Integration Trick \([f(x)]^{-a} = \frac{1}{\Gamma(a)} \int_{0}^{\infty} z^{a-1} \mathrm{e}^{-f(x)z} \mathrm{d}z\)

A useful integration trick to convert the reciprocal of a function into an integral is
\begin{equation}
\frac{1}{[f(x)]^{a}} = \frac{1}{\Gamma(a)} \int\limits_{0}^{\infty} z^{a-1} \mathrm{e}^{-f(x)z} \mathrm{d}z
\end{equation}

The derivation is straightforward, let \(y=f(x)z\) and use the usual integral definition of the gamma function. However, one must be careful calculating the new limits of integration. If they differ from 0 and/or \(\infty\) then the incomplete gamma function must be used and the resulting integral may cause other difficulties.

See here for an excellent example of how this trick can greatly simplify the evaluation of a difficult integral.

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