# Show $\int_0^\infty \ x^{\alpha-1}e^{-x/\beta} \ dx = \beta^\alpha\Gamma(\alpha).$

I am doing some proofs related to the gamma distribution, and one of the results I need to show is that if $$\alpha > 1$$, $$\int_0^\infty \ x^{\alpha-1}e^{-x/\beta} \ dx = \beta^\alpha\Gamma(\alpha).$$

Applying integration by parts (with $$u = x^{\alpha-1}$$, $$dv = e^{-x/\beta} dx$$), I got $$\beta (\alpha-1)\int_0^\infty \ x^{\alpha-2}e^{-x/\beta} \ dx,$$

after all of the stuff in the other term from IBP evaluated to zero.

Now, just from looking at this integral, it looks like IBP is the way to go, but I don't see how this gets me any closer to the result I want. In order to get a $$\beta^\alpha$$ term, do I have to just keep doing integration by parts over and over? Since all I know is that $$\alpha > 1$$, how do I know when I am done doing integration by parts?

I also know that the definition of the gamma function is $$\Gamma(\alpha) = \int_0^\infty \ x^{\alpha-1}e^{-x} \ dx,$$ so I don't know what to do with the exponential in my integral to get to a gamma function.

If $$\alpha$$ had to be an integer, I would start at $$\alpha = 2$$ and try induction, but it can be any real number, so I am not sure how to deal with this.

Since you know that $$\Gamma(\alpha) = \int_0^\infty \ x^{\alpha-1}e^{-x} \ dx$$ considering $$I=\int_0^\infty \ x^{\alpha -1} e^{-\frac{x}{\beta }}\ dx$$ make $$x=\beta t$$ to get $$I=\beta^\alpha \int_0^\infty \ t^{\alpha-1}e^{-t} \ dt$$ and you are done.
Let $$\alpha,\beta\in\mathbb{R}$$ such that $$\alpha >1$$. Then substituting \small\left\lbrace\begin{aligned}y&=\frac{x}{\beta}\\ \mathrm{d}y&=\frac{\mathrm{d}x}{\beta}\end{aligned}\right., gives : $$\int_{0}^{+\infty}{x^{\alpha-1}\mathrm{e}^{-\frac{x}{\beta}}\,\mathrm{d}x}=\beta^{\alpha}\int_{0}^{+\infty}{y^{\alpha-1}\mathrm{e}^{-y}\,\mathrm{d}y}=\beta^{\alpha}\Gamma\left(\alpha\right)$$