For what value of $c$ is $f$ a probability density function?

Question

Consider the function $f(x)$ defined as $f(x)=ce^{-x^4},\; x\in \mathbb{R}$. For what value of $c$ is $f$ a p.d.f?

Solution: Using the p.d.f conditions:

$c\int_{-\infty}^{\infty}e^{-x^4}dx=1$

Set $x^4=u$ then, $dx=\frac{u^{-3/4}}{4} $. So, integration now changes to: $$\frac{c}{4}\int_{0}^{\infty}e^{-u}{u^{-3/4}}du=1 \implies\frac{c}{4}\Gamma(1/4)=1 \implies c=\frac{4}{\Gamma(1/4)}$$.

But the answer given is : $c=\frac{2}{\Gamma(1/4)}$. What have I done wrong here?

Answer 1

Note that since $x$ changes from $-\infty$ to $\infty$, then $u=x^4$ changes from $0$ to $\infty$ twice (once because $x$ changes from $0$ to $\infty$ and the other time because $x$ changes from $-\infty$ to $0$). Hence you must first write $$2c\int_0^\infty e^{-x^4}dx=1$$then $$\frac{c}{2}\int_{0}^{\infty}e^{-u}{u^{-3/4}}du=1$$and continue.

Answer 2

Observe that it must be

$$c\int_{-\infty}^\infty e^{-{x^4}}dx=2c\int_0^\infty e^{-{x^4}}dx$$

and now you do your substitution, otherwise that substitution is illegal as the function is then not injective...