The problem I'm trying gives the range $0< \alpha< 2$. What is the contour I should choose?
I think it should depend on $\alpha$ too. I tried using integration by parts to make it more manageable, but it doesn't seem to be helping much.
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$\begingroup$ At first glance I would guess you should try an indented semicircle (upload.wikimedia.org/wikibooks/en/9/9e/ContourSinzz.gif) $\endgroup$– Ethan AlwaiseDec 9, 2016 at 6:48
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$\begingroup$ @EthanAlwaise No. see RobertZ's answer. $\endgroup$– reunsDec 9, 2016 at 8:34
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$\begingroup$ Take a look at this Felix Marin's answer. You can easily adapt his proof for your case. math.stackexchange.com/a/1097463/186170 $\endgroup$– Marco CantariniDec 9, 2016 at 8:52
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1 Answer
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Hint. By integrating by parts we obtain $$I:=\int_{0}^{\infty}\frac{\log(1+x^2)}{x^{1+\alpha}} dx=\frac{2}{a}\int_{0}^{\infty}\frac{x^{1-\alpha}}{1+x^2}\,dx.$$ Then by using a keyhole contour and the Cauchy Theorem we get $$I:=\frac{4\pi i}{\alpha(1-e^{2\pi i(1-\alpha)})}\left(\text{Res} \left(\frac{z^{1-\alpha}}{1+z^2},i\right) +\text{Res}\left(\frac{z^{1-\alpha}}{1+z^2},-i\right)\right)= \frac{\pi}{\alpha\sin(\pi\alpha/2)}.$$