Evaluate the improper integral $\int_0^\infty x^{-1/2}e^{-(x-a)^2}dx$ Here is an integral I encountered in the wild:

$$\int_0^\infty x^{-1/2}e^{-(x-a)^2}dx .$$

If we substitute $t = x^{1/2}, dt = \frac{1}{2}x^{-1/2}dx$, we have
$$\int_0^\infty 2e^{-(t^2-a)^2}dt .$$
But how can I continue from here (or is it a good substitution to begin with)?
Edit Some comments below correspond to $\int_0^\infty x^{-1/2}e^{(x-a)^2}dx$, which indeed blows up.
 A: Denote the value of the definite integral by $$P(a) := \int_0^\infty \frac{\exp[-(x - a)^2] \,dx}{\sqrt{x}} .$$ For general $a$ there's not a nice, closed-form expression for $P(a)$ in terms of elementary functions.
That said, note that for $a = 0$, the substitution transforms $u = x^2$ transforms the integral to
$$P(0) = \frac{1}{2} \int_0^\infty u^{-\frac{3}{4}} e^{-u} du = \frac{1}{2} \int_0^\infty u^{\frac{1}{4} - 1} e^{-u} du = \frac{1}{2} \operatorname{\Gamma}\left(\frac{1}{4}\right) = 1.812804954\ldots ,$$
where $\Gamma$ is the Gamma function. We can also write $P(0)$ as $$P(0) = \frac{\pi^{3/4} G^{1 / 2}}{2^{1/4}},$$ where $G = 0.8346268415\ldots$ is Gauss' constant.
For general $a$, appealing to a c.a.s. gives $$P(a) = \frac{\pi}{2} \sqrt{|a|} \exp(-a^2) \left[I_{-1/4}\!\left(\frac{a^2}{2}\right) + \operatorname{sgn}(a) I_{1/4}\!\left(\frac{a^2}{2}\right)\right]$$ where $I_m$ is the modified Bessel function of the first kind. For $a < 0$ we can write the value using a single instance of the modified Bessel function of the second kind, $K_m$:
$$P(a) = \sqrt{-\frac{a}{2}} \exp\left(-\frac{a^2}{2}\right) K_{1/4}\!\left(\frac{a^2}{2}\right) .$$
We can also describe some asymptotic behavior of $P$: As $a \to -\infty$,
$$P(a) \sim \exp(-a^2) \sqrt{-\frac{\pi}{2 a}},$$
as $a \to 0$,
$$P(a) \sim \frac{1}{2}\operatorname{\Gamma}\left(\frac{1}{4}\right) + \operatorname{\Gamma}\left(\frac{3}{4}\right) a,$$
and as $a \to +\infty$,
$$P(a) \sim \sqrt{\frac{\pi}{a}} .$$
A: Almost the same as in the answer by Travis Willse, but with a derivation.
\begin{align*}
I(a)&=e^{-a^2}\int_0^\infty e^{2ax-x^2}x^{-1/2}\,dx
\\&=e^{-a^2}\sum_{n=0}^\infty\frac{(2a)^n}{n!}\int_0^\infty x^{n-1/2}e^{-x^2}\,dx
\\&=\frac{e^{-a^2}}{2}\sum_{n=0}^\infty\frac{(2a)^n}{n!}\Gamma\left(\frac n2+\frac14\right)
\\&=e^{-a^2}\frac{\sqrt\pi}2\sum_{n=0}^\infty a^n\frac{\Gamma\left(\frac n2+\frac14\right)}{\Gamma\left(\frac{n+1}2\right)\Gamma\left(\frac n2+1\right)}
\end{align*}
where the last step uses the duplication formula for $n!=\Gamma(n+1)$.
Summing separately over $n=2k$ and $n=2k+1$ for $k\geqslant 0$, we get $$I(a)=e^{-a^2}\frac{\sqrt\pi}2\left[F\left(\frac14,\frac12;a^2\right)+aF\left(\frac34,\frac32;a^2\right)\right],$$ where $F(\alpha,\beta;z)=\sum_{k=0}^\infty\frac{\Gamma(\alpha+k)}{\Gamma(\beta+k)}\frac{z^k}{k!}=\frac{\Gamma(\alpha)}{\Gamma(\beta)}{}_1F_1(\alpha;\beta;z)$ as seen here.
$F(\nu+1/2,2\nu+1;z)=\sqrt\pi e^{z/2}z^{-\nu}I_\nu(z/2)$ (for $\Re\nu>-1/2$) can be deduced from $$\int_0^1 t^{\alpha-1}(1-t)^{\beta-\alpha-1}e^{zt}\,dt=\Gamma(\beta-\alpha)F(\alpha,\beta;z);\\\int_{-1}^1(1-t^2)^{\nu-1/2}e^{zt}\,dt=\sqrt\pi\Gamma(\nu+1/2)(z/2)^{-\nu}I_\nu(z)$$ (here $0<\Re\alpha<\Re\beta$), both shown using the power series of $e^{zt}$ and termwise integration.
Note that $z\mapsto z^{-\nu}I_\nu(z)$ is considered an entire function here, so that likewise $$I(a)=\frac\pi2e^{-a^2/2}a^{1/2}\big(I_{-1/4}(a^2/2)+I_{1/4}(a^2/2)\big)$$ and $I(-a)=(-a/2)^{1/2}e^{-a^2/2}K_{1/4}(a^2/2)$ as given in the mentioned answer.
