# Integral involving Gamma distribution

I need some help with an integral. This is the solution to one of the problems I had to do. Everything is fine, but I don't understand one step:

Now how is

$$\int_0^\infty \frac{\beta_n^{\alpha_n+k}}{\Gamma(\alpha_n+k)}\theta^{\alpha_n+k-1}e^{-\beta_n\theta}d\theta=1$$

Could someone explain to me where did they take this result from? It is probably something standard that I should know, but I don't.

The usual Integral definition of the Gamma function makes this trivial if you substitute $$t=\beta_n\theta$$.

The Gamma Function can be defined as parametric integral

$$\Gamma(z)~=~\int_0^\infty t^{z-1}e^{-t}\mathrm dt\tag1$$

Now we may substitute $$z=\alpha_n+k$$ and $$t=\beta_n\theta$$ to obtain your integral. So we get that

$$\Gamma(\alpha_n+k)=\int_0^\infty t^{\alpha_n+k-1}e^{-t}\mathrm dt\stackrel{t=\beta_n\theta}=\int_0^\infty (\beta_n\theta)^{\alpha_n+k-1}e^{-\beta_n\theta} [\beta_n\mathrm d\theta]=\beta_n^{\alpha_n+k}\int_0^\infty \theta^{\alpha_n+k-1}e^{-\beta_n\theta}\mathrm d\theta$$

Now it is quite easy to conclude that

$$\frac{\beta_n^{\alpha_n+k}}{\Gamma(\alpha_n+k)}\int_0^\infty \theta^{\alpha_n+k-1}e^{-\beta_n\theta}\mathrm d\theta=1$$

$$\therefore~\int_0^\infty \frac{\beta_n^{\alpha_n+k}}{\Gamma(\alpha_n+k)}\theta^{\alpha_n+k-1}e^{-\beta_n\theta}\mathrm d\theta=1$$