How do I find a closed form of the characteristic function of Gamma distribution? It’s written in wikipedia that $(1-i \beta t)^{-\alpha}$ is the characteristic function of the Gamma distribution $Gamma(\alpha,\beta)$. I tried to prove this but I failed. So I googled it for a proof, but surprisingly there is no written proof on google. Even in this website, there are some questions asking for it, but there is no answer. How do I prove it?
That is,

Let $\alpha,\beta>0$ and $t\in \mathbb{R}$.
Then, how do I show that $\int_0^\infty x^{\alpha -1}e^{-x(1-i\beta t)/\beta} dx =\Gamma(\alpha)\beta^\alpha (1-i\beta t)^{-\alpha}$?

Could someone prove this as an answer or tell me a refenrence which deals with this.
Thank you in advance!
 A: Edit: As pointed out in a comment, this is not quite right. Will be fixed soon, after class...
I said this:
This is just the change of variable $x'=x(1-i\beta t)/\beta$ plus the definition of $\Gamma(\alpha)$:
$$\int_0^\infty x^{\alpha -1}e^{-x(1-i\beta t)/\beta} dx 
=((1-i\beta t)/\beta)^{-\alpha}\int_0^\infty x^{\alpha -1}e^{-x} dx =\Gamma(\alpha)\beta^\alpha (1-i\beta t)^{-\alpha}.$$
And of course that's nonsense, it's not just a "change of variable" because $(1-i\beta t)/\beta$ is not real. It's a simple application of Cauchy's Theorem; some of you can stop reading at this point.
The corrected version:
Let $V=\Bbb C\setminus(-\infty,0]$ be the slit plane, and define $f\in H(V)$ by $$f(z)=z^{\alpha-1}e^{-z},$$where $z^{\alpha-1}$ is defined using the principal-value logarithm: $$z^{\alpha-1}=e^{(\alpha-1)\log(z)}.$$Note that if $z\in  V$ and $x>0$ then $\log(xz)=\log(x)+\log(z)$, hence$$(xz)^{\alpha-1}=x^{\alpha-1}z^{\alpha-1}.$$
Also note that if $z$ has positive real  part then $f(rz)\to0$ exponentially as $r>0$ tends to $+\infty$.
Set $\omega=(1-i\beta t)/\beta$ and define $\gamma_1,\gamma_2:[0,R]\to\Bbb C$ by $$\gamma_1(x)=x,$$ $$\gamma_2(x)=\omega x.$$. Let $\gamma_3$ be the straight line from $R$ to $\omega R$.
It follows easily from Cauchy's Theorem that $$\left(\int_{\gamma_1}+\int_{\gamma_3}-\int_{\gamma_2}\right)f(z)\,dz=0.$$
Detail: That's not quite just a special case of CT, since out triangle does not lie in $V$, passing through the origin as it does. That's easily fixed: Consider a contour consisting of that triangle except with a little detour near the origin, so it lies in $V$. Take a limit. (Note that the condition $\alpha-1>-1$ is needed to show that the error tends to $0$.)
Sine $f$ dies exponentially along rays in the right half-plane it follows that $$\lim_{R\to\infty}\int_{\gamma_3}f(z)\,dz=0,$$hence $$\lim_{R\to\infty}\int_{\gamma_2}f(z)\,dz=\lim_{R\to\infty}\int_{\gamma_1}f(z)\,dz.$$But $$\lim_{R\to\infty}\int_{\gamma_1}f(z)\,dz=\int_0^\infty x^{\alpha-1}e^{-x}\,dx=\Gamma(\alpha),$$while, recalling that $(\gamma_2(x))^{\alpha-1}=\omega^{\alpha-1}x^{\alpha-1}$, the definition of $\int_\gamma f(z)\,dz$ shows that $$\lim_{R\to\infty}\int_{\gamma_2}f(z)\,dz=\omega^\alpha\int_0^\infty x^{\alpha-1}e^{-x\omega}\,dx.$$
