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I want to solve the integral $$\int_{\mathbb R} \frac{(x^2+a^2)^{1-2s}}{(x^2+1)^{1-s}} dx$$ for $a>1$ whenever it is defined (I guess this is the case for $\Re(s)>1/2$).

You may use a CAS to provide the answer (I don't have any powerful enough at the moment), but of course I'm also interested in a proof.

principal-ideal-domain
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    How's your complex analysis knowledge? – Michael Seifert Sep 01 '21 at 17:46
  • (For what it's worth, Mathematica returns an abomination involving regularized hypergeometric functions. So there may not be a "nice" answer.) – Michael Seifert Sep 01 '21 at 17:51
  • @MichaelSeifert My complex analysis knowledge is quite okay, I guess. – principal-ideal-domain Sep 01 '21 at 18:28
  • @MichaelSeifert I guess there might also be a solution involving the Gaussian hypergeom. function (not the regularized one) if you make use of the condition $a>1$. Compare with https://math.stackexchange.com/questions/4237285/solving-int-mathbb-r-fracx2a-sx21-dx (read the comments). It seems like Maple tells us those solution. I'd be very interested in the one with the Gaussian hypergeom. function. – principal-ideal-domain Sep 01 '21 at 18:31
  • I tried playing around with contours to see if I could get anywhere with it, but I wasn't able to immediately make progress. I think I've seen similar integrals done via contour integration before, but they're in notes at my office. If I have time tomorrow (and it hasn't been solved by then) I'll see if I can get anything out of them. – Michael Seifert Sep 01 '21 at 18:31
  • @MichaelSeifert Wow, that would be great! Thanks a lot! – principal-ideal-domain Sep 01 '21 at 18:32
  • In terms of the Gauss hypergeometric function, the integral can be expressed as $\operatorname{B}{\left(s-\frac12,\frac12\right)},{_2F_1}{\left(2s-1,s-\frac12;s;1-a^{2}\right)}$ for $s>\frac12,a>0$. – David H Sep 01 '21 at 20:04
  • @DavidH Hi David, thx for the result but that looks a little bit strange to me since GHGF is only defined for arguments with absolute value $<1$. However $1-a^2$ has arbitary large absolute value. – principal-ideal-domain Sep 01 '21 at 20:45
  • @principal-ideal-domain You can uniquely extend the definition to all real arguments less than one by taking Euler's integral representation as your definition of the Gauss hypergeometric function. – David H Sep 01 '21 at 21:12
  • @DavidH Thanks for pointing that out. I didn't know that. But is there another way to write the integral with the GHGF applied within its disk of convergens? – principal-ideal-domain Sep 01 '21 at 21:24
  • @principal-ideal-domain Yes, using Pfaff's transformation: https://functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/16/01/01/0002/ – David H Sep 01 '21 at 21:30
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    @DavidH At https://mathworld.wolfram.com/PfaffTransformation.html it says this transformation holds only for $|x|<1/2$, on your side it's only restricted to $x \notin (1,\infty)$ which looks strange to me. For $x=1$ we devide by zero when we compute $x/(x-1)$. – principal-ideal-domain Sep 01 '21 at 22:29
  • Unfortunately, the techniques in the notes that I was remembering don't apply to this integrand, so I'm stumped. – Michael Seifert Sep 02 '21 at 14:53
  • @MichaelSeifert No problem and thanks a lot for looking into it! – principal-ideal-domain Sep 02 '21 at 15:13

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