# Evaluating $\int_0^\infty \sqrt x e^{-x^3} dx$

I want to evaluate:

$$\int_0^\infty \sqrt x e^{-x^3} dx$$

I guess I should use integration by parts but as I'm not good at it, I cannot go any further.

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You can't be "not good" at using integration by parts ; it's just differentiation... unless you don't know your differentiation formulas ; but that's a problem if you're up to integrating. If you don't know how to compute an integral, it doesn't mean you're "not good", it just means you don't see the right way to go. Don't burden yourself! – Patrick Da Silva Feb 15 '12 at 21:34
It is very satisfying to know as I have an exam tomorrow. Thank you for your nice comment @PatrickDaSilva. – Gigili Feb 15 '12 at 21:38
In this kind of problem (integrating), if you don't have time, then it's awful ; but if you do, the right way to naively find the answer is just to try every way of doing it that you know, without doing mistakes (calculation ones). After some time, you'll be able to "guess" which ways will work better than others ; that's just experience. – Patrick Da Silva Feb 15 '12 at 21:41

Better use substitution:

$$x^{\frac{3}{2}} = u$$

$$\frac{3}{2}x^{\frac{1}{2}}dx = du$$

This gives

$$\int\limits_0^\infty {{e^{ - {{\left( {{x^{3/2}}} \right)}^2}}}\sqrt x dx}$$

$$\frac{2}{3}\int\limits_0^\infty {{e^{ - {u^2}}}du} = \frac{2}{3}\frac{{\sqrt \pi }}{2} = \frac{1}{3}\sqrt \pi$$

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Using a slightly different substitution than Peter did, $u = x^3$, we get an answer in terms of $\Gamma$: \begin{align} \int_0^\infty \sqrt x e^{-x^3} dx &=\int_0^\infty u^{1/6}e^{-u}\frac13u^{-2/3}\;\mathrm{d}u\\ &=\frac13\int_0^\infty u^{-1/2} e^{-u}\;\mathrm{d}u\\ &=\frac13\Gamma\left(\frac12\right)\\ &=\frac{\sqrt{\pi}}{3} \end{align}

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