Rewrite integral $\frac{1}{\pi}\int_0^{\infty}\frac{\sqrt{x}e^{-xt}}{a^2 + x}\text{d} x$ in terms of the complementary error function. I am told by various sources that

$$\frac{1}{\pi}\int_0^{\infty}\frac{\sqrt{x}e^{-xt}}{a^2 + x}\text{d} x = \frac{1}{\sqrt{\pi t}} - ae^{a^2t}\text{Erfc}(a\sqrt{t}),\quad a > 0,\, t > 0, $$

where $\displaystyle\text{Erfc}(x) = \frac{2}{\sqrt{\pi}}\int_x^{\infty}e^{-t^2}\text{d} t$ is the complementary error function, but I am unable to reformulate the left hand side to match the right, or vice versa. Does anyone know how to do this/can point me in the direction of a proof or similar proof?.
 A: Let $I$ be the integral in question.  Then, after following the hint I gave above, we can start with
$$\begin{align}I &= \frac1{\sqrt{\pi t}} - \frac{a^2}{\pi}  \int_{-\infty}^{\infty} du \, \frac{e^{-t u^2}}{a^2+u^2} \end{align}$$
Parseval's theorem states that the integral of the product of two functions is equal to the integral of the product of of the Fourier transforms of those functions (times $1/(2 \pi)$).  Now, the FT's we need are
$$\int_{-\infty}^{\infty} du \, e^{-t u^2} e^{i k u} = \sqrt{\frac{\pi}{t}} e^{-k^2/(4 t)}$$
$$\int_{-\infty}^{\infty} du \, \frac{e^{i k u}}{a^2+u^2} = \frac{\pi}{a} e^{-a |k|} \qquad (a \gt 0)$$
Then
$$\begin{align} \int_{-\infty}^{\infty} du \, \frac{e^{-t u^2}}{a^2+u^2} &= \frac1{2 \pi} \sqrt{\frac{\pi}{t}} \frac{\pi}{a} \int_{-\infty}^{\infty} dk \, e^{-k^2/(4 t)} e^{-a |k|} \\ &= \frac{\pi}{a \sqrt{\pi t}} \int_0^{\infty} dk \, e^{-k^2/(4 t)} e^{-a k} \\ &= \frac{\pi}{a \sqrt{\pi t}} e^{a^2 t} 2 \sqrt{t} \int_{a \sqrt{t}}^{\infty} dv \, e^{-v^2} \\ &= \frac{\pi}{a} e^{a^2 t} \operatorname{erfc}{\left ( a \sqrt{t} \right )} \end{align} $$
Therefore, as was to be shown,

$$I =  \frac1{\sqrt{\pi t}} - a \, e^{a^2 t} \operatorname{erfc}{\left ( a \sqrt{t} \right )} $$

For large $t$, 
$$I = \frac1{2 \sqrt{\pi} a^2} t^{-3/2} + O \left ( t^{-5/2} \right ) $$
