Messing with summations

Note that $\sum_{r=0}^{\infty} \left(\frac {(-1)^r x^{2r+1}}{2^r r! }\right)=x\sum_{r=0}^{\infty} \frac {(\frac {-x^2}{2})^r}{r!}=xe^{\frac {-x^2}{2}}$

Hence integral evaluates to $\sum_{k=0}^{\infty} \frac {1}{4^k(k!)^2}\int_0^{\infty} x^{2k+1}e^{\frac {-x^2}{2}}dx$

On substituting $\frac {x^2}{2} =t$ the integral turns to $\sum_{k=0}^{\infty} \frac {1}{2^k(k!)^2}\int_0^{\infty} t^k e^{-k} dt$

And using well known result $\int_0^{\infty} t^k e^{-k} dt=k!$

Hence our integral turns to solving the summation $\sum_{k=0}^{\infty} \frac {\left(\frac 12\right)^k}{k!}=e^{\frac 12}$