Step 1: Evaluate I = ∫₋∞^∞ dx/(x² + 1) using the residue theorem. Step 2: Consider f(z) = 1/(z² + 1) = 1/[(z−i)(z+i)]. It has simple poles at z = i and z = −i. Step 3: Close the contour with a large semicircle in the upper half-plane. The arc contribution vanishes as the radius → ∞ (Jordan's lemma / ML estimate), so I = ∮f(z)dz over the closed contour. Step 4: Only the pole at z = +i lies inside the upper-half-plane contour. Step 5: Residue at z = i: Res = lim_{z→i} (z−i)/[(z−i)(z+i)] = 1/(2i). Step 6: By the residue theorem, I = 2πi × (1/2i) = π. Result: ∫₋∞^∞ dx/(x²+1) = π. The same method with a keyhole contour handles branch cuts, e.g. ∫_0^∞ x^(−1/2)/(1+x)dx = π.