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!
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}