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.