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I have tried hard to estimate this definite integral $$I:=\int_0^1 x^2 \arctan(e^{-x}) dx,$$ but all in vain. This exercise is from Page 204, 4.115, of Giaquinta & Modica's Book Mathematical Analysis, Foundations of One Variable. Actually I have tried using substitution $t=arctan(e^{-x})$, or $s=e^{-x}$, and with the help of integration by parts, but can not get the result. Even by using Maple, what I got is just the following: $$1/2\,i \left( -2\,{\it polylog} \left( 4,i \right) +2\,{\it polylog} \left( 4,-i \right) +{\it polylog} \left( 2,{\frac {i}{e}} \right) \left( \ln \left( e \right) \right) ^{2}+2\,{\it polylog} \left( 3, {\frac {i}{e}} \right) \ln \left( e \right) +2\,{\it polylog} \left( 4,{\frac {i}{e}} \right) -{\it polylog} \left( 2,{\frac {-i}{e}} \right) \left( \ln \left( e \right) \right) ^{2}-2\,{\it polylog} \left( 3,{\frac {-i}{e}} \right) \ln \left( e \right) -2\,{\it polylog} \left( 4,{\frac {-i}{e}} \right) \right) \left( \ln \left( e \right) \right) ^{-3}. $$

Can anyone help me? Thanks in advance.

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1 Answer 1

up vote 8 down vote accepted

Use the following identity: $$ \arctan\left(\mathrm{e}^{-x}\right) = \frac{\pi}{4} - \frac{1}{2} \int_0^{x} \frac{\mathrm{d}t}{\cosh(t)} $$ giving the following series expansion: $$ \arctan\left(\mathrm{e}^{-x}\right) = \frac{\pi}{4} -\frac{1}{2} \sum_{n=1}^\infty E_{n-1} \frac{x^n}{n!} $$ where $E_{n-1}$ stands for Euler number. Thus, integrating term-wise: $$ \int_0^1 x^2 \arctan(\mathrm{e}^{-x}) \mathrm{d} x = \frac{\pi}{12} - \frac{1}{12} \sum_{n=1}^\infty \frac{E_{n-1}}{n+3} \frac{1}{n!} $$ Numerical verification in Mathematica:

In[30]:= Pi/4 1/3 - 1/2 Sum[EulerE[n - 1]/n! 1/(n + 3), {n, 1, 250}] //
  N[#, 50] &

Out[30]= 0.14856508694081562686677775980253267668127537253127

In[31]:= % - 
 NIntegrate[x^2 ArcTan[Exp[-x]], {x, 0, 1}, WorkingPrecision -> 200]

Out[31]= 0.*10^-51
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Nice solution. Do you think that a closed form solution does not exist? –  Seyhmus Güngören Sep 13 '12 at 10:23

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