**1 Answer**

written 3.8 years ago by | modified 3.6 years ago by |

Consider the contour C consisting of large semi circle and diameter on the real axis with centre at the origin lying in the upper half of the plane.

Consider $ \frac{1}{(z^2+a^2)^3}$ , poles are given by putting $z^2+a^2=0$ i.e. z = +ai, -ai, both are poles of order 2.

Out of these poles z= ai lies in upper half of the plane.

Residue at z= ai

$R = lim_{z→ai} \ \frac{1}{2 !} \frac{d^2}{dz^2}(z-ai)^3 \frac{1}{(z-ai)^3 (z+ai)^3}$

= $ lim_{z→ai} \ \frac{1}{2} \frac{d^2}{dz^2} \frac{1}{(z+ai)^3} = \frac{3}{16a^5 i} $

$∫_{-∞}^∞ \frac{dx}{(x^2+a^2)^3} = 2πi[\frac{3}{(16a^5 i )}] = \frac{3π}{(8a^5)}$

$∫_0^∞ \frac{dx}{(x^2+a^2)^3} = \frac{1}{2} ∫_{-∞}^∞ \frac{dx}{(x^2+a^2)^3} = \frac{1}{2} . \frac{3π}{(8a^5 )} = \frac{3π}{16a^5} $