# Given conditional probability, find Marginal PMF

Suppose $$Y \sim gamma(\alpha= 2; \beta= 1)$$ and the conditional distribution of $$X$$ given Y=y is Poisson with mean = y. Find the marginal probability mass function of $$X$$.

So far I've done:

$$f(x,y) = f(X|Y=y)f_Y(y)$$

$$= \frac{y^xe^{-y}}{x!}ye^{-y}$$

$$f_X(x) = \int_o^\infty \frac{y^{x+1}e^{-2y}}{x!} dy$$

but then after that I'm not sure if there's a special identity related to the Gamma distribution or something that will help with integration. The answer is supposed to be $$\frac{x+1}{2^{x+2}}$$

Note: Integrate by Parts to show for all natural numbers, $$n$$: $$\int_0^\infty y^{n+1} e^{-2y}\mathsf d y = \tfrac 12(n+1)\int_0^\infty y^ne^{-2y}\mathsf d y$$
Also $$\int\limits_0^\infty e^{-2y}\mathsf d y=\tfrac 12$$.