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Question: Prove that $\int_{0}^{\infty}\frac{e^{-x}-e^{-ax}}{x\,secx}dx\,=\frac{1}{2}log\left(\frac{a^2+1}{2}\right)\,\,\,,\,a\geq0$ using DUIS .
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Subject : Applied Mathematics 2

Topic : Differentiation under the integral sign(DUIS)

Difficulty : Medium

am2(82) • 207 views
 modified 6 months ago by awari.swati831 • 140 written 8 months ago by
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$Let I(a) = \int_0^{\infty} \frac{ (e^{-x} - e^{-ax}) }{xsecx} dx$

$\frac{dI}{da} = \int_0^{\infty} \frac{\partial}{\partial a} \frac{ (e^{-x} - e^{-ax}) }{xsecx} dx$

$= \int_0^{\infty} \frac{ x e^{-ax} }{xsecx} dx$

$= \int_0^{\infty} e^{-ax} cosx dx$

$= \left [ \frac{ e^{-ax} }{a^2 + 1}(-acosx + sinx) \right ]_0^{\infty}$

$= \frac{a }{a^2 + 1}$

$\text{I(a)} = \frac{1}{2} log(a^2+1) - \frac{1}{2} log(2) + C$

$\text{When a = 1, I(1) = } \frac{1}{2} log(a^2+1) + C$

$0 = \frac{1}{2} log(a^2+1) + C$

$C = - \frac{1}{2} log(2)$

$I (a) = \frac {1}{2} log\frac{a^2+1}{2} -\frac{1}{2}log2 \\$

$= \frac{1}{2} log \frac{(a^2 + 1)}{2} \\$