Mumbai University > COMPS > Sem 4 > Applied Mathematics 4

Marks : 06

Year : MAY 2015

S1: Consider the contour of a large semi- circle with diameter on real axis; center at origin & above the real axis.

S2: Let $f(z)=\dfrac 1{(z^2+a^2)^3}$

as $z\to\infty ,zf(z)\to 0$

S3: for singularity

$$\therefore (z^2+a^2)^3=0$$

$\therefore z^2=-a^2=i^2a^2\\ \therefore z=\pm ai,\pm ai$

Here, $z_0=-ai$ lies outside while $z_0=ai$ lies inside the contour.

$z_0=ai$ is a pole of order 3

S 4: R1= Residue of F(z) at $z=ai$

$$=\dfrac 1{(n-1)!}\lim\limits_{z\to z_0}\dfrac {d^{n-1}}{dz^{n-1}}(z-z_0)^n\times f(z)$$

$=\dfrac 1{(3-1)!}\lim\limits_{z\to ai}\dfrac {d^2}{dz^2}(z-ai)^3\times \dfrac 1{(z-ai)^3(z+ai)^3}\\ =\dfrac 1{2!}\lim\limits_{z\to ai}\dfrac {d^2}{dz^2}(z+ai)^{-3}\\ =\dfrac 12\lim\limits_{z\to ai} \dfrac d{dz}[-3.(z+ai)^{-4}.1]\\ =\dfrac 12.(-3)\lim\limits_{z\to ai} \dfrac d{dz}[-4.(z+ai)^{-5}.1]\\ =\dfrac {-3}2\times -4\times \dfrac 1{(ai+ai)^5}\\ =\dfrac 6{2^5a^5i^5}\\ =\dfrac 3{16a^5i}$

By cauchy's Residue theorem

$$\int f(z).dz=2\pi i [R_1+R_2+...]$$

$\therefore \int\dfrac 1{(z^2+a^2)^3}.dz=2\pi i[\dfrac 3{16a^5i}]\\ \therefore \int\limits_{\infty}^{\infty}\dfrac {dx}{(x^2+a^2)^3}.dx^1=\dfrac {3\pi}{8a^5}\\ \therefore 2\int\limits_0^{\infty}\dfrac {dx}{(x^2+a^2)^3}.dx=\dfrac {3\pi}{8a^5}$ {Since even function]

$$\therefore \int\limits_0^{\infty}\dfrac {dx}{(x^2+a^2)^3}.dx=\dfrac {3\pi}{16a^5}$$