Contour integral ∫dx/(1+x²)²

Method. Let I = ∫_{−∞}^{∞} dx/(1 + x²)². Take f(z) = 1/(z² + 1)² = 1/((z − i)²(z + i)²) and integrate around the closed contour C formed by the segment [−R, R] and the upper semicircle Γ: |z| = R, R > 1.

Inside C there is a single pole, z = i, of order 2.

Residue at z = i:

Res = d/dz [ (z − i)² f(z) ] at z = i = d/dz [ (z + i)⁻² ] = −2 (z + i)⁻³ at z = i

= −2/(2i)³ = −2/(−8i) = 1/(4i)

Residue theorem:

∮_C f dz = 2πi · 1/(4i) = π/2.

Arc integral: on Γ, |f(z)| ≤ 1/(R² − 1)², so |∫_Γ f dz| ≤ πR/(R² − 1)² → 0 as R → ∞.

Therefore ∫_{−∞}^{∞} dx/(1 + x²)² = π/2.

Answer: I = π/2

(Check: ∫ dx/(1 + x²)² = ½[x/(1 + x²) + tan⁻¹x], which over (−∞, ∞) gives ½ · π = π/2.)