Gamma Distribution Integral

probability distributions

Without appealing to any Fourier inversion results, I'd like to show
\begin{align*}
\frac{1}{2\pi} \int_{\mathbb{R}} e^{-ity} (1-it)^{-\alpha} dt = \frac{1}{\Gamma(\alpha)} y^{\alpha-1} e^{-y},
\end{align*}
for $\alpha > 0$, starting from the left hand side. The function $(1-it)^{-\alpha}$ is the characteristic function of a Ga($\alpha,1)$ random variable, and the left hand side amounts to the fourier inversion formula for integrable characteristic functions. However, I'd like to understand how to get from the left hand side to the right hand side just by computing the integral.

I've fiddled around with expanding things in terms of their power series and integrating by parts but didn't get very far. I've tried some simple cases (e.g. taking $\alpha = 1$) but I don't see how a Gamma function is going to appear.

Best Answer

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{{1 \over 2\pi}\int_{\large\mathbb{R}}\expo{-\ic ty} \pars{1 - \ic t}^{-\alpha}\,\dd t = {1 \over \Gamma\pars{\alpha}}\, y^{\alpha - 1}\expo{-y}:\ {\LARGE ?}}$

Set $\ds{x = 1 -\ic t \iff t = \pars{x - 1}\ic}$: \begin{align} &\bbox[10px,#ffd]{{1 \over 2\pi}\int_{\large\mathbb{R}}\expo{-\ic ty} \pars{1 - \ic t}^{-\alpha}\,\dd t} = -\,{\expo{-y} \over 2\pi}\,\ic \int_{1 - \infty\ic}^{1 + \infty\ic}\expo{yx}x^{-\alpha} \,\dd x \end{align}

I'll consider the branch-cut
$\ds{z^{-\alpha} = \verts{z}^{-\alpha}\exp\pars{-\ic\alpha\arg\pars{z}}\,,\quad\arg\pars{z} \in \pars{-\pi,\pi}\,,\quad z \not = 0}$

  • When $\ds{\color{red}{y < 0}}$, I "close" the contour "to the right of the complex plane" such that the integration vanishes out.
  • When $\ds{\color{red}{y > 0}}$, I "close" the contour "to the left of the complex plane". The whole contour is a key-hole one which takes care of the above described branch-cut.

Then, \begin{align} &\bbox[10px,#ffd]{{1 \over 2\pi}\int_{\large\mathbb{R}}\expo{-\ic ty} \pars{1 - \ic t}^{-\alpha}\,\dd t} \\[8mm] = &\ -\,{\expo{-y} \over 2\pi}\,\ic\bracks{y > 0}\ \times \\[2mm] &\ \bracks{% -\int_{-\infty}^{0}\expo{yx}\pars{-x}^{-\alpha}\expo{-\ic\alpha\pi} \,\dd x - \int_{0}^{-\infty}\expo{yx}\pars{-x}^{-\alpha}\expo{\ic\alpha\pi} \,\dd x} \\[8mm] = &\ -\,{\expo{-y} \over 2\pi}\,\ic\bracks{y > 0}\ \times \\[2mm] &\ \bracks{% -\expo{-\ic\alpha\pi}\int_{0}^{\infty}\expo{-yx}x^{-\alpha}\,\dd x + \expo{\ic\alpha\pi}\int_{0}^{\infty}\expo{-yx}x^{-\alpha} \,\dd x} \\[8mm] = &\ -\,{\expo{-y} \over 2\pi}\,\ic\bracks{y > 0}\bracks{% 2\ic\sin\pars{\pi\alpha}\int_{0}^{\infty}x^{-\alpha}\expo{-yx}\,\dd x} \\[5mm] = &\ {\expo{-y} \over \pi}\bracks{y > 0}\sin\pars{\pi\alpha} \bracks{y^{\alpha - 1}\int_{0}^{\infty}x^{-\alpha} \expo{-x}\,\dd x} \\[5mm] = & {y^{\alpha - 1}\expo{-y} \over \pi} \bracks{y > 0}\sin\pars{\pi\alpha}\Gamma\pars{-\alpha + 1} \qquad\pars{~\Gamma:\ Gamma\ Function~} \\[5mm] = &\ {y^{\alpha - 1}\expo{-y} \over \pi}\bracks{y > 0}\ \underbrace{\pi \over \Gamma\pars{1 - \alpha}\Gamma\pars{\alpha}} _{\ds{\sin\pars{\pi\alpha}}}\ \Gamma\pars{-\alpha + 1}\label{1}\tag{1} \\[5mm] = &\ \bbx{\bracks{y > 0}\, {y^{\alpha - 1}\expo{-y} \over \Gamma\pars{\alpha}}} \end{align}

In \eqref{1}, I used the Euler Reflection Formula.

Related Question