UPSC 2021 Maths Optional Paper 2 Q4b — 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

Using contour integration, evaluate $\int_{-\infty}^\infty\dfrac{\sin x}{x(x^2+a^2)}\,dx$ , $a>0$ .

Technique

Indented contour to handle $1/x$ at origin; residue at $ia$ in upper half-plane; small-semicircle contribution = $-\pi i\cdot\text{Res}$ at simple pole.

Solution

Strategy. $\sin x=\Im e^{ix}$ . Consider $\displaystyle\int_C\dfrac{e^{iz}}{z(z^2+a^2)}\,dz$ on contour avoiding pole at $z=0$ (semicircular indent).

Step 1 — Contour setup

$C=$ real axis from $-R$ to $-\epsilon$ , small semicircle (above) from $-\epsilon$ to $\epsilon$ , real axis from $\epsilon$ to $R$ , large semicircle (above) from $R$ back to $-R$ .

Poles of integrand $f(z)=\dfrac{e^{iz}}{z(z^2+a^2)}=\dfrac{e^{iz}}{z(z-ia)(z+ia)}$ :

$z=0$ : simple (avoided by small indent).
$z=ia$ : simple (inside contour).
$z=-ia$ : simple (outside contour).

Step 2 — Residues

$\text{Res}_{z=ia}f=\dfrac{e^{i\cdot ia}}{ia\cdot(2ia)}=\dfrac{e^{-a}}{-2a^2}$ .

$\text{Res}_{z=0}f=\dfrac{e^0}{(0-ia)(0+ia)}=\dfrac{1}{a^2}$ .

Step 3 — Apply residue theorem with indentation

$\int_C f\,dz=\int_{-R}^{-\epsilon}+\int_{\text{small}}+\int_{\epsilon}^R+\int_{\text{large}}=2\pi i\cdot\text{Res}_{z=ia}f$ .

Large semicircle: by Jordan’s lemma, $\to 0$ as $R\to\infty$ .

Small semicircle around $z=0$ : classical result: the integral along a small semicircle of radius $\epsilon$ around a simple pole, in the upper half-plane (going CCW from $-\epsilon$ to $\epsilon$ , passing through $\epsilon e^{i\pi/2}\cdot$ wait the standard indented contour goes above the singularity, traversing $\theta=\pi\to 0$ ):

The small semicircle traversed from $-\epsilon$ to $+\epsilon$ in upper half-plane means $\theta$ goes from $\pi$ to $0$ (clockwise as viewed). So the integral contributes $-\pi i\cdot\text{Res}_{z=0}f=-\pi i\cdot 1/a^2$ .

Step 4 — Combine

$\int_{-\infty}^\infty f(x)\,dx-\pi i/a^2+0=2\pi i\cdot e^{-a}/(-2a^2)=-\pi i e^{-a}/a^2$ .

So $\int_{-\infty}^\infty f(x)\,dx=\pi i/a^2-\pi i e^{-a}/a^2=\dfrac{\pi i(1-e^{-a})}{a^2}$ .

Step 5 — Take imaginary part

$\int_{-\infty}^\infty\dfrac{\sin x}{x(x^2+a^2)}\,dx=\Im\!\left[\dfrac{\pi i(1-e^{-a})}{a^2}\right]=\dfrac{\pi(1-e^{-a})}{a^2}$ .

Answer

\boxed{\;\int_{-\infty}^\infty\dfrac{\sin x}{x(x^2+a^2)}\,dx=\dfrac{\pi(1-e^{-a})}{a^2}.\;}