I have to somehow show that: $$ \int_0^\infty{\frac{e^{-a t}}{t^{1/2}(t+x)}} \textrm{d}t=\frac{\pi}{\sqrt{x}} e^{a x} \textrm{erfc}(\sqrt{ax}) $$
I've tried substituting $u=\sqrt{t}$ to get $$ 2 \int_0^\infty{\frac{e^{-a u^2}}{u^2+x}} \textrm{d}u $$ which at least has the Gaussian but I don't know where to go from here.
I've also tried looking at the problem backwards but I don't seem to be able to get the polynomial in the denominator.
What's the approach for these problems involving the error function?
Thank you!
Consider the integral
$$\int_0^{\infty} du \frac{e^{-a u^2}}{u^2+x} = e^{a x} \int_0^{\infty} du \frac{e^{-a (u^2+x)}}{u^2+x} = e^{a x} J(a)$$
Then
$$J'(a) = \int_0^{\infty} du e^{-a (u^2+x)} = \sqrt{\pi} e^{-a x} a^{-1/2}$$
Integrating
$$J(a) = \sqrt{\pi} \int_0^a da' \, a'^{-1/2}\, e^{-a' x} = \sqrt{\frac{\pi}{x}}\int_0^{\sqrt{a x}} dt\,e^{-t^2} = \frac{\pi}{2 \sqrt{x}} \operatorname*{erf}{\left ( \sqrt{ax}\right)}$$