---
title: "Half range sine series: eˣ"
subject: "Mathematics III"
chapter: 4
key: "semester_3/mathematics_iii:half-range-sine-series-e-two"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/half-range-sine-series-e-two
site: BCT Engineering
---

# Half range sine series: eˣ

Mathematics III · Chapter 4: Fourier Series

- Asked in: 2067 Magh
- Video: [Half Range sine series f(x)=e^x](https://www.youtube.com/watch?v=BwjvnPvIdWg) — Btech Maths Hub
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/half-range-sine-series-e-two
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-4

## Question

Find the half range sine series for the function f(x) = eˣ for 0 < x < π.

## Short answer

e^x = (2/π) Σ_{n=1}^∞ { n[1−(−1)ⁿ e^π]/(n²+1) } sin nx. Coefficient bₙ = (2/π) n[1−(−1)ⁿ e^π]/(n²+1) found by integration.

## Answer

The half range sine series on (0, π) is f(x) = Σ bₙ sin nx, with bₙ = (2/π)∫₀^π f(x) sin nx dx.

**Coefficient** for f(x) = eˣ. Use ∫ eᵃˣ sin nx dx = eᵃˣ(a sin nx − n cos nx)/(a² + n²) with a = 1:

```text
bₙ = (2/π)∫₀^π eˣ sin nx dx
   = (2/π)[ eˣ(sin nx − n cos nx)/(1 + n²) ]₀^π
   = (2/π)·[ −n(−1)ⁿ e^π + n ]/(1 + n²)
   = (2/π)·n[1 − (−1)ⁿ e^π]/(1 + n²)
```

**Series**

eˣ = (2/π) Σ (n=1 to ∞) { n[1 − (−1)ⁿ e^π] / (n² + 1) } sin nx,  0 < x < π

= (2/π)[ (1 + e^π)/2 · sin x + 2(1 − e^π)/5 · sin 2x + 3(1 + e^π)/10 · sin 3x + … ]

(The first terms follow by putting n = 1, 2, 3.) A numerical check at x = 1 with many terms gives 2.718, matching e¹.

## Check yourself

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

1. bn on (0, π) is:
   A. (2/π)∫0^π f(x) sin nx dx
   B. (1/π)∫ f cos nx dx
   C. (2/π)∫ f cos nx dx
   D. ∫ f sin nx dx
   Answer: A. Standard. (Key point: Formula)

2. ∫ e^x sin nx dx equals:
   A. e^x(sin nx + n cos nx)/(1 + n^2)
   B. e^x(sin nx − n cos nx)/(1 + n^2)
   C. e^x(cos nx + n sin nx)/(1 + n^2)
   D. e^x sin nx/n
   Answer: B. a = 1. (Key point: Integral)

3. bn is:
   A. (2/π) n[1 + (−1)^n e^π]/(1 + n^2)
   B. (2/π)[1 − (−1)^n e^π]/(1 + n^2)
   C. (2/π) n[1 − (−1)^n e^π]/(1 + n^2)
   D. 0
   Answer: C. At π, cos nπ = (−1)^n; at 0, cos 0 = 1. (Key point: Result)

## Close variants set in other exams

- [Half range sine series: e^(2x)](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Ahalf-range-sine-series-e-2x) — Mathematics III, chapter 4, asked 6×
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