Contour integral ∫x²/((x²+a²)(x²+b²))

Method. Let I = ∫_{−∞}^{∞} x²/((x² + a²)(x² + b²)) dx, with a > 0, b > 0, a ≠ b. Take f(z) = z²/((z² + a²)(z² + b²)) around the closed contour C made of [−R, R] and the upper semicircle Γ: |z| = R (R > max(a, b)).

Simple poles inside C: z = ia and z = ib.

Residues:

• At z = ia: Res = (ia)² / ( 2ia (b² − a²) ) = −a² / ( 2ia (b² − a²) ) = −a / ( 2i (b² − a²) )
• At z = ib: Res = (ib)² / ( (a² − b²) · 2ib ) = −b / ( 2i (a² − b²) ) = b / ( 2i (b² − a²) )

Sum = (b − a) / ( 2i (b² − a²) ) = 1 / ( 2i (a + b) ).

Residue theorem:

∮_C f dz = 2πi · 1 / (2i (a + b)) = π/(a + b).

Arc: |f(z)| ≤ R²/((R² − a²)(R² − b²)) ~ 1/R², so ∫_Γ f dz ≤ πR/R² → 0 as R → ∞.

Answer: I = π/(a + b)

Special case a = b gives π/(2a). (Check: a = 1, b = 2 gives π/3, which agrees with a direct partial-fraction evaluation.)