← 2013 Paper 2

UPSC 2013 Maths Optional Paper 2 Q4b — Step-by-Step Solution

15 marks · Section A

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

Question

Using Cauchy’s residue theorem, evaluate the integral

Technique

$z=e^{i\theta}$ on $[0,2\pi]$ converts trig integrals to contour integrals; residue at origin from binomial expansion.

Solution

Strategy. Extend to $[0,2\pi]$ using symmetry ($\sin^4\theta$ has period $\pi$, so $\int_0^{2\pi}=2\int_0^\pi$); then apply the standard $z=e^{i\theta}$ contour-integral trick on $[0,2\pi]$.

Step 1 — Extend to $[0,2\pi]$

$\sin(\theta+\pi)=-\sin\theta$, so $\sin^4(\theta+\pi)=\sin^4\theta$ — period $\pi$. Hence

Step 2 — Contour-integral form of the $[0,2\pi]$ integral

Substitute $z=e^{i\theta}$ (unit circle traversed once counter-clockwise as $\theta:0\to 2\pi$). Then $\sin\theta=\dfrac{z-z^{-1}}{2i}$ and $d\theta=\dfrac{dz}{iz}$. Compute:

(Using $i^{4}=1$.)

Expand $(z-z^{-1})^{4}$ by the binomial theorem:

So

Step 3 — Identify residue at $z=0$

The integrand has a single singularity at $z=0$ inside $|z|=1$. By inspection of the Laurent series, the residue (coefficient of $1/z$) is $6$.

(The other negative powers $-3,\,-5$ are not residue contributions; only the $z^{-1}$ coefficient is.)

Step 4 — Apply residue theorem

Therefore

Answer