Differentiation Under the Integral Sign

Some integrals resist every standard substitution or integration-by-parts trick, yet become easy once you notice they're really one member of a whole family of integrals indexed by a parameter. Differentiating that family with respect to the parameter can produce a much simpler integral, which you solve directly and then integrate back with respect to the parameter to recover the original answer.

The method

Given a hard integral $I = \int f(x)\, dx$, find a natural parameter $t$ to insert so the integral becomes $I(t) = \int f(x, t)\, dx$, with the original problem recovered at a specific value of $t$. Differentiate under the integral sign, $\frac{dI}{dt} = \int \frac{\partial f}{\partial t}\, dx$, swapping the order of differentiation and integration (valid under mild regularity conditions). If the resulting integral is easy, solve it to get $\frac{dI}{dt}$ as a function of $t$, then integrate that back over $t$, using a known boundary value of $I$ at some convenient $t$ to fix the constant of integration.

Worked example

To evaluate $I = \int_0^\infty \frac{e^{-x}\sin x}{x}\, dx$, introduce a parameter as $I(t) = \int_0^\infty \frac{e^{-tx}\sin x}{x}\, dx$ with $I(1)$ the target and $I(\infty) = 0$ as the known boundary. Differentiating gives $I'(t) = -\int_0^\infty e^{-tx}\sin x\, dx = -\frac{1}{1+t^2}$, an elementary Laplace-transform integral. Integrating $I'(t)$ from $t$ to infinity (where $I$ vanishes) gives $I(t) = \frac{\pi}{2} - \arctan t$, so $I(1) = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$, a closed-form answer to an integral with no elementary antiderivative.