0
$\begingroup$

How can I solve this improper integral?

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

$\endgroup$
  • 5
    $\begingroup$ Sub $x=u^2$ and integrate by parts. $\endgroup$ – Ron Gordon Oct 6 '13 at 21:44
  • $\begingroup$ 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? $\endgroup$ – Shan Oct 6 '13 at 22:13
  • 1
    $\begingroup$ Yes it is...but then you did not use the hint Ron gave you... $\endgroup$ – DonAntonio Oct 6 '13 at 23:19
4
$\begingroup$

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

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

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

$\endgroup$
  • $\begingroup$ I think the integral I has the wrong power $2y^2$ should be 2y because we have $\sqrt(x)$ not x $\endgroup$ – sophie-germain Jan 17 '17 at 0:06
  • 1
    $\begingroup$ @arcolombo It's usually hard to remember after all this time...but in this case it is not: you're forgetting the $\;2ydy\;$ : $$\sqrt x\,dx\to y\cdot2y\,dy=2y^2\,dy$$ $\endgroup$ – DonAntonio Jan 17 '17 at 6:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.