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

Marks : 4

Year : 2015

Let I(a) be the given integral then by the rule of differentiation under the integral sign

$\dfrac {dI}{da}=\int\limits_0^{\infty}\dfrac {\partial f}{\partial a}dx=\int\limits_0^{\infty}\dfrac 1{x^2}\times \dfrac 1{1+ax^2}dx$

$\dfrac {dI}{da}=\int\limits_0^{\infty}\dfrac {dx}{1+ax^2}=\dfrac 1a\int\dfrac {dx}{x^2+\dfrac 1a}$

$=\dfrac 1a(\sqrt a)[\tan^{-1}x \sqrt a]^\infty_0$

$\dfrac {dI}{da}=\dfrac 1{\sqrt a}\dfrac \pi2$

Integrating both sides

$I=\dfrac \pi2\int a ^{\dfrac {-1}2}da$

$=\dfrac \pi2\Bigg[\dfrac {a^{1/2}}{\dfrac 12}\Bigg] + C$

$=\pi\sqrt a+ c$

To find c put a=0 I(a)=c

But $I(0)=\int\limits_0^{\infty}0 dx=0$

$C=0\space \space \space I=\pi\sqrt a$