I've worked out the solution to an integral which involves the following expression:
$$\sum_{n=0}^\infty \frac{(-\beta)^n}{n!}\Gamma\left(\frac{n+1}{2},\alpha\right)$$
Where $\Gamma(s,z)$ is the upper incomplete gamma function. My question is: is there a convenient non-series expression for this (i.e. an expression in terms of known special functions?). Note that it is very similar to the multiplication theorem given on the Wikipedia page (unfortunately I couldn't find a better source) for the incomplete gamma function, which states:
$$\Gamma(s,z) =\frac{1}{t^s}\sum_{n=0}^\infty \frac{\left(1-\frac{1}{t}\right)^n}{n!}\Gamma\left(s+n,tz\right)$$
If that helps at all. Thanks!
Using: $$\Gamma (a,z)=\int_z^{\infty } x^{a-1} \exp (-x) \, dx$$
we have:
$$\color{blue}{\sum _{n=0}^{\infty } \frac{(-\beta )^n \Gamma \left(\frac{n+1}{2},\alpha \right)}{n!}}=\\\sum _{n=0}^{\infty } \frac{(-\beta )^n \int_{\alpha }^{\infty } x^{\frac{n+1}{2}-1} \exp (-x) \, dx}{n!}=\\\int_{\alpha }^{\infty } \left(\sum _{n=0}^{\infty } \frac{(-\beta )^n x^{\frac{n+1}{2}-1} \exp (-x)}{n!}\right) \, dx=\\\int_{\alpha }^{\infty } \frac{e^{-x} \sum _{n=0}^{\infty } \frac{x^{n/2} (-\beta )^n}{n!}}{\sqrt{x}} \, dx=\\\int_{\alpha }^{\infty } \frac{e^{-x-\sqrt{x} \beta }}{\sqrt{x}} \, dx=\\\int_{\sqrt {\alpha}}^{\infty }\!2\,{{\rm e}^{-t \left( t+\beta \right) }}\,{\rm d}t =\color{blue}{\\e^{\frac{\beta ^2}{4}} \sqrt{\pi } \text{erfc}\left(\sqrt{\alpha }+\frac{\beta }{2}\right)}$$