## Another Definite Integral

My students claimed they were doing a calculation that required

$\int_{-\infty}^\infty e^{-x^4}dx$.

I’m not sure what physical situation brought up such a question, but we can find the answer anyway. Let’s kill infinity birds with one stone by evaluating

$\int_{0}^\infty x^{\alpha} e^{-x^\beta} dx$.

and treating my students’ problem as a special case.

First define the gamma function by

$\Gamma(n) \equiv \int_0^\infty t^{n-1}e^{-t}dt$.

I have never understood why $\Gamma(n)$ involves $(n-1)$ as the power of $t$, rather than just $n$. It makes even less sense when you consider $\Gamma(n) = (n-1)!$ for natural numbers $n$.

In the definition of the gamma function, make the substitution

$t = x^\beta, t^{n-1} = x^{\beta n-\beta}, dt = \beta x^{\beta-1}$

$\begin{array}{rcl}\Gamma(n) & = & \int_0^\infty x^{\beta n-\beta}e^{-x^\beta}\beta x^{\beta-1}dx \\ { } & = & \int_0^\infty x^{n\beta-1}e^{-x^\beta}dx\end{array}$

We can choose whatever we want for $n$, as long as we think we can find $\Gamma(n)$. So let’s turn this into the original problem by substituting

$\begin{array}{rcl}n\beta-1 & = & \alpha \\ n & = & \frac{\alpha+1}{\beta} \end{array}$

Putting it all together:

$\frac{1}{\beta}\Gamma\left(\frac{\alpha+1}{\beta}\right) = \int_0^\infty x^\alpha e^{-x^\beta}dx$.

So we understand these seemingly-more-complicated definite integrals equally as well as we understand the gamma function.

For the special case my students were interested in, which has $\alpha = 0, \beta = 4$, the integral goes from $-\infty$ to $\infty$, so we need to multiply by two to get

$\int_{-\infty}^{\infty} e^{-x^4}dx = \frac{1}{2}\Gamma(1/4)$.

A computer tells me this evaluates to about 1.8.