Integral in terms of gamma I am trying to rewrite the following in terms of the $\Gamma$ function.
$$\large{\int_0^\infty} x^{- \alpha -1} e^{-\frac{\beta}{x}}dx$$
$\alpha$ does not pose a problem. To remove $\frac{\beta}{x}$ from the exponential I think we need to do a substitution. If we use $u=\frac{\beta}{x}$ then it looks like there is an issue with the lower value of the integral as we can't divide by $0$ when doing the change of variable. What is the best way to approach this problem?
Any help is appreciated, 
Thanks!
 A: If you want to be careful, let's look at
$$ \int_a^b x^{-\alpha-1} e^{-\beta/x} \, dx, $$
then take the limit as $a \downarrow 0$, $b \uparrow \infty$. The integrand is well-defined everywhere on the finite interval $[a,b]$, as is the substitution $u=\beta/x$. Then $dx/x=-du/u$ and the limits transform to $\beta/a$ and $\beta/b$. So the integral becomes
$$ -\int_{\beta/a}^{\beta/b} \left( \frac{\beta}{u} \right)^{-\alpha}u^{-1} e^{-u} \, du = \beta^{-\alpha} \int_{\beta/b}^{\beta/a} u^{\alpha-1} e^{-u} \, du \to \frac{1}{\beta^{\alpha}} \int_0^{\infty} u^{\alpha-1} e^{-u} \, du = \frac{\Gamma(\alpha)}{\beta^{\alpha}} $$
on taking the limits $a \downarrow 0$, $b \uparrow \infty$.
A: Notice, 
let $\frac{\beta}{x}=u\implies -\frac{\beta}{x^2}dx=du \iff \frac{dx}{x^2}=\frac{-du}{\beta}$
$$\int_{0}^{\infty}x^{-\alpha-1}e^{-\frac{\beta}{x}}dx=\int_{0}^{\infty}\frac{e^{-\frac{\beta}{x}}}{x^2(x^{\alpha-1})}dx$$
$$=\int_{\infty}^{0}\frac{e^{-u}}{\left(\frac{\beta}{u}\right)^{\alpha-1}}\left(\frac{-du}{\beta}\right)$$
$$=-\int_{\infty}^{0}\frac{e^{-u}}{\left(\frac{\beta}{u}\right)^{\alpha-1}}\frac{du}{\beta}$$  $$=\frac{1}{\beta^{\alpha}}\int_{0}^{\infty}e^{-u}u^{\alpha-1}du=\frac{\Gamma{(\alpha)}}{\beta^{\alpha}}$$
