Mumbai University > First Year Engineering > sem 2 > Applied Maths 2

Marks : 8

Year : 2014

Let I (a) be the given integral sign

$\dfrac {dI}{da}=\int\limits_0^{\infty}\dfrac {\partial f}{\partial a}dx=\int\limits_0^{\infty}\dfrac {\cos\lambda x}x(e^{-ax})x(-x)dx$

$= -\int\limits_0^{\infty}e^{-ax}\cos\lambda x dx$

$= \Bigg[\dfrac{e^{-ax}}{a^2+\lambda^2}(-a\cos\lambda x+\lambda\sin\lambda x)\Bigg]_0^{\infty}$

$\therefore \dfrac {dI}{da}=-\Bigg[0-\dfrac 1{a^2+\lambda^2}(-a)\Bigg]$

$\therefore \dfrac {dI}{da}= \dfrac{-a}{a^2+\lambda^2}$

Integrating both sides we get

$I=\dfrac {-1}2\log(a^2+\lambda^2)+ c \hspace {2 cm}\text{Direct Formula}$

To find c we put a = b

$\therefore I(b)=\dfrac {-1}2\log(b^2+\lambda^2)+ c$

But $I(b)=\int\limits_0^{\infty}\dfrac {\cos\lambda x}x(e^{-bx}-e^{-bx})$ dx

$=\int\limits_0^{\infty}0dx$

$=0$

$c=\dfrac 12\log(b^2+\lambda^2)$

$I=\dfrac {-1}2\log(a^2+\lambda^2)+\dfrac 12\log(b^2+\lambda^2)$

$=\dfrac 12\log\Bigg(\dfrac{b^2+\lambda^2}{a^2+\lambda^2}\Bigg)$