Proof of a formula containing double factorial

Question

How can I prove the formula: $$\sum_{k=0}^\infty\frac{x^k}{k!!}=\frac{1}{2}e^{x^2/2}\left[2+\sqrt{2\pi}\operatorname{erf}\frac{x}{\sqrt\pi}\right]?$$ Thanks

Answer 1

I believe our definitions of the error function differ by a constant, but the following approach works anyway:

$$\begin{eqnarray*}\sum_{k\geq 0}\frac{x^k}{k!!}&=&\sum_{n\geq 0}\frac{x^{2n}}{2^n n!}+\sum_{n\geq 0}\frac{x^{2n+1}}{(2n+1)!!}\\&=&e^{x^2/2}+\sum_{n\geq 0}\frac{2^n x^{2n+1}}{(2n+1)!}n!\\&=&e^{x^2/2}+\int_{0}^{+\infty}e^{-z}\sum_{n\geq 0}\frac{2^n x^{2n+1} z^n}{(2n+1)!}\,dz\\&=&e^{x^2/2}+\int_{0}^{+\infty}e^{-u^2}\sum_{n\geq 0}\frac{2^{n+1} x^{2n+1} u^{2n+1}}{(2n+1)!}\,du\\ &=&e^{x^2/2}+\int_{0}^{+\infty}\sqrt{2}\,e^{-u^2}\sinh\left(\sqrt{2}\, u x\right)\,du\\&=&\color{red}{e^{x^2/2}+e^{x^2/2}\sqrt{\frac{\pi}{2}}\,\text{Erf}\left(\frac{x}{\sqrt{2}}\right)}.\end{eqnarray*}$$