Integral of $2\int_0^{\infty} e^{-2y}e^{-y}{{(y)^x}\over{x!}} dy$ Is it possible to take this integral?
$2\int_0^{\infty} e^{-2y}e^{-y}{{(y)^x}\over{x!}} dy$
I know I can use the fact that: $\int_0^{\infty} y^k e^{−y} dy = k!$
But I'm basically stuck on how to do this by parts.
I'm ending up with (by parts):
$p_1(x) = 2(x! {{e^{-2y}\over{x!}}} - \int ({{e^{-2y}\over{x!}}})' x!dx)$
I'm not sure how to handle that last integral. It looks like it needs to be broken down by parts as well?
This isn't quite right:
$p_1(x) = 4(e^{-2y} - \int e^{-2y} )$
I should end up with:
${1\over3}({2\over3})^x$
Thanks..

We have
\begin{align*}
2\int_0^\infty  {e^{ - 2y} e^{ - y} \frac{{y^x }}{{x!}}dy} & = \frac{2}{{x!}}\int_0^\infty  {e^{ - 3y} y^x dy}  = \frac{2}{{3^{x + 1} x!}}\int_0^\infty  {e^{ - 3y} (3y)^x d(3y)} \\ & = \frac{2}{{3^{x + 1} x!}}\int_0^\infty  {e^{ - t} t^x dt}  = \frac{2}{{3^{x + 1} }}
\end{align*}
provided $\Re x>-1$.
