Compute $\int_0^\infty e^{-az} \sum_{n=0}^\infty \frac{z^{n+2}(u+z)^n}{(n+1)!(n+2)!} dz$ I got stuck in this problem:
$$\int_0^\infty e^{-az} \sum_{n=0}^\infty \frac{z^{n+2}(u+z)^n}{(n+1)!(n+2)!} dz ~~(1)$$ 
where $a>0$.
My thoughts: By Binomial expansion we have
$$ \int_0^\infty e^{-az} z^{n+2}(u+z)^ndz\\
= \sum_{k=0}^n \frac{n!u^k}{k!(n-k)!} \int_0^\infty z^{2n-k+2}e^{-az}dz\\
= \sum_{k=0}^n \frac{n!u^k}{k!(n-k)!} \frac{(2n-k+2)!}{a^{2n-k+3}}.~~~~~~~~
$$
So 
$$(1)=\sum_{n=0}^\infty \frac{1}{(n+1)!(n+2)!} \sum_{k=0}^n \frac{n!u^k}{k!(n-k)!} \frac{(2n-k+2)!}{a^{2n-k+3}}.
$$
Then I got lost. Please let me know if you have any idea or direction about computing $(1)$. Any help would be greatly appreciated, thanks!
 A: An explicit form for the series can be found. From the series expansion for the modified Bessel function
\begin{equation}
I_{1}\left(z\right)=\sum_{k=0}^{\infty}\frac{(\tfrac{z}{2})^{2k+1}}{k!\left(k+1\right)!}
\end{equation} 
we have
\begin{align}
\sum_{n=0}^\infty \frac{z^{n+2}(u+z)^n}{(n+1)!(n+2)!} &=\frac{\sqrt{z}}{\left( u+z \right)^{3/2}}\sum_{n=0}^\infty \frac{\left[ z(z+u) \right]^{n+3/2}}{(n+1)!(n+2)!}\\
&=\frac{\sqrt{z}}{\left( u+z \right)^{3/2}}\sum_{k=1}^\infty \frac{\left[ z(z+u) \right]^{k+1/2}}{k!(k+1)!}\\
&=\frac{\sqrt{z}}{\left( u+z \right)^{3/2}}\sum_{k=1}^\infty \frac{\left[ \sqrt{z(z+u) }\right]^{2k+1}}{k!(k+1)!}\\
&=\frac{\sqrt{z}}{\left( u+z \right)^{3/2}}\left[I_1\left( 2\sqrt{z(z+u) } \right)-\sqrt{z(z+u) }\right]\\
&=\frac{\sqrt{z}}{\left( u+z \right)^{3/2}}I_1\left( 2\sqrt{z(z+u) } \right)-\frac{z}{z+u}
\end{align} 
To calculate the Laplace transform, we can use an identity tabulated in Ederlyi TI 4.17.15:
\begin{align}
\int_0^\infty &t^{\mu-1}\left( t+\beta \right)^{-\mu}I_{2\nu}\left( \alpha\left( t^2+\beta t \right)^{1/2} \right)e^{-pt}\,dt\\
&=\frac{2\Gamma\left( \mu+\nu \right)e^{\beta p/2}}{\alpha\beta\Gamma(2\nu+1)}M_{1/2-\mu,\nu}\left( \frac{\alpha^2\beta}{2p+\sqrt{p^2-\alpha^2}} \right)W_{1/2-\mu,\nu}\left( \frac{\beta\left( p+\sqrt{p^2-\alpha^2} \right)}{2} \right)
\end{align} 
(with $\mu=3/2, \nu=1/2,\beta=u,\alpha=2$) and the integral representation of the exponential integral (DLMF)
\begin{align}
\int_0^\infty\frac{ze^{-az}}{z+u}\,dz&=\int_0^\infty e^{-az}\,dz-u\int_0^\infty\frac{e^{-az}}{z+u}\,dz\\
&=\frac{1}{a}-ue^{au}E_{1}\left( au \right)
\end{align}
