Given that $\displaystyle \int\limits_{0}^{\infty} e^{-x^{2}}dx$ = $\displaystyle
\frac{\sqrt{\pi}}{2}
$;
find $\displaystyle
\int\limits_{0}^{\infty} x^{- \frac{1}{2}} e^{-2}dx
$
The term $\displaystyle e^{-2}$ is a constant so that the integral you propose is...
$\displaystyle \int_{0}^{\infty} \frac{e^{-2}}{\sqrt{x}}\cdot dx = e^{-2}\cdot \int_{0}^{\infty} \frac{dx}{\sqrt{x}}$ (1)
The problem is that the integral in (1) diverges ... May be that the integral you intended to propose us was...
$\displaystyle \int_{0}^{\infty} \frac{e^{-2x}}{\sqrt{x}}\cdot dx$ (2)
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
As chisigma pointed out (and mr fantastic 2 months ago...), the integral you posted (highlighted in red) is divergent (since it appears to be incomplete)! There is a major mistake in the problem because we can't incorporate the fact that $\displaystyle \int_0^{\infty}e^{-x^2}=\frac{\sqrt{\pi}}{2}$!!
You have been told that if the question you posted is correct, then the integral is divergent. It has also been suggested that the question as posted is very unlikely to be correct.
So there is nothing further do to until as with another post you eventually decide that the question is not correct and post what it is supposed to be, when the odds are on that the method of doing it has already been posted above.
CB
I would guess that our friend meant to write either $\displaystyle \int_0^\infty x^{-1/2}\mathrm e^{-x}\mathrm dx$ or $\displaystyle \int_0^\infty x^{-1/2}\mathrm e^{-2x}\mathrm dx$. Assume the former, as the latter is very similar.
Put $\displaystyle x=u^2$, $\displaystyle u>0$. Then ...