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

Marks: 6M

Year: May 2014

Consider the contour consisting of a large semicircle with center at the origin, in the upper half of the plane and its diameter along the real axis

Let us consider $f(z) = \int\limits_{0}^{\infty} \frac{z^2 d^{iz}}{(z^2 + a^2)^2}dz...............(1)$

Here f(z) has singular points z = $\pm$ai, poles of order 2.

Of which z = ai lies in the upper half plane

Using Integrals Round the large semi-circle

$\int\limits_{-\infty}^{\infty} f(x)dx = \oint f(z) dz = 2\pi i (sum \hspace{0.25cm} of \hspace{0.25cm} the \hspace{0.25cm} residues)................(2)$

$\therefore$ Let $r_{1} = Res(z = ai) = \lim_{z \rightarrow ai} \frac{d}{dz} (z - ai)^2 \cdot \frac{z^3 e^{iz}}{(z - ai)^2 (z + ai)^2} \\ = \frac{d}{dz} \frac{z^3 e^{iz}}{(z + ai)^2} \\ = \lim_{z \rightarrow ai} \frac{[(z + ai)^2 (z^3 \cdot ie^{iz} + 3z^2 \cdot e^{iz} - (z^3 e^{iz}) (2 \cdot (z + ai))]}{(z + ai)^4} \\ = \frac{[(-4a^2)(a^3 \cdot e^{-a} - 3a^2 \cdot e^{-a}) - ( -a^3 i \cdot e^{-a}) (4ai)]}{16a^4} \\ = \frac{e^{-a} [(-4a^5 + 12a^4) - 4a^4]}{16a^4} \\ = \frac{E^{-a} (-4a + 8)}{16} \\ = \frac{e^{-a} (2 - a)}{4}$

$\therefore \oint f(z)dz = 2\pi i (r_{1}) \\ = 2 \pi i \cdot \frac{e^{-a} (2 - a)}{4} \\ = \frac{\pi i e^{-a} (2 - a)}{2}$

Hence,

$$\int\limits_{0}^{\infty} \frac{x^3 sinx}{(x^2 + a^2)^2}dx = \frac{\pi i e^{-a} (2 - a)}{2}$$