Show that $\int_0^\infty e^{-x}{\sqrt x}dx=\frac{\sqrt\pi}{2}$ by using $\int_0^\infty e^{-x^2}dx=\frac{\sqrt\pi}{2}$

I'm trying to do integration by parts to be able to use $\int_0^\infty e^{-x^2}dx=\frac{\sqrt\pi}{2}$, but is not working.

• Try the substitution $u^2=x$. – Alex S Jul 7 '15 at 2:18

First do an integration by parts to get $\int_0^\infty e^{-x}\frac{\mathrm{d}x}{2\sqrt{x}}$, then try the substitution $u = \sqrt{x}$. It would be easier this way.

• It worked! I had tried this before for some reason had failed. Thank you, guys. – Let DC Jul 7 '15 at 2:24

Let $x=u^2$. Then $dx=2udu$ and then the integral becomes

$$\int\limits_{0}^{\infty}e^{-u^2}u(2udu)=\int\limits_{0}^{\infty}2e^{-u^2}u^2du$$

Then you can use integration by parts to get

$$u^2|_0^\infty \int\limits_{0}^{\infty}e^{-u^2}du-\int\limits_0^\infty 4e^{-u^2}udu$$

and you will plug in the values and use integration by parts again. I think you can take if from here.