← 2024 Paper 2

UPSC 2024 Maths Optional Paper 2 Q1c — Step-by-Step Solution

10 marks · Section A

Improper integrals (analysis perspective) · Real Analysis · asked 3× in 13 yrs · Read the full method →

Question

Test the convergence of $\displaystyle\int_0^2\frac{\log x}{\sqrt{2-x}}\,dx$.

Technique

Split the integral and check each endpoint separately; at $x=0$ use $\lim_{x\to 0^+}x\log x=0$; at $x=2$ use limit comparison with $1/\sqrt{2-x}$.

Solution

The integrand has singularities at both endpoints. Examine each.

Endpoint $x=0$

Near $x=0$, $\sqrt{2-x}$ is bounded (between $1$ and $\sqrt{2}$), so $1/\sqrt{2-x}$ is bounded. The integrand behaves like $\log x$, which is integrable near $0$:

and $\lim_{x\to 0^+}x\log x=0$, so this is finite. No issue at $x=0$.

Endpoint $x=2$

Near $x=2$, $\log x\to\log 2$ (finite and non-zero). The dominant singularity is $1/\sqrt{2-x}$:

which converges (substituting $u=2-x$: $\int_0\epsilon u^{-1/2}du$ converges since $-1/2>-1$).

By limit comparison: $\dfrac{\log x/\sqrt{2-x}}{1/\sqrt{2-x}}=\log x\to\log 2\ne 0$ as $x\to 2$. The two integrals converge or diverge together, and the comparison integral converges. No issue at $x=2$.

Answer