Evaluation of $int_{0}^{infty}frac{x^{a-1}-x^{b-1}}{1-x}dx$












2












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For $0<a,b<1.$
Evaluation of $$int_{0}^{infty}frac{x^{a-1}-x^{b-1}}{1-x}dx$$




$bf{My; Try::}$ Let $$I = int_{0}^{1}frac{x^{a-1}-x^{b-1}}{1-x}dx+int_{1}^{infty}frac{x^{a-1}-x^{b-1}}{1-x}dx$$



Now how can i proceed further, Help required, Thanks










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  • 1




    $begingroup$
    Are you familiar with techniques from complex analysis?
    $endgroup$
    – Michael Seifert
    Jan 30 '17 at 20:06










  • $begingroup$
    To Michael Saifert. not to much, elementry knowledge.
    $endgroup$
    – juantheron
    Jan 31 '17 at 3:08


















2












$begingroup$



For $0<a,b<1.$
Evaluation of $$int_{0}^{infty}frac{x^{a-1}-x^{b-1}}{1-x}dx$$




$bf{My; Try::}$ Let $$I = int_{0}^{1}frac{x^{a-1}-x^{b-1}}{1-x}dx+int_{1}^{infty}frac{x^{a-1}-x^{b-1}}{1-x}dx$$



Now how can i proceed further, Help required, Thanks










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Are you familiar with techniques from complex analysis?
    $endgroup$
    – Michael Seifert
    Jan 30 '17 at 20:06










  • $begingroup$
    To Michael Saifert. not to much, elementry knowledge.
    $endgroup$
    – juantheron
    Jan 31 '17 at 3:08
















2












2








2





$begingroup$



For $0<a,b<1.$
Evaluation of $$int_{0}^{infty}frac{x^{a-1}-x^{b-1}}{1-x}dx$$




$bf{My; Try::}$ Let $$I = int_{0}^{1}frac{x^{a-1}-x^{b-1}}{1-x}dx+int_{1}^{infty}frac{x^{a-1}-x^{b-1}}{1-x}dx$$



Now how can i proceed further, Help required, Thanks










share|cite|improve this question









$endgroup$





For $0<a,b<1.$
Evaluation of $$int_{0}^{infty}frac{x^{a-1}-x^{b-1}}{1-x}dx$$




$bf{My; Try::}$ Let $$I = int_{0}^{1}frac{x^{a-1}-x^{b-1}}{1-x}dx+int_{1}^{infty}frac{x^{a-1}-x^{b-1}}{1-x}dx$$



Now how can i proceed further, Help required, Thanks







definite-integrals






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asked Jan 30 '17 at 19:39









juantheronjuantheron

34.2k1147142




34.2k1147142








  • 1




    $begingroup$
    Are you familiar with techniques from complex analysis?
    $endgroup$
    – Michael Seifert
    Jan 30 '17 at 20:06










  • $begingroup$
    To Michael Saifert. not to much, elementry knowledge.
    $endgroup$
    – juantheron
    Jan 31 '17 at 3:08
















  • 1




    $begingroup$
    Are you familiar with techniques from complex analysis?
    $endgroup$
    – Michael Seifert
    Jan 30 '17 at 20:06










  • $begingroup$
    To Michael Saifert. not to much, elementry knowledge.
    $endgroup$
    – juantheron
    Jan 31 '17 at 3:08










1




1




$begingroup$
Are you familiar with techniques from complex analysis?
$endgroup$
– Michael Seifert
Jan 30 '17 at 20:06




$begingroup$
Are you familiar with techniques from complex analysis?
$endgroup$
– Michael Seifert
Jan 30 '17 at 20:06












$begingroup$
To Michael Saifert. not to much, elementry knowledge.
$endgroup$
– juantheron
Jan 31 '17 at 3:08






$begingroup$
To Michael Saifert. not to much, elementry knowledge.
$endgroup$
– juantheron
Jan 31 '17 at 3:08












3 Answers
3






active

oldest

votes


















5












$begingroup$

$newcommand{bbx}[1]{,bbox[8px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$

begin{align}
int_{0}^{infty}{x^{a - 1} - x^{b - 1} over 1 - x},dd x & =
int_{0}^{1}{x^{a - 1} - x^{b - 1} over 1 - x},dd x +
int_{1}^{infty}{x^{a - 1} - x^{b - 1} over 1 - x},dd x
end{align}

In the RHS second integral I'll perform the change $ds{x mapsto 1/x}$:
begin{align}
&bbox[10px,#ffd]{int_{0}^{infty}{x^{a - 1} - x^{b - 1} over 1 - x},dd x}
\[5mm] = &
int_{0}^{1}{x^{a - 1} - x^{b - 1} over 1 - x},dd x +
int_{1}^{0}{pars{1/x}^{a - 1} - pars{1/x}^{1 - b} over 1 - 1/x}
,{dd x over -x^{2}}
\[5mm] = &
int_{0}^{1}{x^{a - 1} - x^{b - 1} over 1 - x},dd x -
int_{0}^{1}{x^{-a} - x^{-b} over 1 - x},dd x
\[5mm] & =
bbx{ds{H_{b - 1} - H_{a - 1} + H_{-a} - H_{-b}}}
quadmbox{with} Repars{a},, Repars{b} in pars{0,1}
end{align}

$ds{H_{n}}$ is a Harmonic Number and I used a well known identity ( as given by Euler ): $ds{H_{z} = int_{0}^{1}{1 - t^{z} over 1 - t},dd t}$ with $ds{Repars{z} > -1}$.




Note that $ds{H_{b - 1} - H_{-b} = -picotpars{pi b}}$ such that:
$$
bbx{ds{int_{0}^{infty}{x^{a - 1} - x^{b - 1} over 1 - x},dd x =
picotpars{pi a} - picotpars{pi b}}}
$$





share|cite|improve this answer











$endgroup$













  • $begingroup$
    I knew the result MyGlasses derived looked familiar!
    $endgroup$
    – J.G.
    Jan 30 '17 at 22:52










  • $begingroup$
    @J.G. Yes. It's true. Thanks.
    $endgroup$
    – Felix Marin
    Jan 30 '17 at 23:15










  • $begingroup$
    Thanks Felix Marin, plz explain me how you get $H_{b-1}-H_{-b} = -pi cot(pi b)$
    $endgroup$
    – juantheron
    Jan 31 '17 at 3:11






  • 1




    $begingroup$
    @juantheron $H_{m}$ is related to the Digamma Function $Psi$. Namely, $H_{m} = Psileft(m + 1right) + gamma$. $gamma$ is the Euler-Mascheroni Constant. Then, $H_{b - 1} - H_{-b} = Psileft(bright) - Psileft(-b + 1right) = -picotleft(pi bright)$. The last one is the Euler Reflection Formula.
    $endgroup$
    – Felix Marin
    Jan 31 '17 at 3:20












  • $begingroup$
    Thanks Felix Marin got it.
    $endgroup$
    – juantheron
    Jan 31 '17 at 4:33



















1












$begingroup$

For $0<a,b<1$, with $x=e^u$
begin{eqnarray}
int_{0}^{infty}frac{x^{a-1}-x^{b-1}}{1-x}dx
&=&
int_{-infty}^{infty}frac{e^{au}-e^{bu}}{1-e^u}dx\
&=&
int_{0}^{infty}frac{e^{au}-e^{bu}}{1-e^u}dx+int_{0}^{infty}frac{e^{(1-a)u}-e^{(1-b)u}}{1-e^u}dx\
&=&
int_{0}^{infty}frac{-e^{-u}}{1-e^{-u}}Big(e^{au}-e^{bu}+e^{(1-a)u}-e^{(1-b)u}Big)dx\
&=&
int_{0}^{infty}sum_{n=1}^infty e^{-nu}Big(e^{au}-e^{bu}+e^{(1-a)u}-e^{(1-b)u}Big)dx\
&=&
sum_{n=1}^inftyint_{0}^{infty}Big(e^{(a-n)u}-e^{(b-n)u}+e^{(1-a-n)u}-e^{(1-b-n)u}Big)dx\
&=&
sum_{n=1}^inftyBig(frac{1}{n-a}-frac{1}{n-b}+frac{1}{n+a-1}-frac{1}{n+b-1}Big)
end{eqnarray}






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  • $begingroup$
    which gives us what?
    $endgroup$
    – tired
    Jan 30 '17 at 21:17










  • $begingroup$
    I'm stuck here. maybe diverge!
    $endgroup$
    – Nosrati
    Jan 30 '17 at 21:21






  • 1




    $begingroup$
    no the divergencies of the single terms are all of the same magnitude ($sim log(n)$)and therefore cancel out...what is left is most likely a so called digamma function
    $endgroup$
    – tired
    Jan 30 '17 at 21:29





















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You have to study what happens near $x=1$ and when $xto infty$, as I suppose you have notice.



Defining $f(x)=frac{x^{a-1}-x^{b-1}}{1-x}$, $lim_{xto 1} f(x)$ exists and it's finite (why?). So you can define $f(1)$ by continuity, and $int_0^1 f(x) dx$ exists and it's a number.



I do not know how to proceed from here. Any help would be appreciated.






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    3 Answers
    3






    active

    oldest

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    3 Answers
    3






    active

    oldest

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    active

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    active

    oldest

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    5












    $begingroup$

    $newcommand{bbx}[1]{,bbox[8px,border:1px groove navy]{displaystyle{#1}},}
    newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
    newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
    newcommand{dd}{mathrm{d}}
    newcommand{ds}[1]{displaystyle{#1}}
    newcommand{expo}[1]{,mathrm{e}^{#1},}
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    newcommand{verts}[1]{leftvert,{#1},rightvert}$

    begin{align}
    int_{0}^{infty}{x^{a - 1} - x^{b - 1} over 1 - x},dd x & =
    int_{0}^{1}{x^{a - 1} - x^{b - 1} over 1 - x},dd x +
    int_{1}^{infty}{x^{a - 1} - x^{b - 1} over 1 - x},dd x
    end{align}

    In the RHS second integral I'll perform the change $ds{x mapsto 1/x}$:
    begin{align}
    &bbox[10px,#ffd]{int_{0}^{infty}{x^{a - 1} - x^{b - 1} over 1 - x},dd x}
    \[5mm] = &
    int_{0}^{1}{x^{a - 1} - x^{b - 1} over 1 - x},dd x +
    int_{1}^{0}{pars{1/x}^{a - 1} - pars{1/x}^{1 - b} over 1 - 1/x}
    ,{dd x over -x^{2}}
    \[5mm] = &
    int_{0}^{1}{x^{a - 1} - x^{b - 1} over 1 - x},dd x -
    int_{0}^{1}{x^{-a} - x^{-b} over 1 - x},dd x
    \[5mm] & =
    bbx{ds{H_{b - 1} - H_{a - 1} + H_{-a} - H_{-b}}}
    quadmbox{with} Repars{a},, Repars{b} in pars{0,1}
    end{align}

    $ds{H_{n}}$ is a Harmonic Number and I used a well known identity ( as given by Euler ): $ds{H_{z} = int_{0}^{1}{1 - t^{z} over 1 - t},dd t}$ with $ds{Repars{z} > -1}$.




    Note that $ds{H_{b - 1} - H_{-b} = -picotpars{pi b}}$ such that:
    $$
    bbx{ds{int_{0}^{infty}{x^{a - 1} - x^{b - 1} over 1 - x},dd x =
    picotpars{pi a} - picotpars{pi b}}}
    $$





    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I knew the result MyGlasses derived looked familiar!
      $endgroup$
      – J.G.
      Jan 30 '17 at 22:52










    • $begingroup$
      @J.G. Yes. It's true. Thanks.
      $endgroup$
      – Felix Marin
      Jan 30 '17 at 23:15










    • $begingroup$
      Thanks Felix Marin, plz explain me how you get $H_{b-1}-H_{-b} = -pi cot(pi b)$
      $endgroup$
      – juantheron
      Jan 31 '17 at 3:11






    • 1




      $begingroup$
      @juantheron $H_{m}$ is related to the Digamma Function $Psi$. Namely, $H_{m} = Psileft(m + 1right) + gamma$. $gamma$ is the Euler-Mascheroni Constant. Then, $H_{b - 1} - H_{-b} = Psileft(bright) - Psileft(-b + 1right) = -picotleft(pi bright)$. The last one is the Euler Reflection Formula.
      $endgroup$
      – Felix Marin
      Jan 31 '17 at 3:20












    • $begingroup$
      Thanks Felix Marin got it.
      $endgroup$
      – juantheron
      Jan 31 '17 at 4:33
















    5












    $begingroup$

    $newcommand{bbx}[1]{,bbox[8px,border:1px groove navy]{displaystyle{#1}},}
    newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
    newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
    newcommand{dd}{mathrm{d}}
    newcommand{ds}[1]{displaystyle{#1}}
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    newcommand{mrm}[1]{mathrm{#1}}
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    newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
    newcommand{verts}[1]{leftvert,{#1},rightvert}$

    begin{align}
    int_{0}^{infty}{x^{a - 1} - x^{b - 1} over 1 - x},dd x & =
    int_{0}^{1}{x^{a - 1} - x^{b - 1} over 1 - x},dd x +
    int_{1}^{infty}{x^{a - 1} - x^{b - 1} over 1 - x},dd x
    end{align}

    In the RHS second integral I'll perform the change $ds{x mapsto 1/x}$:
    begin{align}
    &bbox[10px,#ffd]{int_{0}^{infty}{x^{a - 1} - x^{b - 1} over 1 - x},dd x}
    \[5mm] = &
    int_{0}^{1}{x^{a - 1} - x^{b - 1} over 1 - x},dd x +
    int_{1}^{0}{pars{1/x}^{a - 1} - pars{1/x}^{1 - b} over 1 - 1/x}
    ,{dd x over -x^{2}}
    \[5mm] = &
    int_{0}^{1}{x^{a - 1} - x^{b - 1} over 1 - x},dd x -
    int_{0}^{1}{x^{-a} - x^{-b} over 1 - x},dd x
    \[5mm] & =
    bbx{ds{H_{b - 1} - H_{a - 1} + H_{-a} - H_{-b}}}
    quadmbox{with} Repars{a},, Repars{b} in pars{0,1}
    end{align}

    $ds{H_{n}}$ is a Harmonic Number and I used a well known identity ( as given by Euler ): $ds{H_{z} = int_{0}^{1}{1 - t^{z} over 1 - t},dd t}$ with $ds{Repars{z} > -1}$.




    Note that $ds{H_{b - 1} - H_{-b} = -picotpars{pi b}}$ such that:
    $$
    bbx{ds{int_{0}^{infty}{x^{a - 1} - x^{b - 1} over 1 - x},dd x =
    picotpars{pi a} - picotpars{pi b}}}
    $$





    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I knew the result MyGlasses derived looked familiar!
      $endgroup$
      – J.G.
      Jan 30 '17 at 22:52










    • $begingroup$
      @J.G. Yes. It's true. Thanks.
      $endgroup$
      – Felix Marin
      Jan 30 '17 at 23:15










    • $begingroup$
      Thanks Felix Marin, plz explain me how you get $H_{b-1}-H_{-b} = -pi cot(pi b)$
      $endgroup$
      – juantheron
      Jan 31 '17 at 3:11






    • 1




      $begingroup$
      @juantheron $H_{m}$ is related to the Digamma Function $Psi$. Namely, $H_{m} = Psileft(m + 1right) + gamma$. $gamma$ is the Euler-Mascheroni Constant. Then, $H_{b - 1} - H_{-b} = Psileft(bright) - Psileft(-b + 1right) = -picotleft(pi bright)$. The last one is the Euler Reflection Formula.
      $endgroup$
      – Felix Marin
      Jan 31 '17 at 3:20












    • $begingroup$
      Thanks Felix Marin got it.
      $endgroup$
      – juantheron
      Jan 31 '17 at 4:33














    5












    5








    5





    $begingroup$

    $newcommand{bbx}[1]{,bbox[8px,border:1px groove navy]{displaystyle{#1}},}
    newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
    newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
    newcommand{dd}{mathrm{d}}
    newcommand{ds}[1]{displaystyle{#1}}
    newcommand{expo}[1]{,mathrm{e}^{#1},}
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    newcommand{root}[2]{,sqrt[#1]{,{#2},},}
    newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
    newcommand{verts}[1]{leftvert,{#1},rightvert}$

    begin{align}
    int_{0}^{infty}{x^{a - 1} - x^{b - 1} over 1 - x},dd x & =
    int_{0}^{1}{x^{a - 1} - x^{b - 1} over 1 - x},dd x +
    int_{1}^{infty}{x^{a - 1} - x^{b - 1} over 1 - x},dd x
    end{align}

    In the RHS second integral I'll perform the change $ds{x mapsto 1/x}$:
    begin{align}
    &bbox[10px,#ffd]{int_{0}^{infty}{x^{a - 1} - x^{b - 1} over 1 - x},dd x}
    \[5mm] = &
    int_{0}^{1}{x^{a - 1} - x^{b - 1} over 1 - x},dd x +
    int_{1}^{0}{pars{1/x}^{a - 1} - pars{1/x}^{1 - b} over 1 - 1/x}
    ,{dd x over -x^{2}}
    \[5mm] = &
    int_{0}^{1}{x^{a - 1} - x^{b - 1} over 1 - x},dd x -
    int_{0}^{1}{x^{-a} - x^{-b} over 1 - x},dd x
    \[5mm] & =
    bbx{ds{H_{b - 1} - H_{a - 1} + H_{-a} - H_{-b}}}
    quadmbox{with} Repars{a},, Repars{b} in pars{0,1}
    end{align}

    $ds{H_{n}}$ is a Harmonic Number and I used a well known identity ( as given by Euler ): $ds{H_{z} = int_{0}^{1}{1 - t^{z} over 1 - t},dd t}$ with $ds{Repars{z} > -1}$.




    Note that $ds{H_{b - 1} - H_{-b} = -picotpars{pi b}}$ such that:
    $$
    bbx{ds{int_{0}^{infty}{x^{a - 1} - x^{b - 1} over 1 - x},dd x =
    picotpars{pi a} - picotpars{pi b}}}
    $$





    share|cite|improve this answer











    $endgroup$



    $newcommand{bbx}[1]{,bbox[8px,border:1px groove navy]{displaystyle{#1}},}
    newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
    newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
    newcommand{dd}{mathrm{d}}
    newcommand{ds}[1]{displaystyle{#1}}
    newcommand{expo}[1]{,mathrm{e}^{#1},}
    newcommand{ic}{mathrm{i}}
    newcommand{mc}[1]{mathcal{#1}}
    newcommand{mrm}[1]{mathrm{#1}}
    newcommand{pars}[1]{left(,{#1},right)}
    newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
    newcommand{root}[2]{,sqrt[#1]{,{#2},},}
    newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
    newcommand{verts}[1]{leftvert,{#1},rightvert}$

    begin{align}
    int_{0}^{infty}{x^{a - 1} - x^{b - 1} over 1 - x},dd x & =
    int_{0}^{1}{x^{a - 1} - x^{b - 1} over 1 - x},dd x +
    int_{1}^{infty}{x^{a - 1} - x^{b - 1} over 1 - x},dd x
    end{align}

    In the RHS second integral I'll perform the change $ds{x mapsto 1/x}$:
    begin{align}
    &bbox[10px,#ffd]{int_{0}^{infty}{x^{a - 1} - x^{b - 1} over 1 - x},dd x}
    \[5mm] = &
    int_{0}^{1}{x^{a - 1} - x^{b - 1} over 1 - x},dd x +
    int_{1}^{0}{pars{1/x}^{a - 1} - pars{1/x}^{1 - b} over 1 - 1/x}
    ,{dd x over -x^{2}}
    \[5mm] = &
    int_{0}^{1}{x^{a - 1} - x^{b - 1} over 1 - x},dd x -
    int_{0}^{1}{x^{-a} - x^{-b} over 1 - x},dd x
    \[5mm] & =
    bbx{ds{H_{b - 1} - H_{a - 1} + H_{-a} - H_{-b}}}
    quadmbox{with} Repars{a},, Repars{b} in pars{0,1}
    end{align}

    $ds{H_{n}}$ is a Harmonic Number and I used a well known identity ( as given by Euler ): $ds{H_{z} = int_{0}^{1}{1 - t^{z} over 1 - t},dd t}$ with $ds{Repars{z} > -1}$.




    Note that $ds{H_{b - 1} - H_{-b} = -picotpars{pi b}}$ such that:
    $$
    bbx{ds{int_{0}^{infty}{x^{a - 1} - x^{b - 1} over 1 - x},dd x =
    picotpars{pi a} - picotpars{pi b}}}
    $$






    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 22 '18 at 16:19

























    answered Jan 30 '17 at 21:11









    Felix MarinFelix Marin

    67.8k7107142




    67.8k7107142












    • $begingroup$
      I knew the result MyGlasses derived looked familiar!
      $endgroup$
      – J.G.
      Jan 30 '17 at 22:52










    • $begingroup$
      @J.G. Yes. It's true. Thanks.
      $endgroup$
      – Felix Marin
      Jan 30 '17 at 23:15










    • $begingroup$
      Thanks Felix Marin, plz explain me how you get $H_{b-1}-H_{-b} = -pi cot(pi b)$
      $endgroup$
      – juantheron
      Jan 31 '17 at 3:11






    • 1




      $begingroup$
      @juantheron $H_{m}$ is related to the Digamma Function $Psi$. Namely, $H_{m} = Psileft(m + 1right) + gamma$. $gamma$ is the Euler-Mascheroni Constant. Then, $H_{b - 1} - H_{-b} = Psileft(bright) - Psileft(-b + 1right) = -picotleft(pi bright)$. The last one is the Euler Reflection Formula.
      $endgroup$
      – Felix Marin
      Jan 31 '17 at 3:20












    • $begingroup$
      Thanks Felix Marin got it.
      $endgroup$
      – juantheron
      Jan 31 '17 at 4:33


















    • $begingroup$
      I knew the result MyGlasses derived looked familiar!
      $endgroup$
      – J.G.
      Jan 30 '17 at 22:52










    • $begingroup$
      @J.G. Yes. It's true. Thanks.
      $endgroup$
      – Felix Marin
      Jan 30 '17 at 23:15










    • $begingroup$
      Thanks Felix Marin, plz explain me how you get $H_{b-1}-H_{-b} = -pi cot(pi b)$
      $endgroup$
      – juantheron
      Jan 31 '17 at 3:11






    • 1




      $begingroup$
      @juantheron $H_{m}$ is related to the Digamma Function $Psi$. Namely, $H_{m} = Psileft(m + 1right) + gamma$. $gamma$ is the Euler-Mascheroni Constant. Then, $H_{b - 1} - H_{-b} = Psileft(bright) - Psileft(-b + 1right) = -picotleft(pi bright)$. The last one is the Euler Reflection Formula.
      $endgroup$
      – Felix Marin
      Jan 31 '17 at 3:20












    • $begingroup$
      Thanks Felix Marin got it.
      $endgroup$
      – juantheron
      Jan 31 '17 at 4:33
















    $begingroup$
    I knew the result MyGlasses derived looked familiar!
    $endgroup$
    – J.G.
    Jan 30 '17 at 22:52




    $begingroup$
    I knew the result MyGlasses derived looked familiar!
    $endgroup$
    – J.G.
    Jan 30 '17 at 22:52












    $begingroup$
    @J.G. Yes. It's true. Thanks.
    $endgroup$
    – Felix Marin
    Jan 30 '17 at 23:15




    $begingroup$
    @J.G. Yes. It's true. Thanks.
    $endgroup$
    – Felix Marin
    Jan 30 '17 at 23:15












    $begingroup$
    Thanks Felix Marin, plz explain me how you get $H_{b-1}-H_{-b} = -pi cot(pi b)$
    $endgroup$
    – juantheron
    Jan 31 '17 at 3:11




    $begingroup$
    Thanks Felix Marin, plz explain me how you get $H_{b-1}-H_{-b} = -pi cot(pi b)$
    $endgroup$
    – juantheron
    Jan 31 '17 at 3:11




    1




    1




    $begingroup$
    @juantheron $H_{m}$ is related to the Digamma Function $Psi$. Namely, $H_{m} = Psileft(m + 1right) + gamma$. $gamma$ is the Euler-Mascheroni Constant. Then, $H_{b - 1} - H_{-b} = Psileft(bright) - Psileft(-b + 1right) = -picotleft(pi bright)$. The last one is the Euler Reflection Formula.
    $endgroup$
    – Felix Marin
    Jan 31 '17 at 3:20






    $begingroup$
    @juantheron $H_{m}$ is related to the Digamma Function $Psi$. Namely, $H_{m} = Psileft(m + 1right) + gamma$. $gamma$ is the Euler-Mascheroni Constant. Then, $H_{b - 1} - H_{-b} = Psileft(bright) - Psileft(-b + 1right) = -picotleft(pi bright)$. The last one is the Euler Reflection Formula.
    $endgroup$
    – Felix Marin
    Jan 31 '17 at 3:20














    $begingroup$
    Thanks Felix Marin got it.
    $endgroup$
    – juantheron
    Jan 31 '17 at 4:33




    $begingroup$
    Thanks Felix Marin got it.
    $endgroup$
    – juantheron
    Jan 31 '17 at 4:33











    1












    $begingroup$

    For $0<a,b<1$, with $x=e^u$
    begin{eqnarray}
    int_{0}^{infty}frac{x^{a-1}-x^{b-1}}{1-x}dx
    &=&
    int_{-infty}^{infty}frac{e^{au}-e^{bu}}{1-e^u}dx\
    &=&
    int_{0}^{infty}frac{e^{au}-e^{bu}}{1-e^u}dx+int_{0}^{infty}frac{e^{(1-a)u}-e^{(1-b)u}}{1-e^u}dx\
    &=&
    int_{0}^{infty}frac{-e^{-u}}{1-e^{-u}}Big(e^{au}-e^{bu}+e^{(1-a)u}-e^{(1-b)u}Big)dx\
    &=&
    int_{0}^{infty}sum_{n=1}^infty e^{-nu}Big(e^{au}-e^{bu}+e^{(1-a)u}-e^{(1-b)u}Big)dx\
    &=&
    sum_{n=1}^inftyint_{0}^{infty}Big(e^{(a-n)u}-e^{(b-n)u}+e^{(1-a-n)u}-e^{(1-b-n)u}Big)dx\
    &=&
    sum_{n=1}^inftyBig(frac{1}{n-a}-frac{1}{n-b}+frac{1}{n+a-1}-frac{1}{n+b-1}Big)
    end{eqnarray}






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      which gives us what?
      $endgroup$
      – tired
      Jan 30 '17 at 21:17










    • $begingroup$
      I'm stuck here. maybe diverge!
      $endgroup$
      – Nosrati
      Jan 30 '17 at 21:21






    • 1




      $begingroup$
      no the divergencies of the single terms are all of the same magnitude ($sim log(n)$)and therefore cancel out...what is left is most likely a so called digamma function
      $endgroup$
      – tired
      Jan 30 '17 at 21:29


















    1












    $begingroup$

    For $0<a,b<1$, with $x=e^u$
    begin{eqnarray}
    int_{0}^{infty}frac{x^{a-1}-x^{b-1}}{1-x}dx
    &=&
    int_{-infty}^{infty}frac{e^{au}-e^{bu}}{1-e^u}dx\
    &=&
    int_{0}^{infty}frac{e^{au}-e^{bu}}{1-e^u}dx+int_{0}^{infty}frac{e^{(1-a)u}-e^{(1-b)u}}{1-e^u}dx\
    &=&
    int_{0}^{infty}frac{-e^{-u}}{1-e^{-u}}Big(e^{au}-e^{bu}+e^{(1-a)u}-e^{(1-b)u}Big)dx\
    &=&
    int_{0}^{infty}sum_{n=1}^infty e^{-nu}Big(e^{au}-e^{bu}+e^{(1-a)u}-e^{(1-b)u}Big)dx\
    &=&
    sum_{n=1}^inftyint_{0}^{infty}Big(e^{(a-n)u}-e^{(b-n)u}+e^{(1-a-n)u}-e^{(1-b-n)u}Big)dx\
    &=&
    sum_{n=1}^inftyBig(frac{1}{n-a}-frac{1}{n-b}+frac{1}{n+a-1}-frac{1}{n+b-1}Big)
    end{eqnarray}






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      which gives us what?
      $endgroup$
      – tired
      Jan 30 '17 at 21:17










    • $begingroup$
      I'm stuck here. maybe diverge!
      $endgroup$
      – Nosrati
      Jan 30 '17 at 21:21






    • 1




      $begingroup$
      no the divergencies of the single terms are all of the same magnitude ($sim log(n)$)and therefore cancel out...what is left is most likely a so called digamma function
      $endgroup$
      – tired
      Jan 30 '17 at 21:29
















    1












    1








    1





    $begingroup$

    For $0<a,b<1$, with $x=e^u$
    begin{eqnarray}
    int_{0}^{infty}frac{x^{a-1}-x^{b-1}}{1-x}dx
    &=&
    int_{-infty}^{infty}frac{e^{au}-e^{bu}}{1-e^u}dx\
    &=&
    int_{0}^{infty}frac{e^{au}-e^{bu}}{1-e^u}dx+int_{0}^{infty}frac{e^{(1-a)u}-e^{(1-b)u}}{1-e^u}dx\
    &=&
    int_{0}^{infty}frac{-e^{-u}}{1-e^{-u}}Big(e^{au}-e^{bu}+e^{(1-a)u}-e^{(1-b)u}Big)dx\
    &=&
    int_{0}^{infty}sum_{n=1}^infty e^{-nu}Big(e^{au}-e^{bu}+e^{(1-a)u}-e^{(1-b)u}Big)dx\
    &=&
    sum_{n=1}^inftyint_{0}^{infty}Big(e^{(a-n)u}-e^{(b-n)u}+e^{(1-a-n)u}-e^{(1-b-n)u}Big)dx\
    &=&
    sum_{n=1}^inftyBig(frac{1}{n-a}-frac{1}{n-b}+frac{1}{n+a-1}-frac{1}{n+b-1}Big)
    end{eqnarray}






    share|cite|improve this answer









    $endgroup$



    For $0<a,b<1$, with $x=e^u$
    begin{eqnarray}
    int_{0}^{infty}frac{x^{a-1}-x^{b-1}}{1-x}dx
    &=&
    int_{-infty}^{infty}frac{e^{au}-e^{bu}}{1-e^u}dx\
    &=&
    int_{0}^{infty}frac{e^{au}-e^{bu}}{1-e^u}dx+int_{0}^{infty}frac{e^{(1-a)u}-e^{(1-b)u}}{1-e^u}dx\
    &=&
    int_{0}^{infty}frac{-e^{-u}}{1-e^{-u}}Big(e^{au}-e^{bu}+e^{(1-a)u}-e^{(1-b)u}Big)dx\
    &=&
    int_{0}^{infty}sum_{n=1}^infty e^{-nu}Big(e^{au}-e^{bu}+e^{(1-a)u}-e^{(1-b)u}Big)dx\
    &=&
    sum_{n=1}^inftyint_{0}^{infty}Big(e^{(a-n)u}-e^{(b-n)u}+e^{(1-a-n)u}-e^{(1-b-n)u}Big)dx\
    &=&
    sum_{n=1}^inftyBig(frac{1}{n-a}-frac{1}{n-b}+frac{1}{n+a-1}-frac{1}{n+b-1}Big)
    end{eqnarray}







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 30 '17 at 20:51









    NosratiNosrati

    26.5k62354




    26.5k62354












    • $begingroup$
      which gives us what?
      $endgroup$
      – tired
      Jan 30 '17 at 21:17










    • $begingroup$
      I'm stuck here. maybe diverge!
      $endgroup$
      – Nosrati
      Jan 30 '17 at 21:21






    • 1




      $begingroup$
      no the divergencies of the single terms are all of the same magnitude ($sim log(n)$)and therefore cancel out...what is left is most likely a so called digamma function
      $endgroup$
      – tired
      Jan 30 '17 at 21:29




















    • $begingroup$
      which gives us what?
      $endgroup$
      – tired
      Jan 30 '17 at 21:17










    • $begingroup$
      I'm stuck here. maybe diverge!
      $endgroup$
      – Nosrati
      Jan 30 '17 at 21:21






    • 1




      $begingroup$
      no the divergencies of the single terms are all of the same magnitude ($sim log(n)$)and therefore cancel out...what is left is most likely a so called digamma function
      $endgroup$
      – tired
      Jan 30 '17 at 21:29


















    $begingroup$
    which gives us what?
    $endgroup$
    – tired
    Jan 30 '17 at 21:17




    $begingroup$
    which gives us what?
    $endgroup$
    – tired
    Jan 30 '17 at 21:17












    $begingroup$
    I'm stuck here. maybe diverge!
    $endgroup$
    – Nosrati
    Jan 30 '17 at 21:21




    $begingroup$
    I'm stuck here. maybe diverge!
    $endgroup$
    – Nosrati
    Jan 30 '17 at 21:21




    1




    1




    $begingroup$
    no the divergencies of the single terms are all of the same magnitude ($sim log(n)$)and therefore cancel out...what is left is most likely a so called digamma function
    $endgroup$
    – tired
    Jan 30 '17 at 21:29






    $begingroup$
    no the divergencies of the single terms are all of the same magnitude ($sim log(n)$)and therefore cancel out...what is left is most likely a so called digamma function
    $endgroup$
    – tired
    Jan 30 '17 at 21:29













    0












    $begingroup$

    You have to study what happens near $x=1$ and when $xto infty$, as I suppose you have notice.



    Defining $f(x)=frac{x^{a-1}-x^{b-1}}{1-x}$, $lim_{xto 1} f(x)$ exists and it's finite (why?). So you can define $f(1)$ by continuity, and $int_0^1 f(x) dx$ exists and it's a number.



    I do not know how to proceed from here. Any help would be appreciated.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      You have to study what happens near $x=1$ and when $xto infty$, as I suppose you have notice.



      Defining $f(x)=frac{x^{a-1}-x^{b-1}}{1-x}$, $lim_{xto 1} f(x)$ exists and it's finite (why?). So you can define $f(1)$ by continuity, and $int_0^1 f(x) dx$ exists and it's a number.



      I do not know how to proceed from here. Any help would be appreciated.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        You have to study what happens near $x=1$ and when $xto infty$, as I suppose you have notice.



        Defining $f(x)=frac{x^{a-1}-x^{b-1}}{1-x}$, $lim_{xto 1} f(x)$ exists and it's finite (why?). So you can define $f(1)$ by continuity, and $int_0^1 f(x) dx$ exists and it's a number.



        I do not know how to proceed from here. Any help would be appreciated.






        share|cite|improve this answer











        $endgroup$



        You have to study what happens near $x=1$ and when $xto infty$, as I suppose you have notice.



        Defining $f(x)=frac{x^{a-1}-x^{b-1}}{1-x}$, $lim_{xto 1} f(x)$ exists and it's finite (why?). So you can define $f(1)$ by continuity, and $int_0^1 f(x) dx$ exists and it's a number.



        I do not know how to proceed from here. Any help would be appreciated.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 30 '17 at 19:53

























        answered Jan 30 '17 at 19:47









        A. Salguero-AlarcónA. Salguero-Alarcón

        3,267319




        3,267319






























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