definite integral with exponential

\hspace{-16}$For $\mathbf{a\;,b>0}$. Evaluate the Integral\\\\ $\mathbf{\int_{0}^{\infty}\frac{e^{ax}-e^{bx}}{x.(e^{ax}+1).(e^{bx}+1)}dx}$

2 Answers

262

Aditya Bhutra ·2012-03-11 22:17:16

writing numerator as (e^ax+1) - (e^bx+1) then dividing the integral into two parts ,

I = ₀âˆ«^âˆž dxx(e^bx+1) - ₀âˆ«^âˆž dxx(e^ax+1)

now let f(a) = ₀âˆ«^âˆž dxx(e^ax+1)

f '(a)= ₀âˆ«^âˆž - e^ax.x .dxx(e^ax+1)² = 1a₀âˆ«^âˆž a.e^axdx(e^ax+1)²

f '(a) = -1a

f(a) = - ln (a) +c

now I = f(b) - f(a) = ln ab

1708

man111 singh ·2012-03-12 03:00:00

Very Nice Aditiya.

Same method used by you in Solving \hspace{-16}\mathbf{\int\frac{1}{(x^2+a^2)^2}dx} in goiit.com

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