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How would I evaluate the following integral in terms of Catalan's constant? $$\int_{0}^{\infty} \frac{xe^x}{1+e^{2x}}$$ I am pretty sure you have to turn the integral into a series somehow, because we can write catalans constant as a series. $$G=\sum_{0}^{\infty} \frac{(-1)^n}{(2n+1)^2}$$

I am just not sure how to go about turning the integral into a series, I think I would need to do a $u$ substituion first, but I don't think $u=e^x$ works, and I couldn't think of any other plausible $u$ subs.

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Note that we can write

$$\begin{align} \int_0^\infty \frac{xe^{x}}{1+e^{2x}}\,dx&=\int_0^\infty \frac{xe^{-x}}{1+e^{-2x}}\,dx\\\\ &=\int_0^\infty x\sum_{n=0}^\infty (-1)^n e^{-(2n+1)x}\,dx\\\\ &=\sum_{n=0}^\infty (-1)^n\int_0^\infty x e^{-(2n+1)x}\,dx\\\\ &=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2}\\\\ &=G \end{align}$$

as was to be shown!

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  • $\begingroup$ I'd include why you can change the order of sum and integration. $\endgroup$ – Jakobian Jul 3 '19 at 17:49
  • $\begingroup$ I prefer using $C$ for Catalan's constant; is this also common? $\endgroup$ – Parcly Taxel Jul 3 '19 at 17:49
  • $\begingroup$ @Jakobian - Because of the linearity of continuous integrals. $\endgroup$ – user679268 Jul 3 '19 at 17:57
  • $\begingroup$ @KevinNivek linearity of continuous integrals? $\endgroup$ – Jakobian Jul 3 '19 at 18:04
  • $\begingroup$ @Jakobian - A continuous integral of a sum is equal to the sum of the integrals. $\endgroup$ – user679268 Jul 3 '19 at 18:06

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