--- title: "Fourier integral representation" subject: "Mathematics III" chapter: 4 key: "semester_3/mathematics_iii:fourier-integral-representation-four" source: https://www.bctengineer.com/q/semester_3/mathematics_iii/fourier-integral-representation-four site: BCT Engineering --- # Fourier integral representation Mathematics III · Chapter 4: Fourier Series - Asked in: 2081 Chaitra (new course), 3+1 marks - Video: [Fourier Integral | Fourier Cosine and Sine Integral | Important Problems | Engineering Mathematics](https://www.youtube.com/watch?v=CEgIgY4z66E) — Gautam Varde - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/fourier-integral-representation-four - Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-4 ## Question Find the Fourier integral of f(x) = e^(−x) for x > 0; 0 for x < 0, and hence show that ∫_0^∞ (cos ωx + ω sin ωx)/(1 + ω²) dω = π e^(−x) for x > 0; π/2 for x = 0; 0 for x < 0. ## Short answer The Fourier integral of f(x)=e^(-x) (x>0), 0 (x<0) is f(x)= (1/π)∫_0^∞ (cos ωx+ω sin ωx)/(1+ω²) dω. Hence the given integral equals πe^(-x) (x>0), π/2 (x=0), 0 (x<0). ## Answer **Fourier integral.** f(x) = ∫₀^∞ [A(α) cos αx + B(α) sin αx] dα, where A(α) = (1/π)∫ f(v) cos αv dv and B(α) = (1/π)∫ f(v) sin αv dv. For f(v) = e^(−v) (v > 0), 0 (v < 0): ```text A(α) = (1/π)∫₀^∞ e^(−v) cos αv dv = 1/[π(1 + α²)] B(α) = (1/π)∫₀^∞ e^(−v) sin αv dv = α/[π(1 + α²)] ``` f(x) = (1/π) ∫₀^∞ (cos αx + α sin αx)/(1 + α²) dα **Deduction.** Multiply both sides by π. The integral I(x) = ∫₀^∞ (cos ωx + ω sin ωx)/(1 + ω²) dω = π f(x), where at a jump f is replaced by the mean of its left and right limits. So: - x > 0: f is continuous and equals e^(−x), so I = π e^(−x). - x < 0: f = 0, so I = 0. - x = 0: f jumps from 0 to 1, the integral converges to the mean (0 + 1)/2 = 1/2, so I = π/2. Hence ∫₀^∞ (cos ωx + ω sin ωx)/(1 + ω²) dω = π e^(−x) for x > 0, π/2 for x = 0 and 0 for x < 0. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For the Fourier integral, A(alpha) is given by which expression? A. (1/2pi) integral of f(v) e^(-i alpha v) dv B. (1/pi) integral of f(v) cos(alpha v) dv C. (2/pi) integral of f(v) sin(alpha v) dv D. (1/pi) integral of f(v) sin(alpha v) dv Answer: B. A(alpha) uses cos and B(alpha) uses sin, each with factor 1/pi. (Key point: Fourier integral coefficients) 2. For f = e^(-v) on v > 0, what are A and B? A. A = a/[pi(1+a^2)], B = 1/[pi(1+a^2)] B. A = 1/(1+a^2), B = a/(1+a^2) C. A = 1/[pi(1+a^2)], B = a/[pi(1+a^2)] D. A = 1/[pi(1-a^2)], B = a/[pi(1-a^2)] Answer: C. The integrals of e^(-v)cos(av) and e^(-v)sin(av) over 0 to infinity are 1/(1+a^2) and a/(1+a^2), each divided by pi. (Key point: Evaluating A and B) 3. Why is the integral equal to pi/2 at x = 0? A. f jumps from 0 to 1, and the integral gives the mean 1/2 B. The integral diverges there C. f(0) = 0 so the integral is 0 D. f(0) = 1 so the integral is pi Answer: A. At a jump the Fourier integral converges to the average of the left and right limits, (0+1)/2, then multiplied by pi. (Key point: Value at x = 0) 4. How is the required integral obtained from the Fourier integral of f? A. Set alpha = 0 B. Divide both sides by pi C. Multiply both sides by pi D. Differentiate with respect to x Answer: C. f(x) = (1/pi) times the integral, so the integral equals pi f(x). 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