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I have the following integral

$$\int_0^{\infty}\frac{\ln(x)}{e^x+1}$$

I believe the normal method is to apply Power series expansion on the denominator after multiplying both the top and the bottom by $e^{-x}$. The correct answer should be $-\frac{1}{2}\ln^2(2)$, and I was able to arrive at this answer. I'm wondering if there's a way to solve this problem through contour integration? Usually with this kind of denominator rectangular contour works. I've tried a few, but none works well with the $\ln(x)$ in numerator. Is anyone be able to give me a hint about the choice of contour in this situation?

Infiniticism
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Jessie Christian
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  • Integrate $\frac{z^2}{e^z+1}$ on $[r,R]\cup R e^{[0,2\pi]}\cup[R,r]\cup r e^{[2\pi,0]}$ and let $r\to 0, R\to \infty$. – Infiniticism Jan 10 '22 at 10:39
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    Another way is to evaluate $I(s)=\int_0^\infty \frac{x^{s-1}}{e^x+1} \mathrm dx$ through Zeta functions and differentiate w.r.t. $s$. – Infiniticism Jan 10 '22 at 10:42
  • See also the (unresolved) duplicate: https://math.stackexchange.com/questions/1108312/integral-with-contours – user170231 Jun 23 '25 at 16:02

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