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

# DUIS: ∫ (e^(−ax) − e^(−bx))/x dx

Mathematics I · Chapter 2: Integration and its Applications

- Asked in: 2071 Shrawan, 5 marks
- 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-minus-e-bx
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-2

## Question

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

## Short answer

Treat b as constant, differentiate with respect to a under the integral: dI/da = −∫₀^∞ e^{-ax} dx = −1/a. Integrate: I(a) = −ln a + C. I(b)=0 gives C=ln b, so I(a) = ln(b/a). Hence the integral equals ln(b/a).

## Answer

Let a, b > 0. Define I(a) = ∫₀^∞ (e^(−ax) − e^(−bx))/x dx, treating b as a constant.

**Differentiate with respect to a** (under the integral sign):

dI/da = ∫₀^∞ (−x e^(−ax))/x dx = −∫₀^∞ e^(−ax) dx = −[−e^(−ax)/a]₀^∞ = −1/a

**Integrate:** I(a) = −ln a + C.

**Find C.** When a = b, the integrand is zero, so I(b) = 0. Hence 0 = −ln b + C, C = ln b.

I(a) = ln b − ln a

**Answer: ∫₀^∞ (e^(−ax) − e^(−bx))/x dx = ln(b/a)** (Frullani's integral).

## Check yourself

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

1. dI/da equals
   A. 1/a
   B. -1/a^2
   C. -integral e^(-ax) dx = -1/a
   D. -a
   Answer: C. The factor x cancels the 1/x. (Key point: Derivative)

2. Integrating gives
   A. I(a) = ln a + C
   B. I(a) = 1/a + C
   C. I(a) = -a ln a + C
   D. I(a) = -ln a + C
   Answer: D. Integral of -1/a is -ln a. (Key point: Integration)

3. C is found from
   A. I(b) = 0 because the integrand vanishes at a = b, so C = ln b
   B. I(0) = 0, so C = 0
   C. I(∞)=0, so C = 1
   D. a = 1 gives C = b
   Answer: A. Setting a = b makes the integrand zero. (Key point: Constant)

4. The integral equals
   A. ln(a/b)
   B. ln(b/a)
   C. ln(ab)
   D. b - a
   Answer: B. I = ln b - ln a. (Key point: Result)

## Close variants set in other exams

- [DUIS: ∫ e^(−x) sin bx / x dx](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Aduis-e-x-sin-bx-over-x) — Mathematics I, chapter 2, asked 10×
- [DUIS: ∫ log(1+a²x²)/(1+b²x²) dx](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Aduis-log-1-plus-a2x2) — Mathematics I, chapter 2, asked 4×
- [DUIS: ∫ log((a+b sin x)/(a−b sin x)) dx/sin x](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Aduis-log-a-plus-b-sinx) — Mathematics I, chapter 2, asked 9×

## Related in Mathematics I

- [DUIS: ∫ e^(−ax) sin x / x dx](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Aduis-e-ax-sinx-over-x) — Mathematics I, chapter 2, asked 8×
- [DUIS: ∫ tan⁻¹(ax)/(x(1+x²)) dx](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Aduis-atan-ax-over-x-1-plus-x2) — Mathematics I, chapter 2, asked 2×
- [DUIS: ∫ dx/(a²sin²x+b²cos²x)²](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Aduis-dx-over-a2sin2-b2cos2-squared) — Mathematics I, chapter 2, asked 2×
- [Prove ∫ sin bx / x dx = π/2](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Aintegral-sin-bx-over-x) — Mathematics I, chapter 2
- [Evaluate ∫ dx/x³ from ∞ to α](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Aintegral-dx-over-x3-infinity) — Mathematics I, chapter 2

## Related in other subjects

- [Evaluate ∫∫ e^(x²) dx dy](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Aevaluate-e-x2-dx-dy) — Mathematics, chapter 2
- [Evaluate ∫∫ e^(y/x) dy dx](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Aevaluate-e-y-x-dy) — Mathematics, chapter 2