← 2025 Paper 2

UPSC Maths 2025 Paper 2 Q2c — Solution

20 marks · Section A

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 $f(z) = \frac{z^2 - z + 2}{z^4 + 10z^2 + 9}.$

Step 1 — Factor the denominator and locate poles. $z^4 + 10z^2 + 9 = (z^2+1)(z^2+9) = (z-i)(z+i)(z-3i)(z+3i).$ 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, $\oint_{C_R} f(z)\,dz = 2\pi i\big[\operatorname{Res}_{z=i} f + \operatorname{Res}_{z=3i} f\big].$

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 $\left|\int_{\Gamma_R} f\,dz\right| \le \frac{C}{R^2}\cdot \pi R = \frac{\pi C}{R} \xrightarrow{R\to\infty} 0.$ 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 $\operatorname{Res}_{z=i} f = \frac{1-i}{16i} = \frac{(1-i)(-i)}{16} = \frac{-i + i^2}{16} = \frac{-i-1}{16} = -\frac{1}{16} - \frac{i}{16}.$

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 $\operatorname{Res}_{z=3i} f = \frac{-7-3i}{-48i} = \frac{7+3i}{48i} = \frac{(7+3i)(-i)}{48} = \frac{-7i -3i^2}{48} = \frac{3 - 7i}{48} = \frac{1}{16} - \frac{7i}{48}.$

Step 5 — Sum and multiply by $2\pi i$. The real parts cancel: $\operatorname{Res}_{z=i} f + \operatorname{Res}_{z=3i} f = \left(-\tfrac1{16}+\tfrac1{16}\right) + i\left(-\tfrac1{16} - \tfrac{7}{48}\right) = i\left(-\tfrac{3}{48} - \tfrac{7}{48}\right) = -\frac{10}{48}i = -\frac{5}{24}i.$ Therefore $\int_{-\infty}^{\infty} f(x)\,dx = 2\pi i\cdot\left(-\frac{5}{24}i\right) = 2\pi\cdot\frac{5}{24} = \frac{10\pi}{24} = \frac{5\pi}{12}.$

Answer

$\boxed{\;\int_{-\infty}^{\infty} \frac{x^2 - x + 2}{x^4 + 10x^2 + 9}\,dx = \frac{5\pi}{12}\;}$