Prove ∫ sin bx / x dx = π/2

Let b > 0. We show ∫₀^∞ (sin bx)/x dx = π/2 using differentiation under the integral sign, with the convergence factor e^(−ax), a > 0.

Let I(a) = ∫₀^∞ e^(−ax) (sin bx)/x dx. Differentiate with respect to a:

dI/da = −∫₀^∞ e^(−ax) sin bx dx = −b/(a² + b²)

Integrate with respect to a:

I(a) = −tan⁻¹(a/b) + C

As a → ∞ the integrand e^(−ax) → 0, so I(a) → 0. Hence 0 = −π/2 + C, C = π/2:

I(a) = π/2 − tan⁻¹(a/b).

Now let a → 0⁺, giving

∫₀^∞ (sin bx)/x dx = π/2 (for b > 0).

(For b < 0 the value is −π/2, and for b = 0 it is 0. Quick check of the b-independence: put t = bx, dx/x = dt/t, so the integral is the same for every b > 0.)