How do I show $\int_{-\infty}^{\infty} e^{itw}(a+it)^{-v} dt = \frac{2\pi}{\Gamma(v)}w^{v-1}e^{-wa}$?

Question

I came across the following integral in a paper by R Janik and M Nowak, $\int_{-\infty}^{\infty} e^{itw}(a+it)^{-v}dt = \frac{2\pi}{\Gamma(v)}w^{v-1}e^{-wa}$. They referenced Tables of Integrals by Gradstein and Rhizik, but there was no proof given. Some key assumptions for this integral are $Re (a)>0;w>0;Re (v)>0;|arg(a+it)|<\pi/2$. I am able to calculate the anti-derivative of the integrand, but I am having trouble with the definite integral as the anti-derivative involves the incomplete Gamma function.

Answer 1

Move the $2\pi$ to the LHS, and multiply both sides by $a^v$. What remains on the RHS is the density $f(w)$ of the Gamma(shape=$v$, rate=$a$) distribution: $$ f(w)={a^v\over\Gamma(v)}w^{v-1}e^{-aw}.\tag1 $$ So you should be able to derive (1) by Fourier inversion of the characteristic function $\phi(t)$ of the Gamma distribution: $$ \frac1{2\pi}\int_{-\infty}^\infty e^{-itw}\phi(t)\,dt=f(w)\tag2 $$ We can look up the characteristic function of the Gamma($v$, $a$) distribution, which is $$ \phi(t)=(1-it/a)^{-v}. $$ Plugging this into (2) and changing variables $u\mapsto -t$ should get you the desired result.

Answer 2

if $Re(v) < 1$ : $$\Gamma(1-v) = \int_0^\infty e^{-t} t^{-v} dt = \int_0^\infty e^{it} (it)^{-v} idt = i(iw)^{-v} \int_0^\infty e^{iw t} t^{-v} dt $$ (by a change of contour in the complex plane, $e^{-t} t^{-v}$ being holomorphic for $Re(t) \ge 0, t \ne 0$)

so $$\int_{-\infty}^\infty e^{itw} (a+it)^{-v} dt = e^{-aw}\int_{-\infty}^\infty e^{(a+it)w} (a+it)^{-v} dt= e^{-aw}( i(iw)^{v} -i(-iw)^{v}) \Gamma(1-v) =e^{-aw}w^v i \sin(\pi v/ 2)\Gamma(1-v) =e^{-aw}w^v \frac{i\pi}{\Gamma(v)} $$ by an holomorphic change of contour and the Gamma complement formula. (and perhaps also some mistakes with the constants)

by analiticity of that equality for $Re(v) < 1$, it is true for every $v \in \mathbb{C}$