UPSC 2023 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

Evaluate $\displaystyle\int_0^{2\pi}\frac{\cos 2\theta}{5+4\cos\theta}\,d\theta$ using contour integration.

Technique

$z=e^{i\theta}$ substitution + residue theorem; the integrand has a double pole at the origin requiring a derivative.

Solution

Strategy. Standard $z=e^{i\theta}$ substitution converts the integral to a contour integral on the unit circle $|z|=1$ , then apply the residue theorem.

Step 1 — Substitute. With $z=e^{i\theta}$ , $d\theta=\dfrac{dz}{iz}$ ,

\cos\theta=\tfrac{1}{2}\bigl(z+z^{-1}\bigr),\qquad \cos 2\theta=\tfrac{1}{2}\bigl(z^{2}+z^{-2}\bigr).

Rewriting:

5+4\cos\theta=5+2\bigl(z+z^{-1}\bigr)=\frac{2z^{2}+5z+2}{z}=\frac{(2z+1)(z+2)}{z},

\cos 2\theta=\frac{z^{4}+1}{2z^{2}}.

Therefore

I=\oint_{|z|=1}\frac{(z^{4}+1)/(2z^{2})}{(2z+1)(z+2)/z}\cdot\frac{dz}{iz}=\oint_{|z|=1}\frac{z^{4}+1}{2iz^{2}(2z+1)(z+2)}\,dz.

Step 2 — Locate poles.

The integrand $\Phi(z)=\dfrac{z^{4}+1}{2iz^{2}(2z+1)(z+2)}$ has singularities at $z=0$ (double), $z=-\tfrac{1}{2}$ (simple), $z=-2$ (simple). Inside $|z|=1$ : only $z=0$ and $z=-\tfrac{1}{2}$ .

Step 3 — Residue at $z=-\tfrac{1}{2}$ (simple).

\operatorname*{Res}_{z=-1/2}\Phi=\lim_{z\to-1/2}\bigl(z+\tfrac{1}{2}\bigr)\Phi(z)=\frac{z^{4}+1}{2iz^{2}\cdot 2\cdot(z+2)}\bigg|_{z=-1/2}.

Numerator: $(-\tfrac{1}{2})^{4}+1=\tfrac{1}{16}+1=\tfrac{17}{16}$ . Denominator: $2i\cdot\tfrac{1}{4}\cdot 2\cdot\tfrac{3}{2}=\tfrac{3i}{2}$ .

\operatorname*{Res}_{z=-1/2}\Phi=\frac{17/16}{3i/2}=\frac{17}{24i}=-\frac{17i}{24}.

(Note: the $(2z+1)$ factor contributes a derivative of $2$ , hence the $\cdot 2$ in the denominator.)

Step 4 — Residue at $z=0$ (double pole).

For a pole of order 2,

\operatorname*{Res}_{z=0}\Phi=\lim_{z\to 0}\frac{d}{dz}\bigl[z^{2}\Phi(z)\bigr]=\frac{d}{dz}\bigl[h(z)\bigr]\bigg|_{z=0},\quad h(z)=\frac{z^{4}+1}{2i(2z+1)(z+2)}.

Let $H(z)=\dfrac{z^{4}+1}{(2z+1)(z+2)}$ , so $h=H/(2i)$ . At $z=0$ : $H(0)=\tfrac{1}{1\cdot 2}=\tfrac{1}{2}$ . By the quotient rule

H'(z)=\frac{4z^{3}(2z+1)(z+2)-(z^{4}+1)\bigl[2(z+2)+(2z+1)\bigr]}{[(2z+1)(z+2)]^{2}}.

At $z=0$ : numerator $=0-1\cdot[4+1]=-5$ , denominator $=(1\cdot 2)^{2}=4$ , so $H'(0)=-\tfrac{5}{4}$ . Hence

\operatorname*{Res}_{z=0}\Phi=\frac{H'(0)}{2i}=\frac{-5/4}{2i}=\frac{-5}{8i}=\frac{5i}{8}.

Step 5 — Apply residue theorem.

I=2\pi i\!\left(\operatorname*{Res}_{z=-1/2}+\operatorname*{Res}_{z=0}\right)=2\pi i\!\left(-\frac{17i}{24}+\frac{5i}{8}\right)=2\pi i\!\left(\frac{-17+15}{24}i\right)=2\pi i\cdot\frac{-2i}{24}=\frac{4\pi}{24}=\frac{\pi}{6}.

Answer

\boxed{\;\int_0^{2\pi}\frac{\cos 2\theta}{5+4\cos\theta}\,d\theta=\frac{\pi}{6}.\;}