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UPSC 2018 Maths Optional Paper 2 Q3b — 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

Show by applying the residue theorem that $\displaystyle\int_0^\infty\frac{dx}{(x^2+a^2)^2}=\frac{\pi}{4a^3},\ a>0$.

Technique

Semicircular contour in the upper half-plane; residue at the order-2 pole $z=ia$ via $\frac{d}{dz}[(z-ia)^2 f]$; Jordan-type estimate kills the arc.

Solution

Step 1 — Symmetrize and set up the contour integral

The integrand is even, so

Consider $f(z)=\dfrac{1}{(z^2+a^2)^2}=\dfrac{1}{(z-ia)^2(z+ia)^2}$ and integrate over the contour $C_R=[-R,R]\cup\Gamma_R$, where $\Gamma_R$ is the upper semicircle $|z|=R$. For $a>0$ the poles are at $z=\pm ia$, each of order $2$; only $z=ia$ lies in the upper half-plane (inside $C_R$ for $R>a$).

Step 2 — Vanishing of the arc

On $\Gamma_R$, $|z|=R$ gives $|z^2+a^2|\ge R^2-a^2$, so

since the numerator grows like $R$ while the denominator grows like $R^4$. Hence the arc contributes nothing in the limit.

Step 3 — Residue at the double pole $z=ia$

For a pole of order $2$,

Differentiate:

Evaluate at $z=ia$, where $z+ia=2ia$:

(Using $i^3=-i$.) So $\displaystyle\operatorname*{Res}_{z=ia}f=\frac{1}{4ia^3}=-\frac{i}{4a^3}.$

Step 4 — Apply the residue theorem and take $R\to\infty$

By the residue theorem, for $R>a$,

Letting $R\to\infty$ (the arc term vanishes by Step 2):

Therefore

$\blacksquare$

Verification

• Numerical (SciPy quad, $a=2$): $\displaystyle\int_0^\infty\frac{dx}{(x^2+4)^2}\approx0.0981748$, and $\dfrac{\pi}{4\cdot2^3}=\dfrac{\pi}{32}\approx0.0981748$ — exact agreement. ✓
• Residue computation cross-check: $\operatorname{Res}_{ia}f=-2/(2ia)^3=-2/(-8ia^3)=1/(4ia^3)$; then $2\pi i\cdot\frac{1}{4ia^3}=\frac{\pi}{2a^3}$, halved gives $\frac{\pi}{4a^3}$. ✓
• Beta-function alternative: $\int_0^\infty\frac{dx}{(x^2+a^2)^2}=\frac{1}{a^3}\int_0^\infty\frac{dt}{(t^2+1)^2}=\frac{1}{a^3}\cdot\frac{\pi}{4}=\frac{\pi}{4a^3}$ (sub $x=at$), matching. ✓