Integral in terms of gamma

Question

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!

Answer 1

Notice,

let $\frac{\beta}{x}=u\implies -\frac{\beta}{x^2}dx=du \iff \frac{dx}{x^2}=\frac{-du}{\beta}$ $$\int_{0}^{\infty}x^{-\alpha-1}e^{-\frac{\beta}{x}}dx=\int_{0}^{\infty}\frac{e^{-\frac{\beta}{x}}}{x^2(x^{\alpha-1})}dx$$

$$=\int_{\infty}^{0}\frac{e^{-u}}{\left(\frac{\beta}{u}\right)^{\alpha-1}}\left(\frac{-du}{\beta}\right)$$ $$=-\int_{\infty}^{0}\frac{e^{-u}}{\left(\frac{\beta}{u}\right)^{\alpha-1}}\frac{du}{\beta}$$ $$=\frac{1}{\beta^{\alpha}}\int_{0}^{\infty}e^{-u}u^{\alpha-1}du=\frac{\Gamma{(\alpha)}}{\beta^{\alpha}}$$

Answer 2

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 $a \downarrow 0$, $b \uparrow \infty$.