You can check that both series converge for all real x (say, using the ratio test).
Now,
[tex]\displaystyle e^x = \sum_{n=0}^\infty \frac{x^n}{n!} \implies x e^{-x^2/2} = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{2^n n!}[/tex]
and if you have some familiarity with differential equations, you might recognize that the second series represents a Bessel function, namely
[tex]\displaystyle I_0(x) = \sum_{n=0}^\infty \frac{x^{2n}}{2^{2n} (n!)^2}[/tex]
but we won't bother making use of this.
In the integral, we interchange the integral with the sum
[tex]\displaystyle \int_0^\infty x e^{-x^2/2} \sum_{n=0}^\infty \frac{x^{2n}}{2^{2n} (n!)^2} \, dx = \sum_{n=0}^\infty \frac1{2^{2n} (n!)^2} \int_0^\infty x^{2n+1} e^{-x^2/2} \, dx[/tex]
and consider the sequence of integrals,
[tex]J_n = \displaystyle \int_0^\infty x^{2n} x e^{-x^2/2} \, dx[/tex]
Integrating by parts with
[tex]u = x^{2n} \implies du = 2n x^{2n-1} \, dx[/tex]
[tex]dv = x e^{-x^2/2} \, dx \implies v = -e^{-x^2/2}[/tex]
Then the contribution of uv is 0, so
[tex]J_n = \displaystyle 2n \int_0^\infty x^{2(n-1)} x e^{-x^2/2} \, dx = 2n J_{n-1}[/tex]
We have
[tex]J_0 = \displaystyle \int_0^\infty x e^{-x^2/2} \, dx = 1[/tex]
so that
[tex]J_n = 2n J_{n-1} = 2n (2(n-1)) J_{n-2} = 2n (2(n-1)) (2(n-2)) J_{n-3}[/tex]
and so on; after k steps,
[tex]J_n = 2n (2(n-1)) (2(n-2)) \cdots (2(n-(k-1))) J_{n-k}[/tex]
so that when n = k,
[tex]J_n = 2n (2(n-1)) (2(n-2)) \cdots 2 J_0 = 2^n n![/tex]
Then the integral we want is
[tex]\displaystyle \int_0^\infty x e^{-x^2/2} \sum_{n=0}^\infty \frac{x^{2n}}{2^{2n} (n!)^2} \, dx = \sum_{n=0}^\infty \frac{2^n n!}{2^{2n} (n!)^2} = \sum_{n=0}^\infty \frac1{2^n n!}[/tex]
and knowing the series expansion for [tex]e^x[/tex], it follows that the integral has a value of [tex]e^{1/2} = \boxed{\sqrt{e}}[/tex]
