I want to derive the Poisson Lindley Distribution.
$$ f_x(x|\lambda) = \frac{\lambda^{x-1}}{(x-1)!}e^{-\lambda} $$ $$f_x(x|p) = \frac{p^2}{(p+1)}(\lambda+1)e^{-\lambda p} $$
The Distribution of x, $f_x(x) = \int ^\infty _0 f_x(x|\lambda) f_\lambda(\lambda) \ d\lambda$
Then $$ f_x(x) = \frac{p^2}{(p+1)(x-1)!} \int_0^\infty \lambda^{x-1}(\lambda+1) e^{-\lambda (1+p)} \; d\lambda $$
I broke it as sum of 2 integrals:
$$ f_x(x) = \frac{p^2}{(p+1)(x-1)!}\left[ \int^\infty_0 \lambda^xe^{-\lambda(1+p)} \; d\lambda \ +\int^\infty_0 \lambda^{(x-1)}e^{-\lambda (1+p)} \; d\lambda \right] $$
As I solve $\int ^\infty _0 \lambda^xe^{-\lambda(1+p)} \ d\lambda = \frac{\lambda^x e^{-\lambda(1+p)}}{-(1+p)} +\frac{1}{1+p} \int ^\infty _0 \lambda^{x-1}e^{-\lambda(1+p)} \ d\lambda $
Then rewriting $f_x(x):$ $$f_x(x) = \frac{p^2}{(p+1)(x-1)!} \int^\infty_0 \frac{p+2}{p+1} \lambda^{x-1} e^{-\lambda (1+p)} \; d\lambda $$
I am unable to proceed further to show that the Poisson-Lindley distribution is:
$$\frac{p^2}{(p+1)^{x+3}}(p+2+x)$$
