Integrate exponential over shifted square root

Question

Does anyone know how to compute the following integral?

$$\int^t_0 e^{-a t'}\frac{1}{\sqrt{t-t'}}dt'$$ where $a>0$.

When I plug this into wolfram alpha, I get an imaginary error function term which is both a function of $t$ and $t'$, which doesn't make sense to me.

Answer 1

Set $\sqrt{t-t'}=u$. Then $du=\frac{-dt'}{2\sqrt{t-t'}}$. Hence your integral is equal to $$ 2\int_{0}^{\sqrt{t}}du\ e^{-a (t-u^2) }=2e^{-at}\int_0^\sqrt{t}du\ e^{a u^2}\ , $$ which can be solved in terms of the Dawson function.

Answer 2

It gives to me $${\frac {\sqrt {\pi}{{\rm e}^{-a\,t}}{\it erfi} \left( \sqrt {a\,t} \right) }{\sqrt {a}}}$$ where ${\it erfi} \left( x \right) =2\,{\frac {\int_{0}^{x}\!{{\rm e}^{{t}^{2}}}\,{\rm d}t}{ \sqrt{\pi}}}$