Feynman's integration trick
/ FYNE-man /
Feynman's integration trick is the strategic, problem-solving face of differentiation under the integral sign. The idea is mischievous and effective: faced with a stubborn definite integral, you deliberately smuggle in an extra parameter that is not really there, differentiate with respect to it to get a tamer integral, solve that, then integrate back over the parameter to recover the answer you wanted. Richard Feynman, in his memoir, credits an old calculus text for teaching him this and recalls dazzling colleagues by evaluating integrals they could not.
Concretely you build a family I(a) of integrals so that the one you actually want is some specific value like I(1) or I(0). You pick the parameter so that dI/da is an integral you can do, evaluate dI/da, then integrate that result with respect to a from a known anchor value (where I is obviously something simple, often zero at infinity) up to the target. The art is entirely in where you insert the parameter — a good choice makes the differentiated integrand collapse; a poor one makes things worse.
It is worth saying plainly that this is not cheating or a loophole — it is a legitimate application of a rigorous theorem (the Leibniz rule with constant limits), and the steps can be fully justified by checking the usual convergence and continuity conditions. It shines on integrals over an infinite range or with a parameter-friendly structure, and it overlaps with, but is distinct from, contour integration in the complex plane — both are honest, principled routes to a real answer, just along different roads.
To get the integral from 0 to 1 of (x - 1)/ln(x) dx, set I(a) = integral of (x^a - 1)/ln(x) dx; then dI/da = integral of x^a dx = 1/(a + 1). Integrating gives I(a) = ln(a + 1), so the answer (a = 1) is ln 2.
Inserting the parameter a in the exponent makes the troublesome ln(x) in the denominator cancel on differentiation.
There is no recipe for where the parameter goes — that part is genuine ingenuity, and a clumsy insertion can leave the differentiated integral as hard as the original. The trick assists skill; it does not replace it.