# Bromwich contour integration of the gamma function

$$\frac{1}{2\pi i}\int^{c+i\infty}_{c-i\infty}\Gamma(a+t)\Gamma(b-t) s^{-t}\,dt= \frac{\Gamma(a+b)}{(1+s)^{a+b}}s^a$$

$$\textit{proof}$$

Consider the following function
$$f(z) = \Gamma(z+a)\Gamma(b-z) s^{-z}$$

Suppose that $a,b \in \mathbb{R}$ and $a < b$. Note that the Gamma function has a pole of order 1 at each non-positive integer where we have

$$\mathrm{Res}(\Gamma,-n) = \frac{(-1)^n}{n!}$$

The function $f$ has poles at the following points

$$-n-a,-(n-1)-a,\cdots,-a,b,b+1,\cdots,b+n$$

Notice that the function $f$ is analytic on the region $-a<\mathrm{Re}(z)<b$ , hence consider the following Bromwich contour

By the residue theorem

$$\int_{C_R}f(z)\,dz+\int^{c+iT}_{c-iT}f(z)\,dz = 2\pi i \sum_{k=0}^n \operatorname*{Res}_{z=-a-k }f(z)$$

By taking the limit $\to \infty$ we have

$$\lim_{R \to \infty}\int_{C_R}f(z)\,dz+\int^{c+i\infty}_{c-i\infty}f(t)\,dt = 2\pi i \sum_{k=0}^\infty \operatorname*{Res}_{z=-a-k }f(z)$$

Note that

$$\small \operatorname*{Res}_{z=-a-k }\Gamma(z+a)\Gamma(b-z) s^{-z} = \lim_{z \to -a-k} (b+a+k)\Gamma(-k) s^{a+k} = \frac{(-1)^k\Gamma(a+b+k)}{\Gamma(k+1)}s^{a+k}$$

Note that

$$\small s^a\sum_{k=0}^\infty \frac{\Gamma(a+b+k)}{\Gamma(k+1)} (-s)^k= s^a \Gamma(a+b)\sum_{k=0}^\infty (a+b)_k \frac{(-s)^k}{k!} = {}_2F_1(1;1,a+b,-s)$$

By definition of the Hypergeometric funciton

$$\small s^a \Gamma(a+b)\sum_{k=0}^\infty (a+b)_k \frac{(-s)^k}{k!} =s^a \Gamma(a+b) {}_2F_1(a+b,1;1,-s) = \frac{\Gamma(a+b)}{(1+s)^{a+b}}s^a$$

Also notice that

$$\lim_{R \to \infty}\int_{C_R}f(z)\,dz \to 0$$

So we deduce that

$$\frac{1}{2\pi i}\int^{c+i\infty}_{c-i\infty}\Gamma(a+t)\Gamma(b-t) s^{-t}\,dt= \frac{\Gamma(a+b)}{(1+s)^{a+b}}s^a$$

This entry was posted in Gamma function and tagged , , , , . Bookmark the permalink.