calculus - Evaluating the definite integral $int_0^infty x mathrm{e}^{-frac{(x-a)^2}{b}},mathrm{d}x$

To evaluate $\int_0^\infty x \mathrm{e}^{-\frac{(x-a)^2}{b}}\,\mathrm{d}x$, I have applied the substitution $u=\frac{(x-a)^2}{b}$, $x-a=(ub)^{1/2}$, and $\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{2(x-a)}{b}$. I would first like to ask if $x-a$ should actually equal $\pm (ub)^{1/2}$ (i.e., is it valid to ignore the minus sign, and why?). Applying this substitution,
$$\begin{align} I &=\int_0^\infty x \mathrm{e}^{-\frac{(x-a)^2}{b}}\, \mathrm{d}x \\ &=\int_0^\infty x \mathrm{e}^{-u} \frac{b\,\mathrm{d}u}{2(x-a)} \\ &=\int_{\frac{a^2}{b}}^\infty \left((ub)^{1/2} + a\right)\cdot{}\mathrm{e}^{-u} \frac{b\,\mathrm{d}u}{2(ub)^{1/2}} \\ &=\frac{b}{2}\int_{\frac{a^2}{b}}^\infty \mathrm{e}^{-u}\,\mathrm{d}u + ab^{-1/2}\mathrm{e}^{-u} u^{-1/2}\,\mathrm{d}u \\ &=\frac{b}{2}\int_{\frac{a^2}{b}}^\infty \mathrm{e}^{-u}\,\mathrm{d}u + \frac{a\sqrt{b}}{2}\int_{\frac{a^2}{b}}^\infty \mathrm{e}^{-u} u^{-1/2}\,\mathrm{d}u, \end{align}$$
I find that I am unable to evaluate the second term because the domain is from a non-zero constant to $+\infty$. Were the domain $[0,+\infty)$, the second integral would simply be a $\Gamma$ function. Seeing that the substitution I have attempted has not worked, could someone please propose an alternate route to evaluating this definite integral? Thank you.