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

# DUIS: ∫ e^(−x) sin bx / x dx

Mathematics I · Chapter 2: Integration and its Applications

- Asked in: 2072 Kartik; 2070 Asar; 2069 Chaitra; 2068 Shrawan; 2068 Chaitra (asked 5 times)
- Video: [Differentiation Under Integral Sign (DUIS)](https://www.youtube.com/watch?v=nhqJaV8ZI8c) — Mathematics for Engineers 
- 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-x-sin-bx-over-x
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-2

## Question

Apply differentiation under the integral sign to evaluate ∫₀^∞ e^(−x) sin bx / x dx.

## Short answer

Differentiate I(b)=∫₀^∞ e^{−x} sin(bx)/x dx w.r.t. b to get dI/db = 1/(1+b²). Integrate to I(b)=tan⁻¹ b + C; I(0)=0 gives C=0, so the integral equals tan⁻¹ b.

## Answer

#### Method
Differentiation under the integral sign (Leibnitz rule): if I(b) = ∫₀^∞ f(x, b) dx, with limits independent of b, then dI/db = ∫₀^∞ ∂f/∂b dx.

#### Solution
Let I(b) = ∫₀^∞ e^(−x) (sin bx)/x dx.

Differentiate with respect to b:

dI/db = ∫₀^∞ e^(−x) cos bx dx

Use the standard result ∫₀^∞ e^(−ax) cos bx dx = a/(a² + b²) with a = 1:

dI/db = 1/(1 + b²)

Integrate with respect to b:

I(b) = tan⁻¹ b + C

When b = 0, sin bx = 0, so I(0) = 0. Since tan⁻¹ 0 = 0, this gives C = 0.

**Answer: ∫₀^∞ e^(−x) sin bx / x dx = tan⁻¹ b.**

## Check yourself

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

1. Differentiation under the integral sign allows (limits independent of b)
   A. dI/db = f evaluated at the limits
   B. dI/db = integral of f dx / b
   C. dI/db = integral of d(f)/dx dx
   D. dI/db = integral of df/db dx
   Answer: D. For constant limits, differentiate the integrand with respect to the parameter. (Key point: Rule)

2. dI/db for I(b) = integral e^(-x) sin(bx)/x dx is
   A. integral of e^(-x) cos(bx) dx
   B. integral of e^(-x) sin(bx) dx
   C. integral of -e^(-x) cos(bx)/x dx
   D. integral of x e^(-x) cos(bx) dx
   Answer: A. d/db sin(bx) = x cos(bx), cancelling the 1/x. (Key point: Derivative)

3. The constant C is found using I(0) = 0 because
   A. the integral is symmetric so C = π/2
   B. sin(bx)=0 when b=0, so C = tan^-1(0) = 0
   C. e^(-x) vanishes at infinity so C = 1
   D. tan^-1 b tends to 0 at infinity so C = π
   Answer: B. Setting b=0 gives I = 0 and so C = 0. (Key point: Constant)

4. The value of the integral is
   A. tan^-1(1/b)
   B. log(1+b^2)
   C. tan^-1 b
   D. π/2
   Answer: C. Integrating 1/(1+b^2) gives tan^-1 b. (Key point: Result)

## Close variants set in other exams

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