Unexpected Summation

$\large \int_{0}^{\infty}{e^{-x^2}}xcosx dx \\$ $\ \small \displaystyle \color{#3D99F6}\text{let } x^{2} = u \implies \frac{du}{2} = x dx \\ \implies \displaystyle \int_{0}^{\infty}{e^{-\color{#3D99F6}u}cos{\sqrt{u}}} \frac{du}{2} \\ \text{From taylor series expansion of cosine function, we get -} \\ \implies \displaystyle \frac12 \int_{0}^{\infty}{e^{-u}}\displaystyle {\color{#3D99F6}\sum_{n=0}^\infty \frac{(-1)^{n} (\sqrt{u})^{2n}}{(2n)!}} du \\ \implies {\displaystyle \frac12 {\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}} \int_{0}^{\infty}{e^{-u}}u^n du} \\ \text{The above integral is of the form - } \displaystyle{\Gamma(s+1) = \int_0^\infty e^{-x}x^{s}dx = (s)!} \\ \implies {\displaystyle \frac12 \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} {\color{#3D99F6} \int_{0}^{\infty}{e^{-u}}u^n du}} = \displaystyle \frac12 \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}{\color{#3D99F6}n!} \\ = \boxed{\displaystyle \sum_{n=0}^\infty \frac{(-1)^n n!}{2(2n)!}}, \implies a=-1,\ b=1, \ c=2 \ and \ d=2 \\ \implies a \ + \ b \ + \ c \ + \ d = \boxed{4}$