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I encountered the following identity in a paper on number theory,

$$\int_{-\infty}^{\infty}\frac{dW}{(W+i)^{\frac{3}{2}}(W^2+1)^s}=\frac{e^\frac{-3\pi i}{4}\sqrt{2}\pi \Gamma(2s+\frac{1}{2})}{2^{2s}\Gamma(s+\frac{3}{2})\Gamma(s)},$$

with $Re(s) > 0$ and $i=\sqrt{-1}$.

Since the author did not give the proof for this, maybe it is "well-known", but I failed to gave a proof. Note that the right hand side looks like the Beta function with some multiplier, and we can write $W^2+1=(W+i)(W-i)$ on the left hand side, maybe we need to do some variable substitution.

Does any one know how to prove this identity or have some references? Thank you for any kind of input.

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  • $\begingroup$ I guess you mean $Re(s)>0$ instead of $\gg 0$. $\endgroup$ – Carlo Beenakker Mar 20 '17 at 7:16
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To evaluate $\int_{-\infty}^\infty {dx\over (1+ix)^a\,(1-ix)^b}$, view this as $\int_{-\infty}^\infty \widehat{f_a}(x)\,\overline{\widehat{f_{\overline{b}}}(x)}\,dx$ where $f_c$ are functions whose Fourier transforms are $(1+ix)^{-c}$. Use the Gamma-function identity $$ \int_0^\infty e^{-ty}\,t^c\;{dt\over t}\;=\; y^{-c}\cdot \Gamma(c) $$ at first for real $y>0$, and then for complex $y$ with $\Re(y)>0$. Thus, $(y+ix)^{-c}={1\over \Gamma(c)}\int_0^\infty e^{-t(y+ix)}\,t^c\,{dt\over t}$. To make the right-side look more like a Fourier transform, replace $t$ by $2\pi t$ in the integral: $$ {1\over (1+ix)^c} \;=\; {(2\pi)^c \over \Gamma(c)}\int_0^\infty e^{-2\pi itx}\,e^{-2\pi t}\,t^{c-1}\;dt $$ That is, this is the Fourier transform of the function $f_c$ which is $0$ for $t<0$ and is $e^{-2\pi t}\,t^{c-1}$ for $t>0$. Thus, up to the obvious powers of $i$, by Plancherel the integral is $$ \langle \widehat{f_a},\,\widehat{f_b}\rangle \;=\; \langle f_a,\,f_b\rangle \;=\; {(2\pi)^a\,(2\pi)^b \over \Gamma(a)\,\Gamma(b)}\int_0^\infty e^{-2\pi t}\,t^{a-1}\;e^{-2\pi t}\,t^{b-1}\;dt $$ $$ \;=\;{(2\pi)^a\,(2\pi)^b \over \Gamma(a)\,\Gamma(b)}\int_0^\infty e^{-4\pi t}\,t^{a+b-1}\;{dt\over t} \;=\;{(2\pi)^a\,(2\pi)^b\,(4\pi)^{1-a-b} \Gamma(a+b-1)\over \Gamma(a)\,\Gamma(b)} \;=\; \ldots $$

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  • $\begingroup$ Dear paul garrett, I still have a subproblem, why does the function $f_c$ in the first line exists? And does the second $y$ with $Re(y)>0$ in line four should be $c$? Finally, Thank you for your detailed answer! $\endgroup$ – CHENG Mar 20 '17 at 14:49
  • $\begingroup$ The function $f_c$ in the first line exists at least as an $L^2$ function for $\Re(c)>1/2$. It exists as a tempered distribution for all $c\in\mathbb C$, since $(1+ix)^{-c}$ is a tempered distribution for all $c$. The $\Re(y)>0$ really does refer to $y$ (although, yes, there is an inequality on $c$ for the integrals to make literal sense). That is, we use the identity principle from complex analysis to extend the Gamma identity to non-real $y$, but still with positive real part. $\endgroup$ – paul garrett Mar 20 '17 at 17:01
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Mathematica gives an alternative expression for the same integral valid for ${\rm Re}\,s>-1/4$:

$$ \int_{-\infty}^{\infty}\frac{dW}{(W+i)^{\frac{3}{2}}(W^2+1)^s}=-\frac{1+i}{\sqrt{\pi } (2 s+1)} \sin (2 \pi s) \Gamma (1-2 s) \Gamma \left(2 s+\tfrac{1}{2}\right).$$

Not sure if Mathematica output counts as a "proof", but I would say it qualifies as a "reference" for the identity.

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    $\begingroup$ FullSimplify in Mathematica also produces$$-2\sqrt\pi(1+i)s\frac{\Gamma \left(2 s+\frac12\right)}{\Gamma (2 s+2)}$$ $\endgroup$ – მამუკა ჯიბლაძე Mar 20 '17 at 8:19
  • $\begingroup$ The Mathematica is powerful! $\endgroup$ – CHENG Mar 20 '17 at 13:46

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