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.


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!

| cite | improve this answer | |
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.