm{#1}}
                            newcommand{pars}[1]{left(,{#1},right)}
                            newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
                            newcommand{root}[2]{,sqrt[#1]{,{#2},},}
                            newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
                            newcommand{verts}[1]{leftvert,{#1},rightvert}$




                            $ds{I equiv
                            int_{0}^{pi/2}ln^{2}pars{cos^{2}pars{x}},dd x =
                            {pi^{3} over 6} + 2piln^{2}pars{2}: {LARGE ?}}$                            .




                            begin{align}
                            I & equiv
                            bbox[10px,#ffd]{int_{0}^{pi/2}ln^{2}pars{cos^{2}pars{x}},dd x}
                            ,,,stackrel{x mapsto pi/2 - x}{=},,,
                            4int_{0}^{pi/2}ln^{2}pars{sinpars{x}},dd x
                            \[5mm] & =
                            left.4,Reint_{x = 0}^{x = pi/2}ln^{2}pars{{1 - z^{2} over 2z},ic},{dd z over ic z}
                            ,rightvert_{ z = exppars{ic x}}
                            \[5mm] & =
                            left.4,Imint_{x = 0}^{x = pi/2}ln^{2}pars{{1 - z^{2} over 2z},ic},{dd z over z}
                            ,rightvert_{ z = exppars{ic x}}
                            \[1cm] & stackrel{mrm{as} epsilon to 0^{+}}{sim},,,
                            -, 4,
                            overbrace{Imint_{1}^{epsilon}ln^{2}
                            pars{1 + y^{2} over 2y},{dd y over y}}^{ds{= large 0}}
                            \[2mm] & phantom{stackrel{mrm{as} epsilon to 0^{+}}{sim},,,,,,,} -
                            4,Imint_{pi/2}^{0}bracks{lnpars{1 over 2epsilon} + pars{{pi over 2} - theta}ic}^{2},ic,ddtheta
                            \[2mm] & phantom{stackrel{mrm{as} epsilon to 0^{+}}{sim},,,,,,,,}
                            -4,Imint_{epsilon}^{1}bracks{lnpars{1 - x^{2} over 2x} + {pi over 2},ic}^{2}{dd x over x}
                            \[1cm] & =
                            4int_{0}^{pi/2}bracks{ln^{2}pars{2epsilon} -
                            pars{{pi over 2} - theta}^{2}},ddtheta
                            -4piint_{epsilon}^{1}lnpars{1 - x^{2} over 2x}
                            {dd x over x}
                            \[5mm] & stackrel{mrm{as} epsilon to 0^{+}}{sim}
                            bracks{2piln^{2}pars{2epsilon} - {pi^{3} over 6}} -
                            bracks{4piint_{0}^{1}{lnpars{1 - x^{2}} over x},dd x -
                            4piint_{epsilon}^{1}{lnpars{2x} over x},dd x}
                            \[5mm] & =
                            bracks{2piln^{2}pars{2epsilon} - {pi^{3} over 6}} -
                            bracks{2piint_{0}^{1}{lnpars{1 - x} over x},dd x +
                            2pilnpars{epsilon}lnpars{4epsilon}}
                            \[5mm] & stackrel{mrm{as} epsilon to 0}{to}
                            bbx{{pi^{3} over 6} + 2piln^{2}pars{2}} approx
                            8.1865
                            end{align}




                            Note that
                            $ds{int_{0}^{1}{lnpars{1 - x} over x},dd x =
                            -,mrm{Li}pars{1} = -,{pi^{2} over 6}}$                            .







                            share|cite|improve this answer









                            $endgroup$



                            $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
                            newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
                            newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
                            newcommand{dd}{mathrm{d}}
                            newcommand{ds}[1]{displaystyle{#1}}
                            newcommand{expo}[1]{,mathrm{e}^{#1},}
                            newcommand{ic}{mathrm{i}}
                            newcommand{mc}[1]{mathcal{#1}}
                            newcommand{mrm}[1]{mathrm{#1}}
                            newcommand{pars}[1]{left(,{#1},right)}
                            newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
                            newcommand{root}[2]{,sqrt[#1]{,{#2},},}
                            newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
                            newcommand{verts}[1]{leftvert,{#1},rightvert}$




                            $ds{I equiv
                            int_{0}^{pi/2}ln^{2}pars{cos^{2}pars{x}},dd x =
                            {pi^{3} over 6} + 2piln^{2}pars{2}: {LARGE ?}}$                            .




                            begin{align}
                            I & equiv
                            bbox[10px,#ffd]{int_{0}^{pi/2}ln^{2}pars{cos^{2}pars{x}},dd x}
                            ,,,stackrel{x mapsto pi/2 - x}{=},,,
                            4int_{0}^{pi/2}ln^{2}pars{sinpars{x}},dd x
                            \[5mm] & =
                            left.4,Reint_{x = 0}^{x = pi/2}ln^{2}pars{{1 - z^{2} over 2z},ic},{dd z over ic z}
                            ,rightvert_{ z = exppars{ic x}}
                            \[5mm] & =
                            left.4,Imint_{x = 0}^{x = pi/2}ln^{2}pars{{1 - z^{2} over 2z},ic},{dd z over z}
                            ,rightvert_{ z = exppars{ic x}}
                            \[1cm] & stackrel{mrm{as} epsilon to 0^{+}}{sim},,,
                            -, 4,
                            overbrace{Imint_{1}^{epsilon}ln^{2}
                            pars{1 + y^{2} over 2y},{dd y over y}}^{ds{= large 0}}
                            \[2mm] & phantom{stackrel{mrm{as} epsilon to 0^{+}}{sim},,,,,,,} -
                            4,Imint_{pi/2}^{0}bracks{lnpars{1 over 2epsilon} + pars{{pi over 2} - theta}ic}^{2},ic,ddtheta
                            \[2mm] & phantom{stackrel{mrm{as} epsilon to 0^{+}}{sim},,,,,,,,}
                            -4,Imint_{epsilon}^{1}bracks{lnpars{1 - x^{2} over 2x} + {pi over 2},ic}^{2}{dd x over x}
                            \[1cm] & =
                            4int_{0}^{pi/2}bracks{ln^{2}pars{2epsilon} -
                            pars{{pi over 2} - theta}^{2}},ddtheta
                            -4piint_{epsilon}^{1}lnpars{1 - x^{2} over 2x}
                            {dd x over x}
                            \[5mm] & stackrel{mrm{as} epsilon to 0^{+}}{sim}
                            bracks{2piln^{2}pars{2epsilon} - {pi^{3} over 6}} -
                            bracks{4piint_{0}^{1}{lnpars{1 - x^{2}} over x},dd x -
                            4piint_{epsilon}^{1}{lnpars{2x} over x},dd x}
                            \[5mm] & =
                            bracks{2piln^{2}pars{2epsilon} - {pi^{3} over 6}} -
                            bracks{2piint_{0}^{1}{lnpars{1 - x} over x},dd x +
                            2pilnpars{epsilon}lnpars{4epsilon}}
                            \[5mm] & stackrel{mrm{as} epsilon to 0}{to}
                            bbx{{pi^{3} over 6} + 2piln^{2}pars{2}} approx
                            8.1865
                            end{align}




                            Note that
                            $ds{int_{0}^{1}{lnpars{1 - x} over x},dd x =
                            -,mrm{Li}pars{1} = -,{pi^{2} over 6}}$                            .








                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 2 at 21:40









                            Felix MarinFelix Marin

                            68.2k7109143




                            68.2k7109143






























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