Answer: The Sinc Integral

Key Idea / Intuition

The function $\sin x / x$ has no elementary antiderivative, so direct integration is hopeless. The key trick is to introduce a parameter under the integral sign (Feynman's differentiation trick / Laplace transform): replace the integral by $I(s) = \int_0^\infty e^{-sx} \frac{\sin x}{x}\,dx$, differentiate with respect to $s$ to kill the $x$ in the denominator, evaluate the resulting elementary integral, then integrate back and recover $I(0)$.

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Formal Proof / Solution

Step 1: Introduce a parameter.

Define $I(s) = \int_0^{\infty} e^{-sx} \frac{\sin x}{x}\, dx, \quad s > 0.$

Note $I(0)$ is the desired integral.

Step 2: Differentiate under the integral sign.

$I'(s) = -\int_0^{\infty} e^{-sx} \sin x\, dx.$

This is a standard Laplace transform: $\int_0^{\infty} e^{-sx} \sin x\, dx = \frac{1}{s^2 + 1}.$

(Quick derivation: integrate by parts twice, or use $\sin x = \operatorname{Im}(e^{ix})$ to get $\operatorname{Im}\!\left(\frac{1}{s-i}\right) = \frac{1}{s^2+1}$.)

So: $I'(s) = -\frac{1}{s^2 + 1}.$

Step 3: Integrate back.

$I(s) = -\arctan(s) + C.$

Step 4: Determine the constant.

As $s \to \infty$, $e^{-sx} \to 0$ pointwise and $|\sin x / x| \le 1$, so by DCT, $I(s) \to 0$. Thus: $0 = -\frac{\pi}{2} + C \implies C = \frac{\pi}{2}.$

Step 5: Evaluate at $s = 0$.

$I(0) = -\arctan(0) + \frac{\pi}{2} = \frac{\pi}{2}.$

$\boxed{\int_0^{\infty} \frac{\sin x}{x}\, dx = \frac{\pi}{2}}$

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Why this is beautiful: The integrand $\sin x / x$ is perfectly smooth and bounded, yet resists elementary methods entirely. The parameter trick transforms an impossible integral into a trivial ODE, recovering a clean $\pi/2$ from essentially no information — a hallmark of Feynman's method at its best.

Written to: questions/2026-06-17_pm.md | Answer: answers/2026-06-17_pm.md