# Bayes - Integral to Gamma Function

When trying to understand my professors notes, I came upon this piece and don’t understand the step he took.

$$\frac{(\sum x_i+2)^{n+1}}{\Gamma (n+1)} \int_0^\infty\theta^{n+1} e^{-\theta \sum x_i+2}\text{d}\theta$$

To

$$\frac{(\sum x_i+2)^{n+1}}{\Gamma (n+1)} \frac{\Gamma (n+2)}{(\sum x_i+2)^{n+1}}$$

• Google Gamma distribution and Gamma function. – Xi'an Jan 19 at 16:21
• Thanks, I got the $\Gamma (\Theta) = \Theta^{n+1}*e^{-\Theta}$. But why do we have divided by the term above? – ToTom Jan 19 at 16:34
• That's not what the Gamma function is: in your comment you have written down part of the density for a Gamma$(n+2)$ distribution. – whuber Jan 19 at 17:04
• Hint: try a change of scale in the original integral. – Xi'an Jan 19 at 17:26