← 2014 Paper 2

UPSC 2014 Maths Optional Paper 2 Q3c — 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 the integral ∫0πdθ(1+12cos⁡θ)2\displaystyle\int_0^\pi\dfrac{d\theta}{(1+\tfrac{1}{2}\cos\theta)^{2}} using residues.

Technique

Period-halving via cos⁡\cos symmetry; z=eiθz=e^{i\theta} substitution; double-pole residue via derivative.

Solution

Strategy. Symmetry cos⁡(2π−θ)=cos⁡θ\cos(2\pi-\theta)=\cos\theta ⇒ ∫02π=2∫0π\int_0^{2\pi}=2\int_0^\pi. Compute ∫02π\int_0^{2\pi} via z=eiθz=e^{i\theta} substitution.

Step 1 — Extend to [0,2π][0,2\pi]

∫02πdθ(1+12cos⁡θ)2=2∫0πdθ(1+12cos⁡θ)2.\int_0^{2\pi}\dfrac{d\theta}{(1+\tfrac{1}{2}\cos\theta)^{2}}=2\int_0^\pi\dfrac{d\theta}{(1+\tfrac{1}{2}\cos\theta)^{2}}.

Step 2 — Substitute z=eiθz=e^{i\theta}

cos⁡θ=12(z+z−1)\cos\theta=\tfrac{1}{2}(z+z^{-1}), dθ=dz/(iz)d\theta=dz/(iz). Then

1+12cos⁡θ=1+14(z+z−1)=z2+4z+14z.1+\tfrac{1}{2}\cos\theta=1+\tfrac{1}{4}(z+z^{-1})=\dfrac{z^{2}+4z+1}{4z}. (1+12cos⁡θ)2=(z2+4z+1)216z2.(1+\tfrac{1}{2}\cos\theta)^{2}=\dfrac{(z^{2}+4z+1)^{2}}{16z^{2}}.

So

∫02πdθ(1+12cos⁡θ)2=∮∣z∣=116z2(z2+4z+1)2⋅dziz=16i∮∣z∣=1z dz(z2+4z+1)2.\int_0^{2\pi}\dfrac{d\theta}{(1+\tfrac{1}{2}\cos\theta)^{2}}=\oint_{|z|=1}\dfrac{16z^{2}}{(z^{2}+4z+1)^{2}}\cdot\dfrac{dz}{iz}=\dfrac{16}{i}\oint_{|z|=1}\dfrac{z\,dz}{(z^{2}+4z+1)^{2}}.

Step 3 — Locate poles

z2+4z+1=0z^{2}+4z+1=0 has roots z1=−2+3z_1=-2+\sqrt 3 and z2=−2−3z_2=-2-\sqrt 3.

∣z1∣=3−2≈−0.27|z_1|=\sqrt 3-2\approx -0.27… wait, ∣−2+3∣=2−3≈0.27<1|-2+\sqrt 3|=2-\sqrt 3\approx 0.27<1, so z1z_1 is inside ∣z∣=1|z|=1. ∣z2∣=2+3≈3.73>1|z_2|=2+\sqrt 3\approx 3.73>1, so z2z_2 is outside.

The integrand has a double pole at z1z_1 inside ∣z∣=1|z|=1.

Step 4 — Residue at z1z_1 (double pole)

Factor: (z2+4z+1)2=(z−z1)2(z−z2)2(z^{2}+4z+1)^{2}=(z-z_1)^{2}(z-z_2)^{2}.

Resz=z1z(z−z1)2(z−z2)2=lim⁡z→z1ddz ⁣[z(z−z2)2].\text{Res}_{z=z_1}\dfrac{z}{(z-z_1)^{2}(z-z_2)^{2}}=\lim_{z\to z_1}\dfrac{d}{dz}\!\left[\dfrac{z}{(z-z_2)^{2}}\right].

Compute:

ddz ⁣[z(z−z2)2]=(z−z2)2−z⋅2(z−z2)(z−z2)4=(z−z2)−2z(z−z2)3=−z−z2(z−z2)3.\dfrac{d}{dz}\!\left[\dfrac{z}{(z-z_2)^{2}}\right]=\dfrac{(z-z_2)^{2}-z\cdot 2(z-z_2)}{(z-z_2)^{4}}=\dfrac{(z-z_2)-2z}{(z-z_2)^{3}}=\dfrac{-z-z_2}{(z-z_2)^{3}}.

At z=z1z=z_1:

Resz=z1=4243=163=318.\text{Res}_{z=z_1}=\dfrac{4}{24\sqrt 3}=\dfrac{1}{6\sqrt 3}=\dfrac{\sqrt 3}{18}.

Step 5 — Apply residue theorem

∮∣z∣=1z dz(z2+4z+1)2=2πi⋅318=πi39.\oint_{|z|=1}\dfrac{z\,dz}{(z^{2}+4z+1)^{2}}=2\pi i\cdot\dfrac{\sqrt 3}{18}=\dfrac{\pi i\sqrt 3}{9}. ∫02π=16i⋅πi39=16π39.\int_0^{2\pi}=\dfrac{16}{i}\cdot\dfrac{\pi i\sqrt 3}{9}=\dfrac{16\pi\sqrt 3}{9}.

Halve:

∫0πdθ(1+12cos⁡θ)2=8π39.\int_0^\pi\dfrac{d\theta}{(1+\tfrac{1}{2}\cos\theta)^{2}}=\dfrac{8\pi\sqrt 3}{9}.

Answer

  ∫0πdθ(1+12cos⁡θ)2=8π39.  \boxed{\;\int_0^\pi\dfrac{d\theta}{(1+\tfrac{1}{2}\cos\theta)^{2}}=\dfrac{8\pi\sqrt 3}{9}.\;}
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