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

How can I solve this improper integral?

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

• Sub $x=u^2$ and integrate by parts. – Ron Gordon Oct 6 '13 at 21:44
• I get $$-x^{1/2}e^{-x} - \int_0^\infty -1/2x^{-1/2}e^{-x}$$. I have already solved that inside integral which is $$\sqrt{pi}/2$$ and the limit of the rest of the stuff is 0 making the final answer $$\sqrt{pi}/2$$. Would that be correct? – Shan Oct 6 '13 at 22:13
• Yes it is...but then you did not use the hint Ron gave you... – DonAntonio Oct 6 '13 at 23:19

## 1 Answer

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

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

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

• I think the integral I has the wrong power $2y^2$ should be 2y because we have $\sqrt(x)$ not x – sophie-germain Jan 17 '17 at 0:06
• @arcolombo It's usually hard to remember after all this time...but in this case it is not: you're forgetting the $\;2ydy\;$ : $$\sqrt x\,dx\to y\cdot2y\,dy=2y^2\,dy$$ – DonAntonio Jan 17 '17 at 6:37