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- 2071 Shrawan · 5 marks
DUIS: ∫ (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).
Apply differentiation under integral sign to evaluate ∫₀^∞ (e^(−ax) − e^(−bx))/x dx.
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).
✓ Answer checked by Claude Opus 5.5 (high effort)
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