Integrals I: Feynman's Trick

Why it works: $\dfrac{d}{da}\int f(x,a)\,dx = \int\dfrac{\partial f}{\partial a}\,dx$ whenever $f$ and $\partial_af$ are continuous (Leibniz rule). The parameter is chosen so that $\partial_a f$ is ELEMENTARY even though $f$ is not — a logarithm in a denominator, an oscillating factor, a power that refuses to integrate. Then $I'(a)$ is a routine integral, $I(a)$ is its antiderivative in $a$, and one known value pins the constant. The technique is Richard Feynman's name for a trick already in Leibniz; on the Putnam it appears in some form most years.

Step 1 — differentiate in $a$: $\partial_a\dfrac{x^a - 1}{\ln x} = \dfrac{x^a\ln x}{\ln x} = x^a$ — the logarithm cancels.

Step 2 — $I'(a) = \int_0^1 x^a\,dx = \dfrac{1}{a+1}$.

Step 3 — integrate back: $I(a) = \ln(a + 1) + C$, and $I(0) = 0$ gives $C = 0$.

Step 4 — so $\int_0^1\frac{x^2 - 1}{\ln x}dx = \ln 3$, and in general $\ln(a+1)$. Check: at $a = 1$ the integrand is $\tfrac{x - 1}{\ln x} \approx 1$ near $x = 1$ and $\to 0$ near $0$, so a value near $\ln 2 \approx 0.69$ is plausible ✓.

Step 1 — define $I(t) = \int_0^\infty e^{-tx}\dfrac{\sin x}{x}\,dx$ for $t \ge 0$; the target is $I(0)$, and $I(t) \to 0$ as $t \to \infty$ (the exponential crushes everything).

Step 2 — $I'(t) = -\int_0^\infty e^{-tx}\sin x\,dx = -\dfrac{1}{1 + t^2}$ — a Laplace transform, computable by two integrations by parts or by taking the imaginary part of $\int e^{-(t - i)x}dx = \tfrac{1}{t - i}$.

Step 3 — integrate: $I(t) = -\arctan t + C$, and $I(\infty) = 0$ forces $C = \tfrac\pi2$.

Step 4 — $I(0) = \tfrac\pi2$. (The interchange of limit and integral at $t = 0$ needs care — Abel's theorem or dominated convergence — which is exactly the kind of sentence a Putnam grader wants to see.)

Set $I(a) = \int_0^\infty\frac{e^{-x} - e^{-ax}}{x}dx$ with $I(1) = 0$. Then $I'(a) = \int_0^\infty e^{-ax}dx = \tfrac1a$, so $I(a) = \ln a$ and $I(2) = \ln 2$. (This is also Frullani's theorem, topic 4 — two names for one mechanism.)

Form the difference quotient $\dfrac{I(a + h) - I(a)}{h} = \int_c^d\dfrac{f(x, a+h) - f(x, a)}{h}dx$. By the mean value theorem in $a$, the integrand equals $\partial_af(x, \xi_x)$ for some $\xi_x$ between $a$ and $a + h$. Continuity of $\partial_af$ on the compact rectangle makes it uniformly continuous, so $\partial_af(x, \xi_x) \to \partial_af(x, a)$ UNIFORMLY in $x$ as $h \to 0$ — and uniform convergence lets the limit pass inside the integral. $\blacksquare$ For infinite intervals one needs a dominating integrable function (dominated convergence); Putnam solutions should say so in one sentence.

Integrate by parts twice with $u = \sin x$ then $u = \cos x$: $J = \int_0^\infty e^{-tx}\sin x\,dx = \left[-\tfrac{e^{-tx}}{t}\sin x\right]_0^\infty + \tfrac1t\int_0^\infty e^{-tx}\cos x\,dx = \tfrac1t\left(\left[-\tfrac{e^{-tx}}{t}\cos x\right]_0^\infty - \tfrac1t J\right) = \tfrac{1}{t^2} - \tfrac{1}{t^2}J$. Solve: $J(1 + \tfrac{1}{t^2}) = \tfrac{1}{t^2}$, so $J = \tfrac{1}{1 + t^2}$. $\blacksquare$ Significance: the same "bounce-back and solve" is the engine of $\int e^x\sin x$ from first-year calculus — Feynman's trick is built from techniques you already own.

Step 1 — let $I(b)$ be the integral; $I(0) = \tfrac{\sqrt\pi}{2}$ (half the Gaussian).

Step 2 — differentiate: $I'(b) = -\int_0^\infty 2xe^{-x^2}\sin(2bx)\,dx$.

Step 3 — integrate by parts with $u = \sin 2bx$, $dv = 2xe^{-x^2}dx$ (so $v = -e^{-x^2}$): $I'(b) = \left[e^{-x^2}\sin 2bx\right]_0^\infty - 2b\int_0^\infty e^{-x^2}\cos 2bx\,dx = -2bI(b)$.

Step 4 — a differential equation! $I'/I = -2b$ gives $I(b) = Ce^{-b^2}$, and $I(0)$ fixes $C$: $I(b) = \tfrac{\sqrt\pi}{2}e^{-b^2}$.

Step 5 — the lesson: sometimes $I'$ is expressed back in terms of $I$, and Feynman's trick hands you an ODE instead of a direct integral.

Step 1 — $I'(a) = \int_0^\pi\dfrac{2a - 2\cos x}{1 - 2a\cos x + a^2}dx$.

Step 2 — the Weierstrass substitution (topic 4) or the identity $\dfrac{2a - 2\cos x}{1 - 2a\cos x + a^2} = \dfrac1a\left(1 - \dfrac{1 - a^2}{1 - 2a\cos x + a^2}\right)$ reduces it to the standard $\int_0^\pi\dfrac{dx}{1 - 2a\cos x + a^2} = \dfrac{\pi}{1 - a^2}$.

Step 3 — so $I'(a) = \dfrac1a(\pi - \pi) = 0$ for $|a| < 1$.

Step 4 — $I$ is constant, and $I(0) = \int_0^\pi\ln 1\,dx = 0$: the integral is $0$ for every $|a| < 1$. (For $|a| > 1$, factor out $a^2$ to get $2\pi\ln|a|$.) A whole family of integrals evaluated by showing the derivative vanishes.