How to use this gamma function in this statistics problem?

Question

Can someone explain to me how the steps to get from the top to the bottom (the gamma function)? I do not understand the steps taken in between. I see that the denominator of the fraction gets put in front of the fraction, but I don't understand the rule that allows that.

Answer 1

Suppose $X$ follows gamma distribution with shape parameter $k$ and scale parameter $\theta$, then

$$f_X(x; k, \theta) = \frac{1}{\Gamma(k)\theta^k}x^{k-1}e^{-\frac{x}{\theta}}$$

\begin{align} \int_0^\infty x e^{-\frac{x}{\theta}} \, dx &=\Gamma(2)\theta^2 \int_0^\infty \frac{1}{\Gamma(2)\theta^2} x^{2-1} e^{-\frac{x}{\theta}} \, dx \\ &= \Gamma(2)\theta^2 \end{align}

since $\int_0^\infty \frac{1}{\Gamma(2)\theta^2} x^{2-1} e^{-\frac{x}{\theta}} \, dx=1$ as $\frac{1}{\Gamma(2)\theta^2} x^{2-1} e^{-\frac{x}{\theta}}$ is a pdf.

Answer 2

Make the substitution $x=t \theta $, the limits will not change $dx = \theta dt$ \begin{eqnarray*} \int_{0}^{\infty} x e^{-\frac{x}{\theta}} dx = \int_{0}^{\infty} t \theta e^{-t} \theta dt =\theta^2 \int_{0}^{\infty} t e^{-t} dt . \end{eqnarray*}