This page will be continually under construction as I add new tricks.

## Differentiating Under the Integral

The idea here is to consider a simpler integral without the parameter that renders evaluation of the original integral difficult. Then evaluate the simpler integral and introduce the parameter by differentating both sides of this simpler integral.

This is very useful for integrands with an \(\ln^{n}(x)\) term, as we replace it with \(x^{a}\) and differentiate \(n\) times with respect to \(a\).

Example 1, Example 2, Example 3, Example 4.

## Integrating Under the Integral

The idea here is to consider a simpler integral without the parameter that renders evaluation of the original integral difficult. Then evaluate the simpler integral and introduce the parameter by integrating both sides of this simpler integral.

## Use of Fourier Transforms and Parseval’s Theorem

Parseval’s Theorem

\begin{equation}

\int\limits_{-\infty}^{\infty} f(t)\overline{g(t)} \mathrm{d}t

= \int\limits_{-\infty}^{\infty} \mathrm{F}(s)\overline{\mathrm{G}(s)} \mathrm{d}s

\end{equation}

where \(\mathrm{F}(s) = \mathcal{F}[f(t)]\) and \(\mathrm{G}(s) = \mathcal{F}[g(t)]\) are Fourier transforms.

## Convert the Reciprocal of a Function into an Integral

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.

## Use of the Mellin Transform

Manipulating an integral into a form suitable for evaluation via the Mellin transform, can greatly simplify subsequent analysis.

## Infinite Series Expansion of the Integrand

Here, we expand the integrand in an infinite series and integrate term-by-term. Note that to justify switching the order of integration and summation we invoke the dominated convergence theorem.

## Substitution

Let

\begin{equation}

\frac{\sin(\theta)}{\theta} = \frac{1}{2} \int\limits_{-1}^{1} \mathrm{e}^{i\theta x} dx

\end{equation}

## Frullani’s Integral

\begin{equation}

\int\limits_{0}^{\infty} \frac{f(ax)-f(bx)}{x} dx = [f(\infty)-f(0)] \ln\left(\frac{a}{b}\right)

\end{equation}

In general this is valid if \(f^{\prime}\) is continuous, the integral converges, and the limits of the function at the endpoints of integration exist.

See here for some examples as well as additional conditions for when this trick is applicable.

See here for multiple proofs of this result.