One Step Forward from Gaussian Integral

Question

Now to solve the integral $ \int_0^\infty e^{-x^2} \, dx $ has become a simple task for us. But how can we solve this integral: $$\int_0^\infty e^{-x^3} \, dx $$

Answer 1

Note: Not an Answer but a Generalization

Consider the integral, $\displaystyle I = \int_0^\infty e^{-x^A} \text{ d}x $. Substitute $\displaystyle x = u^{\frac{1}{A}} \Rightarrow \text{ d}x = \frac{1}{A} u^{\frac{1-A}{A}} \text{ d}u $. $$ \therefore I = \int_0^\infty e^{-u} \frac{1}{A} u^{\frac{1-A}{A}} \text{ d}u$$ $\displaystyle I = \frac{1}{A} \int_0^\infty u^{\frac{1}{A} - 1} e^{-u} \text{ d}u $ $\displaystyle \Rightarrow I = \frac{1}{A} \Gamma \left( \frac{1}{A} \right) = \Gamma \left( \frac{1 + A}{A} \right) $

(Using the fact that $ t \Gamma \left(t \right) = \Gamma \left(t + 1 \right) $ ): $$ \therefore \int_0^\infty e^{-x^A} \text{ d}x = \Gamma \left( \frac{1 + A}{A} \right) $$

Answer 2

$$\int_{0}^{+\infty}e^{-x^n}\,dx = \frac{1}{n}\int_{0}^{+\infty}z^{\frac{1}{n}-1}e^{-z}\,dz = \frac{1}{n}\,\Gamma\left(\frac{1}{n}\right) = \Gamma\left(1+\frac{1}{n}\right)$$ through a change of variable and the definition of the $\Gamma$ function.

Answer 3

Notice, the following Laplace transform $$\int_{0}^{\infty}e^{-st}t^ndt=\frac{\Gamma(n+1)}{s^{n+1}}$$

Now, we have $$\int_{0}^{\infty}e^{-x^n}dx$$ Let $x^n=t\implies nx^{n-1}dx=dt\iff dx=\frac{dt}{nt^{(n-1)/n}}$ $$\int_{0}^{\infty}e^{-t}\frac{dt}{nt^{(n-1)/n}}$$ $$=\frac{1}{n}\int_{0}^{\infty}e^{-t}t^{(1-n)/n}dt$$ $$=\frac{1}{n}L[t^{(1-n)/n}]_{s=1}$$ $$=\frac{1}{n}\left[\frac{\Gamma\left(\frac{1-n}{n}+1\right)}{s^{\frac{1-n}{n}+1}}\right]_{s=1}$$ $$=\frac{1}{n}\left[\frac{\Gamma\left(\frac{1}{n}\right)}{1^{\frac{1}{n}}}\right]$$ $$=\frac{1}{n}\Gamma\left(\frac{1}{n}\right)$$

Hence, as per given problem $$\int_{0}^{\infty}e^{-x^3}dx$$ Here, $n=3$ hence we get $$\int_{0}^{\infty}e^{-x^3}dx=\frac{1}{3}\Gamma\left(\frac{1}{3}\right)$$