Mumbai University > Electronics and Telecommunication > Sem 4 > Applied Maths 4

Marks: 6M

Year: Dec 2014

Let us consider $f(z) = \oint \frac{e^{i3z}}{(z^2 + 1)(z^2 + 4)}dz..............(1)$

The poles of f(z) are $z = \pm i$ and $z = \pm 2i$

Of which z = i, 2i lies in the upper half - plane

Using Integral Round the Large Semi-circle

$\int\limits_{\infty}^{\infty} f(x)dx = \oint f(z)dz = 2\pi i$(sum of the residues).......(2)

$\therefore$ Let $r_{1} = Res(z = i) = \lim\limits_{z \rightarrow i} (z - i) \cdot \frac{e^{i3z}}{(z - i)(z + i) (z^2 + 4)} \\ \therefore r_{1} = \frac{e^{-3}}{6i}$

Let $r_{2} = Res(z = 2i) = \lim\limits_{z \rightarrow 2i (z - 2i) \cdot \frac{e^{i3z}}{(z - 2i)(z + 2i)(z^2 + 1)}} \\ r_{2} = \frac{e^{-6}}{-12i} \\ \therefore \oint f(z)dz = 2\pi i (r_{1} + r_{2}) \\ = 2\pi i \bigg(\frac{e^{-3}}{6i} + \frac{e^{-6}}{-12i}) \\ = \frac{\pi}{6} (2e^{-3} - e^{-6})$

Hence $$\int\limits_{-\infty}^{\infty} \frac{cos3x}{(x^2 + 1)(x^2 + 4)}dx = \frac{\pi}{6} (2e^{-3} - e^{-6})$$