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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.

  • 1
    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.2012-02-15

2 Answers 2

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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} $