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Can anyone derive the solution to $$\int_{0}^{+\infty }\left( \frac{1}{1+s^{2}}\right) ^{\nu }ds$$ where $\nu$ is a non-integer positive real? The derivation has been provided under this and this question for integer $\nu$ with the solution

$$\frac{\left( 2\nu -2\right) !\pi }{2^{2\nu -1}\left( \left( \nu -1\right) !\right) ^{2}}$$

Expressed terms of the gamma function, this is,

$$\frac{\Gamma \left( 2\nu -1\right) }{2^{2\nu -1}\left( \Gamma \left( \nu\right) \right) ^{2}}\pi $$

I have spot checked to verify this generalizes the solution for real $\nu$, but this is not a proof.

Elliot Yu
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Edward L. Ruden
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1 Answers1

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Recall that the beta function $\int_0^1 t^{a-1} (1-t)^{b-1} \mathrm{d}t = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$ is known to converge for $a,b >0$. Also recalling that $\Gamma\left(z-\frac{1}{2}\right)\Gamma(z) = 2^{2-2\nu} \sqrt{\pi} \Gamma(2z -1)$ we get \begin{align} \int_{0}^{\infty }\left(\frac{1}{1+s^{2}}\right) ^{\nu }\mathrm{d}s &\overset{\frac{1}{1+s^2} = t}{=} \frac{1}{2}\int_0^{1} t^{\nu- \frac{1}{2}-1}(1-t)^{\frac{1}{2}-1}\mathrm{d}t = \frac{\Gamma\left(\nu - \frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{2\Gamma(\nu)}=\frac{\pi\Gamma(2\nu -1)}{2^{2\nu-1}\Gamma(\nu)^2} \end{align} which converges for real $\nu >\frac{1}{2}$.

Robert Lee
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    Note that the convergence region is for the integral and not for the final expression. – Gary Sep 29 '22 at 18:57
  • @Gary Did I hear analytic continuation ?! – K.defaoite Sep 29 '22 at 20:43
  • Robert, it would be good to mention the use of Legendre's duplication formula in the last step. – K.defaoite Sep 29 '22 at 20:43
  • Robert, apologies for my ignorance. – K.defaoite Sep 29 '22 at 20:44
  • I tracked down your relationships in Grad. and Ryz. ed. 8: 8.380.1, 8.384.1, 8.310.1, and 8.335.1 and it looks tight. I have what I need, thanks. – Edward L. Ruden Sep 29 '22 at 22:43
  • @K.defaoite You did not understand what I meant. – Gary Sep 30 '22 at 07:08
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    @Gary I was just joking around. – K.defaoite Sep 30 '22 at 09:10
  • @Gary, although the answer to "why ... $\Gamma$ so often works" answers the question, it answers it in a much more general and abstract way. Robert Lee's answer is more concise and focused on the specific question, and uses simpler math. He doesn't even have to use complex numbers! It is the elegant solution. – Edward L. Ruden Sep 30 '22 at 18:25
  • @EdwardL.Ruden I thought you were interested in why replacing factorials with gamma functions should give a valid result in the general case. If not, then this answer is indeed better. Of course you may want to prove the formula for the beta function in terms of gamma functions. – Gary Sep 30 '22 at 18:54
  • @Gary, my need is simply to verify the equation for a paper. B in terms of $\Gamma$ is in Grad. and Ryz., so I can just cite it. If Robert Lee can provide me his middle intial, I can acknowedge him for "useful correspondence". – Edward L. Ruden Sep 30 '22 at 20:17