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How can I solve this improper integral?

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

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

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

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

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  • $\begingroup$ 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
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    $\begingroup$ @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
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$$ 2 \big( \frac{ \sqrt[ ]{ \pi }erf( \sqrt[]{x}) }{4}- \frac{\sqrt[]{x} e^{-x} }{2} \big)+c$$ Will be the indefinite integral with erf being the Gauss error function. using the bounds leaves you with $$\frac{ \sqrt[]{ \pi } }{2} $$

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