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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
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modified 6 months ago by gravatar for awari.swati831 awari.swati831140 written 8 months ago by gravatar for smitapn612 smitapn6120
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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} \\ $

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modified 6 months ago by gravatar for awari.swati831 awari.swati831140 written 7 months ago by gravatar for smitapn612 smitapn6120
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