← 2019 Paper 2

UPSC 2019 Maths Optional Paper 2 Q4c — Step-by-Step Solution

15 marks · Section A

Improper integrals (unbounded interval/integrand) · Calculus · asked 5× in 13 yrs · Read the full method →

Question

Discuss the convergence of $\displaystyle\int_1^2 \frac{\sqrt x}{\ln x}\,dx$.

Technique

Identify the endpoint singularity at $x=1$; Taylor $\ln x\sim x-1$ gives a $1/(x-1)$ singularity; limit comparison with the divergent $\int dx/(x-1)$.

Solution

Step 1 — Locate the singularity

The integrand $\dfrac{\sqrt x}{\ln x}$ is continuous on $(1,2]$. The only trouble is at the lower limit $x=1$, where $\ln x\to0^+$, making the integrand $\to+\infty$. So this is an improper integral of the second kind (singularity at $x=1$); convergence is decided by the behaviour as $x\to1^+$.

Step 2 — Asymptotics near $x=1$

Put $x=1+t$, $t\to0^+$. Then $\ln x=\ln(1+t)=t-\tfrac{t^2}{2}+\cdots\sim t$, and $\sqrt x=\sqrt{1+t}\to1$. Hence

Step 3 — Comparison with $\displaystyle\int \frac{dx}{x-1}$

The comparison/limit test: since

the two integrands have the same convergence behaviour at $x=1$. But the reference integral diverges:

Therefore, by the limit comparison test, the given integral diverges.

Answer