How does one compute $I = \int_{0}^{\infty} e^{-2t^{2/3}}dt$?

Question

As stated in the title, I seek an effective way of computing

$$I = \int_{0}^{\infty} e^{-2t^{2/3}}dt.$$

My initial impression was to try to make a transformation to spherical coordinates, but my calculus III (or whatever) is a little rusty, and that didn't turn out to be very useful.

(Note that this isn't a "homework problem," or anything; I'm just curious.)

EDIT: My solution using the Gamma function (for anyone interested):

Let $x = 2t^{2/3}$. Then, $\frac{3}{4}t^{1/3}dx = dt$, and so

$$ I = \frac{3}{4} \int_{0}^{\infty} t^{1/3}e^{-x}dx .$$

And, since $x = 2t^{2/3}$, we have that $\frac{x}{2} = t^{2/3} \implies \left( \frac{x}{2} \right)^{3/2} = t \implies \left( \frac{x}{2} \right)^{1/2} = t^{1/3}$. So,

$$ I = \frac{3}{4} \int_{0}^{\infty} \left( \frac{x}{2} \right)^{1/2} e^{-x}dx = \frac{3}{4\sqrt2} \int_{0}^{\infty} x^{1/2} e^{-x}dx = \frac{3}{4\sqrt2} \int_{0}^{\infty} x^{3/2 - 1} e^{-x}dx .$$

$$ = \frac{3}{4\sqrt2} \Gamma \left( \frac{3}{2} \right) \approx 0.4699928.$$

Answer 1

Try to express it in terms Gamma function $\Gamma(t)=\int^{\infty}_0 x^{t-1}e^{-x}dx$. Use the fact that $\Gamma(t+1)=t \Gamma(t)$ and $\Gamma(\frac{1}{2})=\sqrt{\pi}$.

Answer 2

By setting $t=z^3$, then $z=\sqrt{u}$, we have: $$ I = \int_{0}^{+\infty} 3z^2 e^{-2z^2}\,dz = \frac{3}{2}\int_{0}^{+\infty} \sqrt{u}\, e^{-2u}\,du = \frac{3}{4\sqrt{2}}\int_{0}^{+\infty} \sqrt{s}\, e^{-s}\,ds, $$ hence:

$$ I = \int_{0}^{+\infty} e^{-2t^{2/3}}\,dt = \frac{3\,\Gamma\left(\frac{3}{2}\right)}{4\sqrt{2}} = \color{red}{\frac{3}{8}\sqrt{\frac{\pi}{2}}}.$$