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I want to solve the integral $$\int_{\mathbb R} \frac{(x^2+a)^{-s}}{x^2+1} dx$$ for $a>0$ whenever it is defined (I guess this is the case for $\Re(s)>-1/2$). Here I'm a little bit stuck. Can you help?

principal-ideal-domain
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  • It is probably some hypergeometric function, so with general $s$ and $a$ I don't think you will get closed form in terms of standard functions. Any constraints on the parameters? Large $a$ or integer $s$? – Diger Aug 31 '21 at 12:19
  • @Diger It might be enough for me to get the solution for $a>1$, I'm not sure if that helps. But $s$ shall stay complex valued (integers for $s$ are not enough). A solution with hypergeometric functions would be fine for me as well. – principal-ideal-domain Aug 31 '21 at 12:28
  • Have you thrown it into some CAS? Maple gives me an answer in terms of hypergeom. functions, so you might try that then. – Diger Aug 31 '21 at 12:30
  • @Diger I've only tried WolframAlpha. For integers $s$ it can solve it. – principal-ideal-domain Aug 31 '21 at 12:31
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    For large $a$ you have the expansion $$\frac{\pi}{(a-1)^s} - \frac{2\sqrt{\pi} , \Gamma(s+1/2)}{a^{s+1/2} , \Gamma(s)} , {_2F_1}(1,s+1/2;3/2;1/a)$$ if that helps. – Diger Aug 31 '21 at 12:35
  • @Diger So this solution is exact? How large has $a$ to be for the solution to be true? Maybe $a>1$ is enough since the hypergeometric Gauss function allows arugments with absolute value $<1$. – principal-ideal-domain Aug 31 '21 at 12:38
  • It is exact as long as $a>1$. For $a<1$ there are probably some identities to continue the hypergeometric function analytically. – Diger Aug 31 '21 at 12:39
  • @Diger I guess for me only $a>1$ is relevant. So thank you very much. A proof of course would be nice. By the way: The case $a=1$ is easy, it's just the beta function. I learnt that earlier today: https://math.stackexchange.com/questions/4237208/compute-int-mathbb-r-x2a2-s-dx – principal-ideal-domain Aug 31 '21 at 12:41
  • Mathematica gives $\pi \sec (\pi s) \left((1-a)^{-s}-\frac{\sqrt{\pi } a^{\frac{1}{2}-s}}{\Gamma (s)} , _2\tilde{F}_1\left(\frac{1}{2},1;\frac{3}{2}-s;a\right)\right)\text{ if }\Re(a)>0\land \Re(s)>-\frac{1}{2}$, where $_p\tilde{F}_q$ is the regularised hypergeometric function. – KStarGamer Aug 31 '21 at 14:34
  • @KStarGamer Is this for real $a>1$ just what Diger stated? – principal-ideal-domain Aug 31 '21 at 14:41
  • Yes, for $\Re (a) > 1$, I believe it can be simplified to Diger's expression, however, you wanted it to hold for $\Re (a) > 0$, which Mathematica's expression does. – KStarGamer Aug 31 '21 at 14:43
  • @KStarGamer Do you have $\tilde F = F$ in case the argument has absolute value $<1$? I'm wondering why in the two answers the hypergeometric functions gets different parameters and arguments. – principal-ideal-domain Aug 31 '21 at 14:48

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