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I saw somewhere that the above integral is equal to $\pi/4$ for all real number $m$.

This seems to be surprising. Does anyone have a nice proof?

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marked as duplicate by J.R., Daniel Fischer, user127.0.0.1, Claude Leibovici, Yiorgos S. Smyrlis Mar 2 '14 at 17:12

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    $\begingroup$ This should not be surprising: let $x\mapsto x^{-1}$ and you will get the same integral with an extra factor of $x^m$. Add both and you've gotten rid of $m$. $\endgroup$ – Julien Godawatta Mar 2 '14 at 15:00
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Let

$$\begin{align} I(m) &= \int_0^\infty \frac{dx}{(1+x^m)(1+x^2)}\tag{$x = t^{-1}$}\\ &= \int_0^\infty \frac{t^{-2}\,dt}{(1+t^{-m})(1+t^{-2})}\\ &= \int_0^\infty \frac{dt}{(1+t^{-m})(1+t^2)}\\ &= I(-m). \end{align}$$

But

$$\begin{align} I(m) + I(-m) &= \int_0^\infty \left(\frac{1}{1+x^m} + \frac{1}{1+x^{-m}}\right)\frac{dx}{1+x^2}\\ &= \int_0^\infty \left(\frac{1}{1+x^m} + \frac{x^m}{1+x^{m}}\right)\frac{dx}{1+x^2}\\ &= \int_0^\infty \frac{dx}{1+x^2}\\ &= \frac{\pi}{2}. \end{align}$$

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  • $\begingroup$ Beat me to it :-) (+1) $\endgroup$ – robjohn Mar 2 '14 at 15:17
  • $\begingroup$ You typed too slowly here, it seems. $\endgroup$ – Daniel Fischer Mar 2 '14 at 15:18
  • $\begingroup$ Yeah, I had to look up the prior post. I believe that this question is a duplicate, however. $\endgroup$ – robjohn Mar 2 '14 at 15:21
  • $\begingroup$ Wouldn't surprise me. But I have no idea what terms to search for to find a duplicate. If you do, go ahead. $\endgroup$ – Daniel Fischer Mar 2 '14 at 15:25
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    $\begingroup$ Found it here, and my answer is the same as yours :-) $\endgroup$ – robjohn Mar 2 '14 at 15:27

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