Is there any standard way to evaluate the following summation?
$$ S(x):=\sum_{k=0}^\infty \frac{x^k}{(k+1/2)!} $$
where $(k+1/2)! = (k+1/2)(k-1/2)\ldots (1/2)$.
EDIT : As some of you asked for motivation for this question, the only thing that I can come up with is that I was trying to think of an analogous formula for half integral arguments $s$,
$$ \Gamma(s,x) = (s-1)! e^{-x} \sum_{m=0}^{s-1} \frac{x^m}{m!} $$
$$S(x)=\sum_{k=0}^\infty \frac{x^k}{(k+1/2)!}=\sum_{k=0}^\infty \frac{(2x)^k (2k)!!}{(2k+1)!}=\sum_{k=0}^\infty \frac{(4x)^k k!}{(2k+1)!}=\sum_{k=0}^\infty \frac{(4x)^k }{k!}\frac{(k!)^2}{(2k+1)!}$$ Given that $$\frac{(k!)^2}{(2k+1)!}=\frac{\Gamma(k+1)\Gamma(k+1)}{\Gamma(2k+2)}=B(k+1;k+1)$$ $$S(x)=\sum_{k=0}^\infty \frac{(4x)^k }{k!}\int_0^1t^k(1-t)^kdt$$ Changing the order of summation and integration and summing the series $$S(x)=\int_0^1e^{4xt(1-t)}dt\,=\,\Big(t=z+\frac{1}{2}\Big)\,\,\int_{-\frac{1}{2}}^{\frac{1}{2}}e^{4x(\frac{1}{4}-z^2)}dz=2e^x\int_0^\frac{1}{2}e^{-4xz^2}dz$$ $$S(x)=\frac{e^x}{\sqrt x}\int_0^\sqrt xe^{-t^2}dt=\frac{\sqrt\pi}{2}\frac{e^x}{\sqrt x}\text{erf}(\sqrt x)$$