Gamma function for $\int_{0}^{\infty}x^{n/2-2}e^{-x}dx$

Question

To solve the integral i have made a change of variable to arrive to gamma function. But the exponential value of the new variable x is n/2-2. How can i continue? \begin{equation*} \begin{split} n\theta \int_{0}^{\infty}\frac{1}{t}\frac{(\theta t)^{n/2-1}}{2^{n/2}\Gamma(n/2)}e^{-\theta t/2}dt \end{split} \end{equation*} This is my change of variable: \begin{align*} x=\frac{\theta}{2}t \quad & \frac{2x}{\theta}=t \quad & x=0 \;\; x=\infty \\ dx=\frac{\theta}{2}dt \quad &\frac{2dx}{\theta}=dt \quad & t=0 \;\; t=\infty \end{align*}

\begin{equation*} \begin{split} n\theta \int_{0}^{\infty}\frac{1}{t}\frac{(\theta t)^{n/2-1}}{2^{n/2}\Gamma(n/2)}e^{-\theta t/2}dt &= \frac{n\theta^{2}}{\Gamma(n/2)}\int_{0}^{\infty}\frac{(\theta)^{n/2-2}}{2^{n/2}}t^{n/2-2}e^{-\theta t/2}dt= \frac{n\theta^2}{\Gamma(n/2)}\int_{0}^{\infty}\frac{\theta^{n/2-2}}{2^{n/2}}\frac{2^{n/2-2}}{\theta^{n/2-2}}x^{n/2-2}e^{-x}\frac{2}{\theta}dx \\ &=\frac{n\theta^2}{\Gamma(n/2)}\int_{0}^{\infty}\frac{2}{\theta}\;x^{n/2-2}e^{-x}dx= \frac{2n\theta}{\Gamma(n/2)}\int_{0}^{\infty}x^{n/2-2}e^{-x}dx \end{split} \end{equation*}
But in the last integral, I have been looking for a gamma function for that form of integral but nothing

Answer 1

The last integral is precisely $\Gamma\left(\frac{n}{2}-1\right)$. Do you se why? Then, you have:

$$\frac{2n\theta}{\Gamma(n/2)}\Gamma\left(\frac{n}{2}-1\right)=\frac{2n\theta\cdot\Gamma(n/2-1)}{(n/2-1)\Gamma(n/2-1)}=\frac{4n\theta}{n-2}$$With the derivation valid for any real $n>2$.