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

Question

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.

Answer 1

Better use substitution:

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

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

This gives

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

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

Answer 2

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