**1 Answer**

written 3.8 years ago by |

Cconsider 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{z^2}{(z^2+a^2)(z^2+b^2)}$ , poles are given by putting $(z^2+a^2)(z^2+b^2) =0$ i.e. z = +ai, - ai ,+bi , -bi. Out of these poles z=ai, z=bi lie in upper half of the plane.

Residue at z= ai

$R_1 = lim_{z→ai} \ (z-ai) \frac{z^2}{(z-ai)(z+ai) (z^2+b^2)} = \frac{a}{2i (a^2-b^2)}$

Similarly,

Residue at z= bi

$R_2 = lim_{z→bi} \ (z-bi) \frac{z^2}{(z-bi)(z+bi) (z^2+a^2)} = \frac{(- b)}{2i (a^2-b^2)}$

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