Find the Laplace transform of the Gamma pdf Per wikipedia the Laplace transform of the gamma distribution is
$$L_X(s) = (1+\theta s)^{-k} = \frac{\beta^\alpha}{(s+\beta)^\alpha}$$ As an exercise I would like to show this.The definition I have of the Laplace transform is:
$$L_X(t) = \mathbb{E}[e^{-tX}]=\int_0^\infty e^{-Xt}f(t)\mathrm{dt}$$ where in this case the function $f(t)$ is the pdf of the Gamma distribution. There are two forms, which I guess correspond to the two versions of the Laplace transform on Wikipedia. These are
$$\frac{1}{\Gamma(k)\theta^k}x^{k-1}e^-\frac{x}{\theta} \:\:\:\text{    and }\:\:\:
\frac{\beta^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\beta x}$$ where $\Gamma(m) = \int_0^\infty x^{m-1}e^{-x}\mathrm{dx}$ is the gamma function. I think the second version might be have simpler notation so I combine the above as carefully as I can:
$$L_X(t) = \int_0^\infty e^{-xt} \frac{\beta^\alpha}{\int_0^\infty y^{\alpha-1}e^{-y}\mathrm{dy} }x^{\alpha-1}e^{-\beta x}\mathrm{dt}
= \beta^\alpha\int_0^\infty  \frac{ x^{\alpha-1} e^{-x(t+\beta)}}{\int_0^\infty y^{\alpha-1}e^{-y}\mathrm{dy} }\mathrm{dt} 
$$
I am not sure if I should have, or need, different variables for both each integral. I have no idea how to proceed. Do we have to assume alpha is an integer to get a factorial underneath? Or can we just ignore it? I guess if I am right to use a different variable, then I can just leave it as $\Gamma(\alpha)$ and hope it cancels later?
Trying this: (and noticing something that looks like a Gamma function to hopefully factor...)$$
\frac{\beta^\alpha}{\Gamma(\alpha)} \int_0^\infty   x^{\alpha-1} e^{-x(t+\beta)}\mathrm{dt} = 
\frac{\beta^\alpha}{\Gamma(\alpha)} \int_0^\infty   x^{\alpha-1}e^{-x} e^{(t+\beta)}\mathrm{dt} = \beta^\alpha \int_0^\infty e^{t+\beta} \mathrm{dt}
$$
I am a bit dubious that I can take the gamma function out like that, the integral of a product isn't necessarily the product of the integrals $\int x^2\mathrm{dx}\neq \int x\mathrm{dx} \cdot \int x \mathrm{dx}$.
I am also skeptical that the remaining integral should be wrt x.  My calculus is rusty and I might have lost track of what variable I am using. 
So I think I spotted an error in the integration variable and have edited to be wrt t instead of x. I will try to continue...
$$\beta^\alpha\big[
e^{t+\beta}
\big]_0^\infty$$
So I am definitely stuck now and can not see my mistake, maybe from spending too long looking at it, but I would appreciate any help.
 A: $$\mathbb{E}\left[e^{-sX}  \right] = \int_{\mathbb{R}} e^{-sx} \frac{\beta^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\beta x} \mathrm{d}x = \frac{\beta^\alpha}{\Gamma(\alpha)(\beta + s)^\alpha}\underbrace{\int_{\mathbb{R}}  \left((\beta+s)x\right)^{\alpha-1}e^{-(\beta+s) x} \mathrm{d}\left((\beta+s)x\right)}_{=\Gamma(\alpha), \text{ use the substitution $u = (\beta + s) x $}} = \frac{\beta^\alpha}{(\beta + s)^\alpha}$$
A: If the pdf of the distribution is $$f_{\alpha, \beta}(x) = \frac{\beta^\alpha}{\Gamma(\alpha)} x^{\alpha-1} e^{-\beta x},$$ then by definition, the distribution $\Gamma_{\alpha, \beta}$ defined by $$\Gamma_{\alpha, \beta}(A) = \int_A f_{\alpha, \beta}(x) \: d\lambda(x) = \int_A \frac{\beta^\alpha}{\Gamma(\alpha)} x^{\alpha-1} e^{-\beta x} \: d\lambda(x)$$ is a probability measure on the measurable space $([0,\infty), \mathcal B)$, where $\mathcal B$ is the Borel $\sigma$-algebra on $[0,\infty)$ and $\lambda$ is the one-dimensional Lebesgue measure. In particular, what this implies is $$\int_0^\infty \frac{\beta^\alpha}{\Gamma(\alpha)} x^{r-1} e^{-\beta x} \: dt = \Gamma_{\alpha, \beta}([0,\infty)) = 1$$ for any $\alpha, \beta > 0$. Then, if $X$ is a Gamma-distributed random variable with parameters $\alpha$ and $\beta$, 
\begin{align}
L_X(s) &= \int_0^\infty e^{-sx} f_{\alpha, \beta}(x) \: dx = \int_0^\infty \frac{\beta^\alpha}{\Gamma(\alpha)} x^{\alpha-1} e^{-(s+\beta)x} \: dx \\
&= \frac{\beta^\alpha}{(s+\beta)^\alpha} \int_0^\infty \frac{(\beta+s)^\alpha}{\Gamma(\alpha)} x^{\alpha-1} e^{-(s+\beta)x} \: dx = \frac{\beta^\alpha}{(s+\beta)^\alpha},
\end{align}
where we have used the fact that this integral equals $1$ (using $\beta+s$ instead of $\beta$, for $s \geq 0$, the domain of the Laplace transform). 
