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(Motivation) I made this improper integral for fun:

$$\int_{0}^{\infty}x^{a}e^{-x^{3}}dx = \frac{1}{3}\Gamma\left(\frac{a+1}{3}\right)$$

where $\Re(a) > -1$. After struggling with various integration techniques, I thought of using spherical coordinates.

(Question) Is there a way to prove that equation without enforcing spherical coordinates as I did below?

(Attempt) Let the integral in question be $I$. Then

$$ \begin{align} I =& \int_{0}^{\infty}x^{a}e^{-x^{3}}dx \\ =& \frac{2}{3}\int_{0}^{\infty}x^{\frac{2a-1}{3}}e^{-x^{2}}dx \tag{1}\\ I^3 =& \frac{8}{27}\int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{\infty}e^{-x^{2}-y^{2}-z^{2}}\left(xyz\right)^{\frac{2a-1}{3}}dxdydz \cr =& \frac{8}{27}\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}\int_{0}^{\infty}e^{-\rho^{2}}\left(\rho^{3}\cos\theta\sin\theta\sin^{2}\phi\cos\phi\right)^{\frac{2a-1}{3}}\rho^{2}\sin\phi d\rho d\theta d\phi \tag{2}\\ =& \frac{8}{27}\int_{0}^{\infty}\rho^{2a+1}e^{-\rho^{2}}d\rho\int_{0}^{\frac{\pi}{2}}\left(\cos\theta\right)^{\frac{2a-1}{3}}\left(\sin\theta\right)^{\frac{2a-1}{3}}d\theta \cr &\cdot \int_{0}^{\frac{\pi}{2}}\left(\sin\phi\right)^{\frac{4a+1}{3}}\left(\cos\phi\right)^{\frac{2a-1}{3}}d\phi \tag{3}\\ =& \frac{8}{27}\cdot\frac{\Gamma\left(a+1\right)}{2}\cdot\frac{B\left(\frac{a}{3}+\frac{1}{3},\frac{a}{3}+\frac{1}{3}\right)}{2}\cdot\frac{B\left(\frac{2a}{3}+\frac{2}{3},\frac{a}{3}+\frac{1}{3}\right)}{2} \tag{4}\\ =& \frac{1}{27}\cdot\Gamma\left(a+1\right)\cdot\frac{\Gamma\left(\frac{a}{3}+\frac{1}{3}\right)^{2}}{\Gamma\left(\frac{2a}{3}+\frac{2}{3}\right)}\cdot\frac{\Gamma\left(\frac{2a}{3}+\frac{2}{3}\right)\Gamma\left(\frac{a}{3}+\frac{1}{3}\right)}{\Gamma\left(a+1\right)} \tag{5}\\ =& \frac{1}{27}\Gamma\left(\frac{a}{3}+\frac{1}{3}\right)^{3} \cr \therefore I =& \frac{1}{3}\Gamma\left(\frac{a+1}{3}\right). \cr \end{align} $$ Explanations:

$(1)$: Substitute $x \mapsto x^{2/3}$

$(2)$: Employ spherical coordinates

$(3)$: Product of multiple integrals is iterated integrals

$(4)$: A property of the Beta Function

$(5)$: A property of the Beta Function

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    $\begingroup$ no need for such a roundabout approach. Just make the change of variables $t=x^3$ in the original integral. $\endgroup$
    – peek-a-boo
    Feb 5 at 7:17
  • $\begingroup$ @peek-a-boo That's a much easier approach. I don't know how I didn't think of that. $\endgroup$
    – Accelerator
    Feb 5 at 7:23

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