Improper integral: $\,\frac{1}{\pi}\int^\infty_0 \frac{\sqrt{x}}{1+x}e^{-xt}\,dx$

Question

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 $t>0$.

Answer 1

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

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