Sum involving the hypergeometric function, power and factorial functions

Question

I am finding some trouble in calculating the following sum involving the hypergeometric function, power and factorial functions. $$ \sum_{y=1}^\infty \mathrm{e}^z \cdot {}_1F_1\left(1-y;2;-z\right) \frac{\mu^y}{y!} \mathrm{e}^{-\mu} $$

Could you please provide me some hints to solve this sum? Thank you in advance

Answer 1

1. Multiply your sum by $e^{\mu-z}$ and differentiate with respect to $\mu$ to get $$\sum_{n=0}^{\infty} {}_1F_1(-n,2,-z)\frac{\mu^n}{n!}\tag{1}$$
2. Use the (I hope, accidentally) unaccepted answers to your question here to express $_1F_1$ in the sum as generalized Laguerre polynomials: $$ \frac{_1F_1(-n,2,-z)}{n!}=\frac{L_n^{(1)}(-z)}{(1+1)_n}.\tag{2}$$
3. Use (2) and the generating function of Laguerre polynomials (formula 18.12.14) to sum up the series (1) to $$ \left(-\mu z\right)^{-1/2}e^{\mu}J_1(2\sqrt{-\mu z})=\left(\mu z\right)^{-1/2}e^{\mu}I_1(2\sqrt{\mu z})$$
4. Integrate back with respect to $\mu$ and multiply the result by $e^{z-\mu}$. The final result is $$ e^{z-\mu}\int_0^{\mu}\left(\nu z\right)^{-1/2}e^{\nu}I_1(2\sqrt{\nu z})\,d\nu=z^{-1}e^{z-\mu}\int_0^{2\sqrt{\mu z}}e^{{x^2}/{4z}}I_1(x)\,dx.$$ I don't think the last integral can be expressed in terms of elementary or reasonably simple special functions.