← 2019 Paper 2

UPSC 2019 Maths Optional Paper 2 Q1c — Step-by-Step Solution

10 marks · Section A

Improper integrals (unbounded interval/integrand) · Calculus · asked 5× in 14 yrs · Read the full method →

Question

Evaluate

Technique

Differentiation under the integral sign (Feynman) in the parameter $a$, partial fractions, then integrate $I'(a)$ from $a=0$ using $I(0)=0$.

Solution

Step 1 — Introduce the parameter and differentiate

Let

$I(0)=0$. The integrand is dominated near $x=0$ by $\frac{ax}{x}=a$ (bounded) and decays like $\frac{\pi/2}{x^3}$ at infinity, so the integral converges and differentiation under the integral sign is justified. With $\dfrac{\partial}{\partial a}\arctan(ax)=\dfrac{x}{1+a^2x^2}$,

Step 2 — Partial fractions ($a\ne 1$)

For $a^2\ne 1$,

(Check: $\frac{a^2(1+x^2)-(1+a^2x^2)}{(a^2-1)(1+x^2)(1+a^2x^2)}=\frac{a^2-1}{(a^2-1)(\cdots)}$. ✓)

Step 3 — Integrate term by term

Using $\displaystyle\int_0^\infty\frac{dx}{1+x^2}=\frac\pi2$ and $\displaystyle\int_0^\infty\frac{dx}{1+a^2x^2}=\frac{1}{a}\cdot\frac\pi2$ (sub $u=ax$, $a>0$):

The factor $(a-1)$ cancels, so although the partial fraction split required $a\ne1$, the resulting $I'(a)=\frac{\pi}{2(a+1)}$ is smooth across $a=1$.

Step 4 — Integrate in $a$ with $I(0)=0$

Answer