Evaluation of $int_{frac{1}{2014}}^{2014} frac{tan^{-1}x}{x} dx$












1












$begingroup$


In there any way to calculate Integral





$$int_{frac{1}{2014}}^{2014} frac{tan^{-1}x}{x} dx$$





by using simple properties of definite integration. I used substitution $tan^{-1}x=t $ and applied by parts but then I am stuck at $intint csc(2t)dt$ How should I proceed? What is the most appropriate method to solve this question?






calculus integration definite-integrals







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$endgroup$












  • $begingroup$
    Hint: Try by parts first, so you have $int {1 over x(x^2+1)} $ and then use partial fractions.
    $endgroup$
    – Morgormir
    Feb 18 '16 at 19:23












  • $begingroup$
    @Morgormir Your integration by parts is incorrect. You should instead get $intfrac{ln{x}}{1+x^{2}}$, which isn't any easier. An elementary antiderivative for this function does not exist.
    $endgroup$
    – David H
    Feb 18 '16 at 19:30










  • $begingroup$
    You are correct, I wrote out the integral with $x^2$ as it's fraction, my mistake.
    $endgroup$
    – Morgormir
    Feb 18 '16 at 20:49
















1












$begingroup$


In there any way to calculate Integral





$$int_{frac{1}{2014}}^{2014} frac{tan^{-1}x}{x} dx$$





by using simple properties of definite integration. I used substitution $tan^{-1}x=t $ and applied by parts but then I am stuck at $intint csc(2t)dt$ How should I proceed? What is the most appropriate method to solve this question?






calculus integration definite-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    Hint: Try by parts first, so you have $int {1 over x(x^2+1)} $ and then use partial fractions.
    $endgroup$
    – Morgormir
    Feb 18 '16 at 19:23












  • $begingroup$
    @Morgormir Your integration by parts is incorrect. You should instead get $intfrac{ln{x}}{1+x^{2}}$, which isn't any easier. An elementary antiderivative for this function does not exist.
    $endgroup$
    – David H
    Feb 18 '16 at 19:30










  • $begingroup$
    You are correct, I wrote out the integral with $x^2$ as it's fraction, my mistake.
    $endgroup$
    – Morgormir
    Feb 18 '16 at 20:49














1












1








1


3



$begingroup$


In there any way to calculate Integral





$$int_{frac{1}{2014}}^{2014} frac{tan^{-1}x}{x} dx$$





by using simple properties of definite integration. I used substitution $tan^{-1}x=t $ and applied by parts but then I am stuck at $intint csc(2t)dt$ How should I proceed? What is the most appropriate method to solve this question?






calculus integration definite-integrals







share|cite|improve this question











$endgroup$




In there any way to calculate Integral





$$int_{frac{1}{2014}}^{2014} frac{tan^{-1}x}{x} dx$$





by using simple properties of definite integration. I used substitution $tan^{-1}x=t $ and applied by parts but then I am stuck at $intint csc(2t)dt$ How should I proceed? What is the most appropriate method to solve this question?






calculus integration definite-integrals




calculus integration definite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '18 at 2:23









DavidG

2,060720




2,060720










asked Feb 18 '16 at 19:17









MathematicsMathematics

1,7331030




1,7331030












  • $begingroup$
    Hint: Try by parts first, so you have $int {1 over x(x^2+1)} $ and then use partial fractions.
    $endgroup$
    – Morgormir
    Feb 18 '16 at 19:23












  • $begingroup$
    @Morgormir Your integration by parts is incorrect. You should instead get $intfrac{ln{x}}{1+x^{2}}$, which isn't any easier. An elementary antiderivative for this function does not exist.
    $endgroup$
    – David H
    Feb 18 '16 at 19:30










  • $begingroup$
    You are correct, I wrote out the integral with $x^2$ as it's fraction, my mistake.
    $endgroup$
    – Morgormir
    Feb 18 '16 at 20:49


















  • $begingroup$
    Hint: Try by parts first, so you have $int {1 over x(x^2+1)} $ and then use partial fractions.
    $endgroup$
    – Morgormir
    Feb 18 '16 at 19:23












  • $begingroup$
    @Morgormir Your integration by parts is incorrect. You should instead get $intfrac{ln{x}}{1+x^{2}}$, which isn't any easier. An elementary antiderivative for this function does not exist.
    $endgroup$
    – David H
    Feb 18 '16 at 19:30










  • $begingroup$
    You are correct, I wrote out the integral with $x^2$ as it's fraction, my mistake.
    $endgroup$
    – Morgormir
    Feb 18 '16 at 20:49
















$begingroup$
Hint: Try by parts first, so you have $int {1 over x(x^2+1)} $ and then use partial fractions.
$endgroup$
– Morgormir
Feb 18 '16 at 19:23






$begingroup$
Hint: Try by parts first, so you have $int {1 over x(x^2+1)} $ and then use partial fractions.
$endgroup$
– Morgormir
Feb 18 '16 at 19:23














$begingroup$
@Morgormir Your integration by parts is incorrect. You should instead get $intfrac{ln{x}}{1+x^{2}}$, which isn't any easier. An elementary antiderivative for this function does not exist.
$endgroup$
– David H
Feb 18 '16 at 19:30




$begingroup$
@Morgormir Your integration by parts is incorrect. You should instead get $intfrac{ln{x}}{1+x^{2}}$, which isn't any easier. An elementary antiderivative for this function does not exist.
$endgroup$
– David H
Feb 18 '16 at 19:30












$begingroup$
You are correct, I wrote out the integral with $x^2$ as it's fraction, my mistake.
$endgroup$
– Morgormir
Feb 18 '16 at 20:49




$begingroup$
You are correct, I wrote out the integral with $x^2$ as it's fraction, my mistake.
$endgroup$
– Morgormir
Feb 18 '16 at 20:49










2 Answers
2






active

oldest

votes


















6












$begingroup$

For $x>0$,
$$tan^{-1}(x)=cot^{-1}left(frac1xright)$$
Here,
$$I=int_{frac{1}{2014}}^{2014} frac{tan^{-1}x}{x} dx$$
Let $x=frac1y$. Then,
$$int_{frac{1}{2014}}^{2014} frac{tan^{-1}x}{x} dx=int^{frac{1}{2014}}_{2014} frac{tan^{-1}frac1y}{frac1y}left(-frac1{y^2}right)dy=-int^{frac{1}{2014}}_{2014} frac{tan^{-1}frac1y}{y}dy=-int^{frac{1}{2014}}_{2014} frac{cot^{-1}y}{y}dy=-int^{frac{1}{2014}}_{2014} frac{frac{pi}{2}-tan^{-1}y}{y}dy=int^{frac{1}{2014}}_{2014} frac{tan^{-1}y-frac{pi}{2}}{y}dy=-I-int^{frac{1}{2014}}_{2014} frac{frac{pi}{2}}{y}dy$$
Thus,
$$I=int_{frac{1}{2014}}^{2014} frac{pi}{4y}dy$$
which can be easily solved.






share|cite|improve this answer











$endgroup$





















    6












    $begingroup$

    A straightforward one: use the substitution $u=1/x$ and the formula:
    $$forall xinmathbb{R}_+^*, arctanleft(frac1xright)=fracpi2-arctan(x).$$
    For the sake of generality, we'll compute the integral from $1/a$ to $a$ for some $ainmathbb{R}_+^*$:
    $$I=int_{1/a}^{a}frac{arctan(x)}{x},mathrm{d}x=int_a^{1/a}uarctan(1/u)left(-frac{mathrm{d}u}{u^2}right)=int_{1/a}^afrac{pi/2-arctan(u)}{u},mathrm{d}u=int_{1/a}^afrac{pi}{2u},mathrm{d}u-I.$$
    Hence
    $$2I=int_{1/a}^afrac{pi}{2u},mathrm{d}u,$$
    i.e.,
    $$I=fracpi4int_{1/a}^afrac{mathrm{d}u}u=fracpi4bigl(ln(a)-ln(1/a)bigr)=fracpi2ln(a).$$






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      6












      $begingroup$

      For $x>0$,
      $$tan^{-1}(x)=cot^{-1}left(frac1xright)$$
      Here,
      $$I=int_{frac{1}{2014}}^{2014} frac{tan^{-1}x}{x} dx$$
      Let $x=frac1y$. Then,
      $$int_{frac{1}{2014}}^{2014} frac{tan^{-1}x}{x} dx=int^{frac{1}{2014}}_{2014} frac{tan^{-1}frac1y}{frac1y}left(-frac1{y^2}right)dy=-int^{frac{1}{2014}}_{2014} frac{tan^{-1}frac1y}{y}dy=-int^{frac{1}{2014}}_{2014} frac{cot^{-1}y}{y}dy=-int^{frac{1}{2014}}_{2014} frac{frac{pi}{2}-tan^{-1}y}{y}dy=int^{frac{1}{2014}}_{2014} frac{tan^{-1}y-frac{pi}{2}}{y}dy=-I-int^{frac{1}{2014}}_{2014} frac{frac{pi}{2}}{y}dy$$
      Thus,
      $$I=int_{frac{1}{2014}}^{2014} frac{pi}{4y}dy$$
      which can be easily solved.






      share|cite|improve this answer











      $endgroup$


















        6












        $begingroup$

        For $x>0$,
        $$tan^{-1}(x)=cot^{-1}left(frac1xright)$$
        Here,
        $$I=int_{frac{1}{2014}}^{2014} frac{tan^{-1}x}{x} dx$$
        Let $x=frac1y$. Then,
        $$int_{frac{1}{2014}}^{2014} frac{tan^{-1}x}{x} dx=int^{frac{1}{2014}}_{2014} frac{tan^{-1}frac1y}{frac1y}left(-frac1{y^2}right)dy=-int^{frac{1}{2014}}_{2014} frac{tan^{-1}frac1y}{y}dy=-int^{frac{1}{2014}}_{2014} frac{cot^{-1}y}{y}dy=-int^{frac{1}{2014}}_{2014} frac{frac{pi}{2}-tan^{-1}y}{y}dy=int^{frac{1}{2014}}_{2014} frac{tan^{-1}y-frac{pi}{2}}{y}dy=-I-int^{frac{1}{2014}}_{2014} frac{frac{pi}{2}}{y}dy$$
        Thus,
        $$I=int_{frac{1}{2014}}^{2014} frac{pi}{4y}dy$$
        which can be easily solved.






        share|cite|improve this answer











        $endgroup$
















          6












          6








          6





          $begingroup$

          For $x>0$,
          $$tan^{-1}(x)=cot^{-1}left(frac1xright)$$
          Here,
          $$I=int_{frac{1}{2014}}^{2014} frac{tan^{-1}x}{x} dx$$
          Let $x=frac1y$. Then,
          $$int_{frac{1}{2014}}^{2014} frac{tan^{-1}x}{x} dx=int^{frac{1}{2014}}_{2014} frac{tan^{-1}frac1y}{frac1y}left(-frac1{y^2}right)dy=-int^{frac{1}{2014}}_{2014} frac{tan^{-1}frac1y}{y}dy=-int^{frac{1}{2014}}_{2014} frac{cot^{-1}y}{y}dy=-int^{frac{1}{2014}}_{2014} frac{frac{pi}{2}-tan^{-1}y}{y}dy=int^{frac{1}{2014}}_{2014} frac{tan^{-1}y-frac{pi}{2}}{y}dy=-I-int^{frac{1}{2014}}_{2014} frac{frac{pi}{2}}{y}dy$$
          Thus,
          $$I=int_{frac{1}{2014}}^{2014} frac{pi}{4y}dy$$
          which can be easily solved.






          share|cite|improve this answer











          $endgroup$



          For $x>0$,
          $$tan^{-1}(x)=cot^{-1}left(frac1xright)$$
          Here,
          $$I=int_{frac{1}{2014}}^{2014} frac{tan^{-1}x}{x} dx$$
          Let $x=frac1y$. Then,
          $$int_{frac{1}{2014}}^{2014} frac{tan^{-1}x}{x} dx=int^{frac{1}{2014}}_{2014} frac{tan^{-1}frac1y}{frac1y}left(-frac1{y^2}right)dy=-int^{frac{1}{2014}}_{2014} frac{tan^{-1}frac1y}{y}dy=-int^{frac{1}{2014}}_{2014} frac{cot^{-1}y}{y}dy=-int^{frac{1}{2014}}_{2014} frac{frac{pi}{2}-tan^{-1}y}{y}dy=int^{frac{1}{2014}}_{2014} frac{tan^{-1}y-frac{pi}{2}}{y}dy=-I-int^{frac{1}{2014}}_{2014} frac{frac{pi}{2}}{y}dy$$
          Thus,
          $$I=int_{frac{1}{2014}}^{2014} frac{pi}{4y}dy$$
          which can be easily solved.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 17 '18 at 2:23









          DavidG

          2,060720




          2,060720










          answered Feb 18 '16 at 19:25









          GoodDeedsGoodDeeds

          10.3k31335




          10.3k31335























              6












              $begingroup$

              A straightforward one: use the substitution $u=1/x$ and the formula:
              $$forall xinmathbb{R}_+^*, arctanleft(frac1xright)=fracpi2-arctan(x).$$
              For the sake of generality, we'll compute the integral from $1/a$ to $a$ for some $ainmathbb{R}_+^*$:
              $$I=int_{1/a}^{a}frac{arctan(x)}{x},mathrm{d}x=int_a^{1/a}uarctan(1/u)left(-frac{mathrm{d}u}{u^2}right)=int_{1/a}^afrac{pi/2-arctan(u)}{u},mathrm{d}u=int_{1/a}^afrac{pi}{2u},mathrm{d}u-I.$$
              Hence
              $$2I=int_{1/a}^afrac{pi}{2u},mathrm{d}u,$$
              i.e.,
              $$I=fracpi4int_{1/a}^afrac{mathrm{d}u}u=fracpi4bigl(ln(a)-ln(1/a)bigr)=fracpi2ln(a).$$






              share|cite|improve this answer









              $endgroup$


















                6












                $begingroup$

                A straightforward one: use the substitution $u=1/x$ and the formula:
                $$forall xinmathbb{R}_+^*, arctanleft(frac1xright)=fracpi2-arctan(x).$$
                For the sake of generality, we'll compute the integral from $1/a$ to $a$ for some $ainmathbb{R}_+^*$:
                $$I=int_{1/a}^{a}frac{arctan(x)}{x},mathrm{d}x=int_a^{1/a}uarctan(1/u)left(-frac{mathrm{d}u}{u^2}right)=int_{1/a}^afrac{pi/2-arctan(u)}{u},mathrm{d}u=int_{1/a}^afrac{pi}{2u},mathrm{d}u-I.$$
                Hence
                $$2I=int_{1/a}^afrac{pi}{2u},mathrm{d}u,$$
                i.e.,
                $$I=fracpi4int_{1/a}^afrac{mathrm{d}u}u=fracpi4bigl(ln(a)-ln(1/a)bigr)=fracpi2ln(a).$$






                share|cite|improve this answer









                $endgroup$
















                  6












                  6








                  6





                  $begingroup$

                  A straightforward one: use the substitution $u=1/x$ and the formula:
                  $$forall xinmathbb{R}_+^*, arctanleft(frac1xright)=fracpi2-arctan(x).$$
                  For the sake of generality, we'll compute the integral from $1/a$ to $a$ for some $ainmathbb{R}_+^*$:
                  $$I=int_{1/a}^{a}frac{arctan(x)}{x},mathrm{d}x=int_a^{1/a}uarctan(1/u)left(-frac{mathrm{d}u}{u^2}right)=int_{1/a}^afrac{pi/2-arctan(u)}{u},mathrm{d}u=int_{1/a}^afrac{pi}{2u},mathrm{d}u-I.$$
                  Hence
                  $$2I=int_{1/a}^afrac{pi}{2u},mathrm{d}u,$$
                  i.e.,
                  $$I=fracpi4int_{1/a}^afrac{mathrm{d}u}u=fracpi4bigl(ln(a)-ln(1/a)bigr)=fracpi2ln(a).$$






                  share|cite|improve this answer









                  $endgroup$



                  A straightforward one: use the substitution $u=1/x$ and the formula:
                  $$forall xinmathbb{R}_+^*, arctanleft(frac1xright)=fracpi2-arctan(x).$$
                  For the sake of generality, we'll compute the integral from $1/a$ to $a$ for some $ainmathbb{R}_+^*$:
                  $$I=int_{1/a}^{a}frac{arctan(x)}{x},mathrm{d}x=int_a^{1/a}uarctan(1/u)left(-frac{mathrm{d}u}{u^2}right)=int_{1/a}^afrac{pi/2-arctan(u)}{u},mathrm{d}u=int_{1/a}^afrac{pi}{2u},mathrm{d}u-I.$$
                  Hence
                  $$2I=int_{1/a}^afrac{pi}{2u},mathrm{d}u,$$
                  i.e.,
                  $$I=fracpi4int_{1/a}^afrac{mathrm{d}u}u=fracpi4bigl(ln(a)-ln(1/a)bigr)=fracpi2ln(a).$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Feb 18 '16 at 19:35









                  gniourf_gniourfgniourf_gniourf

                  3,565818




                  3,565818






























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