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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.

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  • $\begingroup$ Not wolfram, but maple 11 on a windos 98 virtual machine: AI := Int(exp(-a*x)/sqrt(t-x), x = 0 .. t); - t - / - | exp(-a x) - AI := | ---------- dx - | 1/2 - / (t - x) - 0 > A := value(AI); - 1/2 1/2 1/2 - erf(t (-a) ) Pi exp(-a t) - A := --------------------------------- - 1/2 - (-a) The use of erfi is by hand for avoid the minus on a, and convert to reals all values. $\endgroup$ – Alvaro Jan 18 '18 at 5:56
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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.

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

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  • $\begingroup$ Out of curiosity, what was the exact expression you used in wolfram? $\endgroup$ – rhz Jan 18 '18 at 0:54

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