Residue Theorem for Gamma Function

Question

I am kinda stuck and not sure what to do at this point of the calculation where:

$$\int_{c\ -\ j\infty}^{c\ +\ j\infty} \left(\,x^{-1}\sigma\,\sqrt{\, 2\,}\,\,\right)^{s}\Gamma\left(\,{s \over 2}\,\right)\,{\rm d}s $$

The Gamma Function produces multiple singularities and I am not sure if the Residue Theorem can be applicable here.

Answer 1

Since the Stirling approximation: $$\log\Gamma(z) = \left(z-\frac{1}{2}\right)\log z-z+\log\sqrt{2\pi}+O\left(\frac{1}{|z|}\right)$$ holds uniformly over $\{z:\pi-|\arg z|\geq\varepsilon\}$, assuming $c\in\mathbb{R}^+$ we have: $$ \int_{c-i\infty}^{c+i\infty} u^s\, \Gamma\left(\frac{s}{2}\right)\,ds = \lim_{\substack{ n\to+\infty\\ T\to +\infty}}\oint_{\gamma_{n,T}}u^s \Gamma\left(\frac{s}{2}\right)\,ds$$ where $\gamma_{n,T}$ is the rectangular contour having vertices in $c+iT,c-iT,$ $-n-\frac{1}{2}+iT,$ $-n-\frac{1}{2}-iT$. Using the residue theorem to evaluate the last integral, we get:

$$ \int_{c-i\infty}^{c+i\infty} u^s\, \Gamma\left(\frac{s}{2}\right)\,ds = \color{red}{2\pi i\cdot\sum_{m=0}^{+\infty}\frac{2(-1)^m}{m!\,u^{2m}}}=\color{blue}{4\pi i\cdot e^{-\frac{1}{u^2}}}.$$

Answer 2

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ With $\quad\ds{\dsc{x^{-1}\sigma\root{2} \equiv \expo{t/2}}\ \imp \dsc{t=2\ln\pars{x^{-1}\sigma\root{2}}}}$.

We deform the contour such that the final integration is reduced to two integrals 'just above and below' the negative real axis. It's taking into account by the small shifts $\ds{\pm\ic 0^{+}}$ in expression $\pars{1}$:

\begin{align}&\color{#66f}{\large% \int_{c\ -\ \infty\ic}^{c\ +\ \infty\ic} \pars{x^{-1}\sigma\root{2}}^{s}\Gamma\pars{s \over 2}\,\dd s} =\int_{c\ -\ \infty\ic}^{c\ +\ \infty\ic}\expo{ts/2}\Gamma\pars{s \over 2}\,\dd s \\[5mm]&=2\int_{c/2\ -\ \infty\ic}^{c/2\ +\ \infty\ic}\expo{ts}\Gamma\pars{s} \,\dd s =-2\int_{-\infty}^{0}\expo{ts}\Gamma\pars{s + \ic 0^{+}}\,\dd s -2\int_{0}^{-\infty}\expo{ts}\Gamma\pars{s - \ic 0^{+}}\,\dd s \\[5mm]&=-2\int_{-\infty}^{0}\expo{ts} \bracks{\Gamma\pars{s + \ic 0^{+}} - \Gamma\pars{s - \ic 0^{+}}}\,\dd s \qquad\qquad\qquad\qquad\qquad\qquad\pars{1} \end{align}

It's well known the Gamma function $\ds{\pars{~\Gamma\pars{z}~}}$ has poles at $\ds{z = 0,-1,-2,\ldots}$ and it can be expressed as a Mittag-Leffler expansion ( a sum over the residues ):

\begin{align}&\color{#66f}{\large% \int_{c\ -\ \infty\ic}^{c\ +\ \infty\ic} \pars{x^{-1}\sigma\root{2}}^{s}\Gamma\pars{s \over 2}\,\dd s} = \\[5mm]&=-2\int_{-\infty}^{0}\expo{ts} \sum_{n\ =\ 0}^{\infty}{\pars{-1}^{n} \over n!}\ \overbrace{% \bracks{\pars{1 \over s + n + \ic 0^{+}} - \pars{1 \over s + n - \ic 0^{+}}}} ^{\dsc{-2\pi\ic\,\delta\pars{s + n}}}\,\dd s\tag{2} \\[5mm]&=4\pi\ic\sum_{n\ =\ 0}^{\infty}\expo{-tn}{\pars{-1}^{n} \over n!} =4\pi\ic\sum_{n\ =\ 0}^{\infty}{\pars{-\expo{-t}}^{n} \over n!} =4\pi\ic\exp\pars{-\expo{-t}} \\[5mm]&=4\pi\ic\exp\pars{-\,{1 \over \bracks{\expo{t/2}}^{2}}} =4\pi\ic\exp\pars{-\,{1 \over 2x^{-2}\sigma^{2}}} =\color{#66f}{\large 4\pi\ic\exp\pars{-\,{x^{2} \over 2\sigma^{2}}}} \end{align}

The above $\ds{\dsc{\mbox{red expression}}}$ ( expression $\pars{2}$ ) is a well known identity such as: $$ {1 \over x \pm \ic 0^{+}}=\,{\rm P.V.}\pars{1 \over x} \mp \ic\,\delta\pars{x} $$ where the $\ds{=}$ sign is symbolic such that the identity holds under an integration procedure.

Indeed, the whole integration can be performed in a closed contour around the negative real axis with parallel lines just above and below the axis. In that case, we can avoid any mention to the Dirac Delta function $\ds{\delta\pars{x}}$.

Answer 3

This is how I answered my question, thanks to the previous 2 answers given:$\newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} $

$${x^{-1}\sigma\root{2}\equiv\expo{t/2}\imp{t=2\ln\pars{x^{-1}\sigma\root{2}}}}$$ $$\int_{c\ -\ j\infty}^{c\ +\ j\infty} \left(\,x^{-1}\sigma\,\sqrt{\, 2\,}\,\,\right)^{s}\Gamma\left(\,{s \over 2}\,\right)\,{\rm d}s=\int_{c\ -\ j\infty}^{c\ +\ j\infty}\expo{ts/2}\Gamma\left(\,{s \over 2}\,\right)\,{\rm d}s$$

let $u=s/2$ and $ds=2du$. This gives:

$$2\int_{c/2\ -\ j\infty}^{c/2\ +\ j\infty}\expo{ut}\Gamma\left(\,{u}\,\right)\,{\rm d}s$$

Using Cauchy's integral theorem:

$$2\int_{c/2\ -\ j\infty}^{c/2\ +\ j\infty}\expo{ut}\Gamma\left(\,{u}\,\right)\,{\rm d}s={4\pi i\sum_{n=0}^{+\infty}\frac{(-\expo{-t})^n}{n!}}$$

Using infinite series and substitution of $t$ into the equation:

$${4\pi i\sum_{n=0}^{+\infty}\frac{(-\expo{-t})^n}{n!}}={4\pi i\cdot e^{-\frac{x^2}{2\sigma^2}}}$$