Prove that $\int_0^\infty e^{-x^4}\ dx = \Gamma\left(\frac{5}{4}\right) = \left(\frac{1}{4}\right)!$ using the gamma function

Question

Prove that $$\int_0^\infty e^{-x^4}\ dx = \Gamma\left(\frac{5}{4}\right) = \left(\frac{1}{4}\right)!$$

using the gamma function

$$\Gamma(z+1) = \int_0^\infty e^{-t}t^{z}\ dt$$

I have a solution in my notes that does:

$$\int_0^\infty e^{-x^4}\ dx = \frac{1}{4}\int_0^\infty e^{-u}u^{-3/4}\ du = \cdots$$

what kind of substitution is this?

Answer 1

The substitution performed is just $u=x^4$. $$u=x^4 \iff \color{blue}{x=u^{1/4}} \iff \color{red}{dx=\frac{1}{4}u^{-3/4}~du}$$ Hence: $$\int_0^{\infty} e^{\color{blue}{-x^4}}~\color{red}{dx}=\int_0^{\infty}e^{\color{blue}{-u}}\cdot \color{red}{\frac{1}{4}u^{-3/4}~du}=\frac{1}{4}\int_0^{\infty} e^{-u}\cdot u^{-3/4}~du$$ The integration limits stay the same during the substitution since $x=0 \implies u=0$ and $x\to \infty \implies u\to \infty$.

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You can continue the problem by identifying the value of $z$ on the definition of the gamma function, then using the following identity: $$\Gamma(x+1)=x\Gamma(x)$$