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I have this integral:

$$\int_0^\infty \sqrt{x} e^{-\frac{x}{a}}dx$$

I am not sure how to solve this. I think it may involve the erf function (courtesy of wolfram) but I am not sure how to appropriately use this. The answer should be:

$$0.5(\sqrt\pi)(a^{3.2})$$

I would appreciate any help

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$$ I = \int_0^{\infty} \sqrt{x} e^{-\frac{x}{a}} \ dx $$

$$ x = au \Rightarrow dx = a \ du $$

$$ \Rightarrow I = a\sqrt{a} \int_0^{\infty} x^{\frac{1}{2}}e^{-x} \ dx $$

$$ \Gamma(t) = (t-1)! = \int_0^{\infty} x^{t-1} e^{-x} \ dx $$

$$ \Rightarrow I = a\sqrt{a} \ \Gamma \left(\frac{3}{2} \right) = a^{\frac{3}{2}}\frac{\sqrt{\pi}}{2} $$

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  • $\begingroup$ Don't use \frac on exponents it makes it impossible to read. Use a/b instead. $\endgroup$ – Ali Caglayan Aug 28 '13 at 21:34
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Hint: (I suppose you mean $a>0$ otherwise the integral diverges)

Make the substitution $t=\sqrt{x/a}$, this will transform the integral into something you should be able to handle.

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  • $\begingroup$ your hint gives a very simple answer . (+1) $\endgroup$ – what'sup Aug 28 '13 at 21:04

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