---
title: "DUIS: ∫ e^(−ax) sin x / x dx"
subject: "Mathematics I"
chapter: 2
key: "semester_1/mathematics_i:duis-e-ax-sinx-over-x"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/duis-e-ax-sinx-over-x
site: BCT Engineering
---

# DUIS: ∫ e^(−ax) sin x / x dx

Mathematics I · Chapter 2: Integration and its Applications

- Asked in: 2074 Asoj; 2074 Chaitra, 5 marks; 2073 Shrawan (asked 3 times)
- Video: [Leibnitz Rule - Evaluate ∫e^(-ax) sin x/x dx=tan^-1(1/a) prove ∫sinx/x=П/2 Partial Differentiation](https://www.youtube.com/watch?v=F7RHC_-2Sqg) — 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/duis-e-ax-sinx-over-x
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-2

## Question

Apply the rule of differentiation under the integral sign to evaluate ∫₀^∞ e^(−ax) sin x / x dx and hence deduce that ∫₀^∞ sin x / x dx = π/2.

## Short answer

Differentiate I(a)=∫₀^∞ e^{-ax} sin x/x dx to get dI/da = -1/(1+a²). Integrate: I(a)= -tan⁻¹ a + C; as a→∞, I→0 gives C=π/2, so I(a)=π/2 − tan⁻¹ a. Setting a=0 gives ∫₀^∞ sin x/x dx = π/2.

## Answer

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.**

## Check yourself

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

1. dI/da for I(a) = integral e^(-ax) sin x / x dx is
   A. integral e^(-ax) cos x dx
   B. integral x e^(-ax) sin x dx
   C. -integral e^(-ax) sin x dx
   D. -integral e^(-ax) sin x / x dx
   Answer: C. Differentiating e^(-ax) gives -x e^(-ax), cancelling 1/x. (Key point: Derivative)

2. Integrating dI/da = -1/(1+a^2) gives
   A. I = tan^-1 a + C
   B. I = -log(1+a^2) + C
   C. I = 1/(1+a^2) + C
   D. I = -tan^-1 a + C
   Answer: D. The integral of -1/(1+a^2) is -tan^-1 a. (Key point: Integration)

3. The constant C is found from
   A. I(a) → 0 as a → ∞, so C = π/2
   B. I(0)=0, so C = 0
   C. I(1)=1, so C = 1
   D. a → 0 gives C = π
   Answer: A. 0 = -π/2 + C. (Key point: Constant)

4. Letting a → 0 gives
   A. = π
   B. integral of sin x / x from 0 to ∞ = π/2
   C. = 0
   D. = 1
   Answer: B. π/2 - tan^-1(0) = π/2. (Key point: Deduction)

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