DUIS: ∫ e^(−ax) sin x / x dx

Let I(a) = ∫₀^∞ e^(−ax) (sin x)/x dx, a > 0.

Differentiate with respect to a (the limits do not depend on a):

dI/da = ∫₀^∞ (−x) e^(−ax) (sin x)/x dx = −∫₀^∞ e^(−ax) sin x dx

Using ∫₀^∞ e^(−ax) sin bx dx = b/(a² + b²) with b = 1:

dI/da = −1/(1 + a²)

Integrate: I(a) = −tan⁻¹ a + C.

Find C. As a → ∞, e^(−ax) → 0 so I(a) → 0. Then 0 = −π/2 + C, hence C = π/2.

I(a) = π/2 − tan⁻¹ a = cot⁻¹ a.

Result: ∫₀^∞ e^(−ax) sin x / x dx = π/2 − tan⁻¹ a.

Deduction. Let a → 0: e^(−ax) → 1 and tan⁻¹ 0 = 0, so

∫₀^∞ (sin x)/x dx = π/2.