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Good evening! How could one evaluate the following integral $$\frac{1}{\pi}\int^\infty_0 \frac{\sqrt{x}}{1+x}e^{-xt}\,dx$$ I have tried the substitution $x\equiv x^2$ but still I could not manage to get to a final result. Any ideas would be really appreciated! Also $t>0$.

amWhy
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Arian
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    You probably mean improper integral, not indefinite. – user208259 Jan 21 '15 at 17:01
  • @user208259:indeed. thnx! – Arian Jan 21 '15 at 17:02
  • My guess is that the complementary error function is involved. See here http://mathworld.wolfram.com/Erfc.html – Wintermute Jan 21 '15 at 17:03
  • @mtiano: the answer is indeed $$\frac{1}{\sqrt{\pi t }}-e^t \text{Erfc}\left(\sqrt{t}\right).$$ It just follows from expressing $\frac{\sqrt{x}}{1+x}$ as a geometric series. – Jack D'Aurizio Jan 21 '15 at 17:04
  • Did this come from an inverse Laplace transform? – Ron Gordon Jan 21 '15 at 18:11
  • @RonGordon: Yes! – Arian Jan 21 '15 at 18:16
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    Thought so. I did a similar one here: http://math.stackexchange.com/questions/504760/inverse-laplace-of-frac1-sqrts-1/504788#504788 in which I came up with that integral and just displayed the answer. My derivation there was using Parseval's theorem, so this is a different, more accessible way. – Ron Gordon Jan 21 '15 at 18:20
  • I was ILT $\frac{1}{1+\sqrt{z}}$, used the same contour as in your provided link but in the end I failed to get a closed form for the integral above. Nevertheless the method below is really nice. – Arian Jan 21 '15 at 18:34

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Yes, sub $x=u^2$ to get

$$\frac{2}{\pi} \int_0^{\infty} du \frac{u^2}{1+u^2} e^{-t u^2} = \frac1{\pi} \int_{-\infty}^{\infty} du \left (1-\frac1{1+u^2} \right )e^{-t u^2}= \frac1{\sqrt{\pi t}} - \frac1{\pi} \int_{-\infty}^{\infty} du \frac{e^{-t u^2}}{1+u^2} $$

The latter integral may be evaluated a few different ways. One way is to multiply and divide by $e^{-t}$, and differentiate:

$$\frac1{\pi} \int_{-\infty}^{\infty} du \frac{e^{-t u^2}}{1+u^2} = \frac{e^t}{\pi} \int_{-\infty}^{\infty} du \frac{e^{-t (1+u^2)}}{1+u^2}$$

and

$$\frac{d}{dt} \int_{-\infty}^{\infty} du \frac{e^{-t (1+u^2)}}{1+u^2} = -e^{-t} \int_{-\infty}^{\infty} du \, e^{-t u^2} = - \frac{\sqrt{\pi}e^{-t}}{\sqrt{ t}} $$

Integrate back with respect to $t$ and get an error function:

$$ \int_{-\infty}^{\infty} du \frac{e^{-t (1+u^2)}}{1+u^2}= -\sqrt{\pi}\int^t dt' \frac{e^{-t'}}{\sqrt{ t'}} = C-\pi \operatorname{erf}{\sqrt{t}} $$

Noting that

$$\int_{-\infty}^{\infty} du \frac{1}{1+u^2} = \pi$$

We finally have the integral taking the value

$$\frac1{\sqrt{\pi t}} - e^t \operatorname{erfc}{\sqrt{t}} $$

Ron Gordon
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