---
title: "Prove ∫ sin bx / x dx = π/2"
subject: "Mathematics I"
chapter: 2
key: "semester_1/mathematics_i:integral-sin-bx-over-x"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/integral-sin-bx-over-x
site: BCT Engineering
---

# Prove ∫ sin bx / x dx = π/2

Mathematics I · Chapter 2: Integration and its Applications

- Asked in: 2069 Chaitra
- Video: [Leibnitz Rule  Evaluate ∫(e^(-ax) sin bx/x dx  Hence prove ∫sinx/x = П/2 Partial Differentiation](https://www.youtube.com/watch?v=c6iK4aRkWdY) — m-easy maths
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/integral-sin-bx-over-x
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-2

## Question

Prove that ∫₀^∞ sin bx / x dx = π/2 (b > 0).

## Short answer

Introduce I(a) = ∫₀^∞ e^{-ax}(sin bx)/x dx. Differentiate: dI/da = −∫₀^∞ e^{-ax} sin bx dx = −b/(a²+b²). Integrate to get I(a)=π/2 − arctan(a/b). Letting a→0 gives ∫₀^∞ sin bx / x dx = π/2.

## Answer

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.)

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. dI/da equals
   A. b/(a^2+b^2)
   B. -a/(a^2+b^2)
   C. -b/(a^2+b^2)
   D. -1/(a^2+b^2)
   Answer: C. Standard Laplace integral of sin bx. (Key point: Derivative)

2. Using I → 0 as a → ∞ gives
   A. C = 0
   B. C = π
   C. C = b
   D. C = π/2
   Answer: D. 0 = -π/2 + C. (Key point: Constant)

3. For b > 0 the integral equals
   A. π/2
   B. π/(2b)
   C. b π/2
   D. 0
   Answer: A. As a → 0, tan^-1(0) = 0. (Key point: Result)

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