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UPSC 2025 Maths Optional Paper 2 Q2c — Step-by-Step Solution

20 marks · Section A

Contour integration of real integrals using residues · Complex Analysis · asked 10× in 14 yrs · Read the full method →

Question

Use the method of contour integration to prove that $\displaystyle\int_{-\infty}^{\infty} \dfrac{x^2 - x + 2}{x^4 + 10x^2 + 9}\,dx = \dfrac{5\pi}{12}$.

Technique

Apply the residue theorem on a semicircular contour in the upper half-plane, summing residues at the enclosed poles, with a bound showing the large semicircle contributes nothing.

Solution

Let

Step 1 — Factor the denominator and locate poles.

The four simple poles are $z = \pm i,\ \pm 3i$. In the upper half-plane the poles are $z=i$ and $z=3i$.

Step 2 — Set up the contour. Take $C_R$ = segment $[-R,R]$ on the real axis together with the semicircle $\Gamma_R$ of radius $R$ in the upper half-plane, traversed counterclockwise, with $R>3$ so that both poles $i,3i$ lie inside. By the residue theorem,

Step 3 — The semicircle vanishes as $R\to\infty$. For $|z|=R$, the numerator is $O(R^2)$ and the denominator $\sim R^4$, so $|f(z)| \le \dfrac{C R^2}{R^4} = \dfrac{C}{R^2}$ for large $R$. Then

Hence $\displaystyle\int_{-\infty}^{\infty} f(x)\,dx = 2\pi i\big[\operatorname{Res}_{z=i} f + \operatorname{Res}_{z=3i} f\big].$

Step 4 — Compute the residues (simple poles; use $\operatorname{Res}_{z=z_0} \frac{p(z)}{q(z)} = \frac{p(z_0)}{q'(z_0)}$ with $q'(z)=4z^3+20z$).

At $z=i$: $\ p(i) = i^2 - i + 2 = 1 - i$, and $q'(i) = 4i^3 + 20i = -4i + 20i = 16i$. Thus

At $z=3i$: $\ p(3i) = (3i)^2 - 3i + 2 = -9 - 3i + 2 = -7 - 3i$, and $q'(3i) = 4(3i)^3 + 20(3i) = -108i + 60i = -48i$. Thus

Step 5 — Sum and multiply by $2\pi i$. The real parts cancel:

Therefore

Answer