Mumbai University > COMPS > Sem 4 > Applied Mathematics 4

Marks : 06

Year : DEC 2014

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

S2 : Let $f(z)=\dfrac {z^2}{(z^2+1)(z^2+4)}$

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

S3: For singularity $,(z^2+1)(z^2+4)=0$

$\therefore z^2=-1=i$ or $z^2=-4=4i^2$

$\therefore z=\pm i$ or $z=\pm 2i$

Here $, z_0=-i ,-2i$ lies outside while $z_0=i ,2i$ lies inside the contour.

$Z_0=i, 2i$ are simple poles.

S4: $R_1=$ Residue of f(z) at $z=1$

$$=\lim\limits_{z\to z_0}(z-z_0)\times f(z)$$

$=\lim\limits_{z\to i} (z-i)\times \dfrac {z^2}{(z-1)(z+i)(z+2i)(z-2i)}\\ =\lim\limits_{z\to i}\dfrac {z^2}{(z+i)(z^2+4)}\\ =\dfrac i{(i+i)[(i)^2+4]}\\ =\dfrac {-1}{6i}$

$R_2=$ Residue of $f(z)$ at $(z=2i)$

$$=\lim\limits_{z\to2i}(z-2i)\times \dfrac {z^2}{(z-1)(z+i)(z+2i)(z-2i)}$$

$=\lim\limits_{z\to2i}\dfrac {z^2}{(z+2i)(z^2+1)}\\ =\dfrac {(2i)^2}{(2i+2i)[(2i)^2+1]}\\ =\dfrac 1{3i}$

By Cauchy's Residue theorem

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

$\therefore \int\dfrac {z^2}{(z^2+1)(z^2+4)}dz\\ =2\pi i \Bigg\{\dfrac {-1}{6i}+\dfrac 1{3i}\Bigg\}\\ \therefore \int\limits_{-\infty}^{\infty}\dfrac {x^2}{(x^2+1)(x^2+4)}dx\\ =2\pi i \times \dfrac 1{6i}\\ =\dfrac \pi3$