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.

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