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I want to evaluate the integral $\displaystyle \int_0^\infty \dfrac{\ln x}{e^x+1}\,{\rm d} x$ using contour integration.

At first I though using a rectangular. Problem is that I cannot establish the vertices there. There is one poles included in the contour, $z=i\pi$ , but then there is a dead end.

Someone suggested me using a Hankel contour. I took a peak online to see what it looks like but I don't know how to use this.. This may actually work.

Of course the integral is evaluated fair and square if we differentiate the Gamma and eta functions and then plug in $s=1$.. but I really need to see (if there is an easy one proof) with complex analysis.

Tolaso
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  • are you sure it is convergent? – Mosk Jan 17 '15 at 21:58
  • Yep, it does converge.. to $-\dfrac{\ln^2 2}{2}$. – Tolaso Jan 17 '15 at 22:38
  • I reckon it must be pretty tough... though... unless someone uses the known formula of Gamma and Eta .. which makes it trivial. – Tolaso Jan 17 '15 at 22:41
  • I don't know if this satisfies you, but if we use the Laplace transform we can see that the integral equals the opposite of $$\sum_{n=1}^{+\infty}\frac{H_{n-1}(-1)^n}{n}$$ that can be evaluated with the residue theorem. – Jack D'Aurizio Jan 17 '15 at 23:28
  • Hang on a minute.. use the laplace transform?? Ok.. of which function? And how can the last sum be evaluated with residues? – Tolaso Jan 17 '15 at 23:29
  • We can compute the residues of $(\gamma+\psi(-x))^2 \cos(\pi x)/x $ and relate them with the previous series. – Jack D'Aurizio Jan 17 '15 at 23:35
  • Ok.. not of my taste... but I'd like to see it, if the question is answer.. and of which function are we evaluating the Laplace Transform ?? – Tolaso Jan 17 '15 at 23:38
  • The logarithm: $$\mathcal{L}(\log x)=-\frac{\gamma+\log t}{t}.$$ – Jack D'Aurizio Jan 17 '15 at 23:40
  • Ok... I am aware of that Laplace.. I'd like to see that method of yours... Thank you..! – Tolaso Jan 17 '15 at 23:42
  • I don't think you have high enough rep to see it, but there is a deleted solution to a similar integral that uses a rectangular contour to compute$$\int_0^\infty\frac{\log x}{\cosh x},dx$$I'm not sure if the same method would apply here to find$$\int_0^\infty\frac{\log x}{e^x+1},dx=\int_0^\infty\frac{\log x}{2e^{x/2}\cosh\frac x2},dx$$OTOH perhaps there's an alternative contour solution that can be applied to$$\int_0^1\frac{\log\left(\log\frac1x\right)}{1+x},dx\quad...$$ – user170231 Jan 25 '24 at 19:37

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