Integrating Powers of Exponentials


March 2, 2021

When working with distributions that were powers of exponentials, of which the normal and exponential distributions are special cases, I had to calculate the integrals of exponentials. It’s possible to transform these into expressions involving the Gamma function. Specifically I found that for all positive p:

\[\int_{0}^{\infty} x^m e^{-x^p}\, \rm{d}x = \frac{\Gamma\left(\frac{m+1}{p}\right)}{p}\]

This is useful for calculating moments of powers of exponentials, namely for positive p and k:

\[\int_{-\infty}^{\infty} (x - c)^m e^{-\left\vert\frac{x-c}{k}\right\vert^p} \, \rm{d}x = \left\{\begin{array}{rl} 0, & \text{if } m = 1,3,5,7,9,\ldots \\ \frac{2 k^{m+1} \Gamma\left(\frac{m+1}{p}\right)}{p}, & \text{if } m = 0, 2, 4, 6, 8, \ldots \end{array}\right.\]

This can all be proved with simple integration by substitution. First starting with the moment integral \(\int_{-\infty}^{\infty} (x - c)^m e^{-\left\vert\frac{x-c}{k}\right\vert^p} \, \rm{d}x\), where m is a non-negative integer and p and k are positive. We can shift the integral so it centred on the midpoint c with the substitution \(u = x - c\), which gives the integral \(\int_{-\infty}^{\infty} x^m e^{-\left\vert\frac{x}{k}\right\vert^p} \, \rm{d}x\).

Now the term \(e^{-\left\vert\frac{x}{k}\right\vert^p}\) is symmetric about the origin, and so when m is odd the integrand is anti-symmetric about the origin, and so the contributions for positive and negative x cancel out, resulting in a zero integral. When m is even the integral is symmetric about the origin and so the contributions for positive and negative x are equal; so we can calculate the integral as \(2 \int_{0}^{\infty} x^m e^{-\frac{x}{k}^p} \, \rm{d}x\), where we have dropped the absolute value because x/k is positive.

Finally we can remove k from the exponent by rescaling the distribution with the substitution \(u = \frac{x}{k}\), giving the integral as \(2 k^{m+1} \int_{0}^{\infty} x^m e^{-x^p} \, \rm{d}x\). So we have reduced the problem to calculating the integral of a power of x multiplied by a power of the exponential.

The integral \(\int_{0}^{\infty} x^m e^{-x^p} \,\rm{d}x\) can be transformed to remove the power from the exponent with the transformation \(u = x^p\); this gives

\[\int_{0}^{\infty} x^m e^{-x^p} \,\rm{d}x = \frac{1}{p} \int_{0}^{\infty} x ^{\frac{m + 1}{p} - 1} e^{-x} \,\rm{d}x\]

and by the definition of the Gamma function this is just \(\frac{\Gamma\left(\frac{m+1}{p}\right)}{p}\).

It’s easy to verify this in simple cases; when \(p = 1\) we get the integral being \(m!\) which can be proved by integration by parts. For \(m = 0\) and \(p = 2\) this gives \(\frac{\sqrt{\pi}}{2}\) which is the Gaussian integral.

These integrals are useful for calculating statistics of different exponential type distributions.