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?.

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    $\begingroup$ Sub $x=u^2$ and use the fact that $$\frac{u^2}{a^2+u^2} = 1-\frac{a^2}{a^2+u^2} $$Then apply Parseval's theorem or differentiation under the integral for the resulting integral. $\endgroup$ – Ron Gordon Dec 5 '17 at 17:47
  • $\begingroup$ Thanks for the response @RonGordon. I am not entirely sure how you would use either differentiation under the integral or Parseval's Theorem here, but I managed to get the result by writing $$\int_0^{\infty}\frac{e^{-u^2t}}{a^2 + u^2}\text{d}u = \int_0^{\infty}e^{-u^2t}\int_0^{\infty}e^{-(a^2 + u^2)x}\text{d} x\text{d} u $$ and then applying Fubini's Theorem plus a few more substitutions. If you have time I would like to see a few more details, say the first step or two, of your proposed methods. $\endgroup$ – Billy Pilgrim Dec 6 '17 at 14:42

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

$$\int_{-\infty}^{\infty} du \, e^{-t u^2} e^{i k u} = \sqrt{\frac{\pi}{t}} e^{-k^2/(4 t)}$$

$$\int_{-\infty}^{\infty} du \, \frac{e^{i k u}}{a^2+u^2} = \frac{\pi}{a} e^{-a |k|} \qquad (a \gt 0)$$


$$\begin{align} \int_{-\infty}^{\infty} du \, \frac{e^{-t u^2}}{a^2+u^2} &= \frac1{2 \pi} \sqrt{\frac{\pi}{t}} \frac{\pi}{a} \int_{-\infty}^{\infty} dk \, e^{-k^2/(4 t)} e^{-a |k|} \\ &= \frac{\pi}{a \sqrt{\pi t}} \int_0^{\infty} dk \, e^{-k^2/(4 t)} e^{-a k} \\ &= \frac{\pi}{a \sqrt{\pi t}} e^{a^2 t} 2 \sqrt{t} \int_{a \sqrt{t}}^{\infty} dv \, e^{-v^2} \\ &= \frac{\pi}{a} e^{a^2 t} \operatorname{erfc}{\left ( a \sqrt{t} \right )} \end{align} $$

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 $t$,

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


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