How to calculate the exp-like series with reciprocal square coefficients?

Question

I am trying to calculate a series $\sum_{n=0}^\infty \frac{1}{(a+1+n)^2} \frac{(-x)^n}{n!}$ for $a \in [0, 100]$ and $x \in [0,150]$. I know that it is related to generalized hypergeometric $_2F_2(a+1,a+1;a+2,a+2;-x)$. But I don't want to use the complicated $_2F_2()$. Any suggestions on how to approximately calculate this series? Such as asymptotes with simple functions like $\exp(),\log(), \Gamma()$, etc. Thanks.

Answer 1

Since $\frac{1}{(a+1+n)^2} = \int_{0}^{1}- z^{n+a}\log(z)\,dz$ the whole series can be represented as

$$ \int_{0}^{1}-z^a \log(z) \sum_{n\geq 0}\frac{(-xz)^n}{n!}\,dz = \int_{0}^{1}-z^a \log(z) e^{-xz}\,dz=\frac{1}{x^{a+1}}\int_{0}^{x}-u^a e^{-u}\log\frac{u}{a}\,du $$ and computed through the incomplete $\Gamma$ function and its derivative, having manageable representations as generalized continued fractions.