Improper integral $\int_0^\infty \sqrt{x}e^{-x}\,dx$

Question

How can I solve this improper integral?

$$ \int_0^\infty \sqrt{x}e^{-x}\,dx $$ using the following result: $$ \int_0^\infty e^{-x^2}\,dx = \frac{\sqrt{\pi}}{2}. $$

Answer 1

$$x=y^2\implies dx=2y\,dy$$

and your integral is

$$I:=\int\limits_0^\infty 2y^2e^{-y^2}dy$$

Now by parts

$$u=y\;\;,\;\;u'=1\\v'=2ye^{-y^2}\;\;,\;\;v=-e^{-y^2}$$

so

$$I=\left.-ye^{-y^2}\right|_0^\infty+\int\limits_0^\infty e^{-y^2}dy=\frac{\sqrt\pi}2$$