Question: Evaluate $ \int_{- \infty}^{\infty} \frac{dx}{(x^2+1)^2} $

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$ f(x) = \frac{1}{(x^2+1)^2} \\ f(z) = \frac{1}{(z^2+1)^2} $

For poles $ z^2 = -1 \implies z = \pm i $

$ f(z) = \frac{1}{[(z-i)(z+i)]^2} = \frac{1}{(z-i)^2(z+i)^2} $

z = i and z = -i are the two poles of which z = i is the only pole which lies inside and if of order 2.

Residue of f(z) at z = i is:

$ \frac{1}{(2-1)!} \lim_{z \to i} \frac{d}{dz}[(z-i)^2 [\frac{1}{(z-i)^2(z+i)^2}]] \\ = \lim_{z \to i} \frac{d}{dz}[\frac{1}{(z+i)^2}] \\ = \lim_{z \to i} [- \frac{2}{(z+i)^3}] = - \frac{2}{(i+i)^3} = \frac{-2}{8i^3} = \frac{-2}{-8i} = \frac{1}{4i} $

$ \int_c \frac{1}{(z^2 +1)^2} dz = 2 \pi i [\frac{1}{4i}] = \frac{\pi}{2} \\ \int_{-\infty}^{\infty} \frac{1}{(x^2+1)^2} dx = \frac{\pi}{2} $

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