I am trying to evaluate the following definite integral: $$ \int\limits_{-\infty}^{+\infty} \frac{xe^{\frac{x}{2}}}{e^{2x}+1}dx $$ I think that answer can be expressed using gamma function, but I can't transform it in an appropriate way. I've tried to substitute $ t = e^{2x} $, change limits to $ 0 $ and $ +\infty $ and then integrate by parts, but it seems doesn't work. Can somebody give me a hint?
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the result should be $$-1/4,\sqrt {2}{\pi}^{2}$$ – Dr. Sonnhard Graubner Feb 27 '17 at 17:22
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You have the right idea: $ du = 2e^{2x} \, dx $. The limits change to $0$ and $\infty$, and the integral becomes $$ I = \int_0^{\infty}\frac{u^{-3/4}\log{u}}{4(1+u)} \, du. $$ We can now do this the easy way by noticing that it is the derivative of $$ \frac{1}{4}\int_0^{\infty} \frac{u^{s-1}}{1+u} \, du $$ evaluated at $s=1/4$. This integral is well-known to evaluate to $\frac{1}{4}\pi\csc{\pi s}$ (see e.g. here), and differentiating gives $ -\frac{1}{4}\pi^2\csc{\pi s}\cot{\pi s} $. Putting $s=1/4$ and evaluating gives $$ I = -\frac{\pi^2}{2\sqrt{2}} $$
