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It's so deceptively simple and none of the usual techniques are working. Any and all insights are welcome.

mathlove
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Archaick
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    WA gives an answer in terms of polylogarithms for the indefinite integral. The definite integral has a factor of $\pi$ in it, which makes me think that WA performs it with complex contour integration. – Ian Apr 18 '15 at 16:16
  • With that and the insights given to me in http://math.stackexchange.com/questions/430218/dogbone-contour-integral-branch-cuts-residue-at-infinity in mind, I think it might be a good idea to start with $u=1/x-1$. – Ian Apr 18 '15 at 16:18
  • $I~=~\dfrac\pi8~\ln2.$ – Lucian Apr 18 '15 at 16:18
  • This is a duplicate. http://math.stackexchange.com/questions/322229/how-to-evaluate-int-01-frac-lnx1x21-dx – user222031 Apr 18 '15 at 16:28

1 Answers1

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Setting $x=\tan(t)$, we obtain the integral to be \begin{align} I & = \int_0^{\pi/4} \log(1+\tan(t))dt = \int_0^{\pi/4}\log(\sin(t)+\cos(t))dt - \int_0^{\pi/4}\log(\cos(t))dt\\ & = \int_0^{\pi/4} \log(\sqrt2) dt + \int_0^{\pi/4} \log(\sin(t+\pi/4))dt - \int_0^{\pi/4}\log(\cos(t))dt\\ & = \int_0^{\pi/4} \log(\sqrt2) dt + \int_{\pi/4}^{\pi/2} \log(\sin(t))dt - \int_0^{\pi/4}\log(\cos(t))dt\\ & = \int_0^{\pi/4} \log(\sqrt2) dt + \int_{\pi/4}^{\pi/2} \log(\cos(\pi/2-t))dt - \int_0^{\pi/4}\log(\cos(t))dt\\ & = \int_0^{\pi/4} \log(\sqrt2) dt + \int_{0}^{\pi/4} \log(\cos(t))dt - \int_0^{\pi/4}\log(\cos(t))dt\\ & = \dfrac{\pi}8\ln(2) \end{align}

Adhvaitha
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