An integral identity relate to the Gamma function or the Beta function

Question

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.

Answer 1

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 $$ \langle \widehat{f_a},\,\widehat{f_b}\rangle \;=\; \langle f_a,\,f_b\rangle \;=\; {(2\pi)^a\,(2\pi)^b \over \Gamma(a)\,\Gamma(b)}\int_0^\infty e^{-2\pi t}\,t^{a-1}\;e^{-2\pi t}\,t^{b-1}\;dt $$ $$ \;=\;{(2\pi)^a\,(2\pi)^b \over \Gamma(a)\,\Gamma(b)}\int_0^\infty e^{-4\pi t}\,t^{a+b-1}\;{dt\over t} \;=\;{(2\pi)^a\,(2\pi)^b\,(4\pi)^{1-a-b} \Gamma(a+b-1)\over \Gamma(a)\,\Gamma(b)} \;=\; \ldots $$

Answer 2

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.