\( \int_{0}^{\infty}xe^{-x^3}\,dx \)

[Tolaso J Kos]

Prove that: \( \displaystyle \int_{0}^{\infty}xe^{-x^3}\,dx=\frac{1}{3}\Gamma \left ( \frac{2}{3} \right ) \).

[Papapetros Vaggelis]

The given integral converges because :

\(\displaystyle{\begin{aligned} \lim_{x\to +\infty}x^2\,x\,e^{-x^3}&=\lim_{x\to +\infty}\dfrac{x^3}{e^{x^3}}\\&=\lim_{x\to +\infty}\dfrac{\left(x^3\right)'}{\left(e^{x^3}\right)'}\\&=\lim_{x\to +\infty}\dfrac{3\,x^2}{3\,x^2\,e^{x^3}}\\&=\lim_{x\to +\infty}\dfrac{1}{e^{x^3}}\\&=0\end{aligned}}\)

By applying the substitution \(\displaystyle{t=x^3\,,t\in\left[0,+\infty\right)}\) , we get

\(\displaystyle{x=t^{1/3}\implies \mathrm{d}x=\dfrac{1}{3}\,t^{-2/3}\,\mathrm{d}t}\) and :

\(\displaystyle{\begin{aligned} \int_{0}^{\infty}x\,e^{-x^3}\,\mathrm{d}x&=\int_{0}^{\infty}t^{1/3}\,e^{-t}\,\dfrac{1}{3}\,t^{-2/3}\,\mathrm{d}t\\&=\dfrac{1}{3}\,\int_{0}^{\infty}t^{-1/3}\,e^{-t}\,\mathrm{d}t\\&=\dfrac{1}{3}\,\int_{0}^{\infty}t^{(2/3)-1}\,e^{-t}\,\mathrm{d}t\\&=\dfrac{1}{3}\,\Gamma\,\left(\dfrac{2}{3}\right)\end{aligned}}\)