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UPSC 2022 Maths Optional Paper 2 Q2c — Step-by-Step Solution

20 marks · Section A

Contour integration of real integrals using residues · Complex Analysis · asked 9× in 13 yrs · Read the full method →

Question

Evaluate $\displaystyle\int_{-\infty}^\infty\dfrac{\cos x\,dx}{(x^2+a^2)(x^2+b^2)}$, $a>b>0$, using calculus of residues.

Technique

Standard contour integration: replace $\cos x$ with $\text{Re}(e^{ix})$; close in upper half-plane (Jordan’s lemma kills semicircle); residues at the two simple poles $z=ia, ib$.

Solution

Strategy. Consider $\displaystyle\int_{-\infty}^\infty\dfrac{e^{ix}\,dx}{(x^2+a^2)(x^2+b^2)}$ and take real part. Close the contour in the upper half-plane.

Step 1 — Set up the contour

Let $f(z)=\dfrac{e^{iz}}{(z^2+a^2)(z^2+b^2)}$. Poles: $z=\pm ia$ and $z=\pm ib$.

In the upper half-plane: poles at $z=ia$ and $z=ib$ (both above real axis since $a,b>0$).

Contour $C_R$: real axis $[-R,R]$ plus upper semicircle of radius $R$.

By Jordan’s lemma, the semicircle contribution vanishes as $R\to\infty$ (the $e^{iz}$ decays in the upper half-plane for $\Im z>0$).

Step 2 — Residue at $z=ia$

$z=ia$ is a simple pole. Residue:

(Note: $(z^2+a^2)=(z-ia)(z+ia)$.)

Since $b<a$: $b^2-a^2<0$, so $\text{Res}_{ia}f=\dfrac{e^{-a}}{2ia(b^2-a^2)}=\dfrac{-e^{-a}}{2ia(a^2-b^2)}$.

Step 3 — Residue at $z=ib$

Step 4 — Apply residue theorem

Step 5 — Take real part

The original integral $\int\cos x/(\cdots)$ is the real part. The expression above is already real (no $i$), so:

Answer