# Evaluating sum over partial gamma function

I am trying to calculate the expectation value of some quantity and it eventually boils down to the following sum (and another similar one): $$\sum_{n=0}^{\infty} \frac{\beta^n (-\alpha)^{n-1/2}\Gamma(1/2-n,-\alpha)}{n!}$$

where $0<\beta<\alpha$ and $\Gamma$ is the partial Gamma function. Does anyone have any ideas of how to evaluate this in terms of known special functions? This would be the best case scenario. I would also be interested in numerical algorithms which approximate it. It seems like I am sampling the partial Gamma function in the most annoying parts of its domain...although the summand is related to $\gamma*$, the usual holomorphic extension function.

I have been trying to express the partial gamma function in terms of various integrals and sums, switch the order, and see what happens, with no success. I have been searching through function handbooks and partial gamma function papers trying to find useful identities, but with no luck.

$$\begin{eqnarray*}\sum_{n\geq 0}\frac{\beta^n}{n!}\cdot\left((-\alpha)^{n-1/2}\int_{-\alpha}^{+\infty}x^{-n-1/2}e^{-x}\,dx\right)&=&\int_{-1}^{+\infty}\sum_{n\geq 0}\frac{\beta^n}{n!}(-1)^{n-1/2}x^{-n-1/2}e^{-\alpha x}\,dx\\&=&\int_{-1}^{+\infty}-\frac{ i e^{-\frac{\beta}{x}-\alpha x}}{\sqrt{x}}\,dx\end{eqnarray*}$$ hence we have a peculiar situation: the real part of the series depends on the contribute given by $\int_{-1}^{0}$, while the imaginary part depends on the contribute given by $\int_{0}^{+\infty}$, that is the value of a Laplace transform. In particular, $$\int_{0}^{+\infty}\exp\left(-\frac{\beta}{x}-\alpha x\right)\frac{dx}{\sqrt{x}}=\sqrt{\frac{\pi}{\alpha}}\exp\left(-2\sqrt{\alpha\beta}\right).$$ So our problem boils down to computing: $$\int_{0}^{1}\exp\left(\frac{\beta}{x}+\alpha x\right)\frac{dx}{\sqrt{x}}=2\left(\frac{\beta}{\alpha}\right)^{1/4}\int_{0}^{\left(\frac{\alpha}{\beta}\right)^{1/4}}\exp\left[\sqrt{\alpha\beta}\left(x^2+\frac{1}{x^2}\right)\right]\,dx$$ that through the substitution $x-\frac{1}{x}=z$ is mapped into an integral of the following kind: $$e^{2\sqrt{\alpha\beta}}\int_{-\infty}^{t}e^{z^2\sqrt{\alpha\beta}}\left(1+\frac{z}{\sqrt{4+z^2}}\right)\,dz$$ with $t>1$.
• can you explain what $PV$ means? – Ian Hincks Apr 19 '16 at 13:48
• Also I am unsure why the very first integral is $\int x^{n-3/2}e^{-x}dx$ rather than $\int x^{-n+1/2}e^{-x} dx$. – Ian Hincks Apr 19 '16 at 14:18