# How did the author find this integral over a Poisson distribution whose parameter has a exponential distribution?

So my text book on probability theory gives an example of a distribution with random parameters, which after a few paragraphs ends up with this integral:

$$\begin{split} \int_0^{\infty}\frac{x^k}{k!}e^{-2x}dx &= \frac{1}{2^{k+1}} \int_0^{\infty}\frac{1}{\Gamma(k+1)}2^{k+1}x^{k+1-1}e^{-2x}dx\\ &= \frac{1}{2^{k+1}} \cdot 1 \end{split}$$

where $$\Gamma$$ is the Gamma function.

My question is this: How in the world did the author calculate this integral?

For those of you who are curious, that integral is the result of calculating $$P(X=k|M=x)$$ with $$M \in \text{Exp}(1)$$ and $$X|M=m \in \text{Po}(m)$$. The example can be found in An intermediate course in Probability, 2nd Ed. (Gut, 2009, pp 39).

Just doing the substitution $$u=2x$$ and using $$\Gamma (t)=\int_0^\infty x^{t-1}e^{-x}\,\mathrm d x.$$
• $$\int_0^\infty x^ke^{-2x}\,\mathrm d x\underset{u=2x}{=}\frac{1}{2^{k+1}}\int_0^\infty u^ke^{-u}\,\mathrm d u=\frac{1}{2^{k+1}}\Gamma (k+1).$$@Mossmyr
• Nooow I see. Then you can cancel the $\frac{1}{\Gamma(k+1)}$ found in the equation in my post. But I think you meant to write $\frac{1}{2^k}$ in your comment. Sep 18, 2019 at 13:11
• No, it's $\frac{1}{2^{k+1}}$ (that come from the fact that $dx=\frac{1}{2}du$)