← 2023 Paper 1

UPSC 2023 Maths Optional Paper 1 Q1d — Step-by-Step Solution

10 marks · Section A

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

Question

Examine the convergence of the integral $\displaystyle\int_0^1\frac{\log x}{1+x}\,dx$.

Technique

Identify the only singularity ($x=0$); compare the integrand against $\log x$ which is integrable on $(0,1]$.

Solution

The only potential singularity is at $x=0$, where $\log x\to-\infty$. Examine integrability there.

Step 1 — Behaviour near $x=0$.

For $x\in(0,1]$, $\dfrac{1}{1+x}$ is continuous and bounded between $\tfrac{1}{2}$ and $1$. So the integrand $\dfrac{\log x}{1+x}$ behaves like $\log x$ up to a bounded factor.

Step 2 — Compare with $\int_0^1\log x\,dx$.

We know $\int_0^1\log x\,dx=[x\log x-x]_0^1=(0-1)-\lim_{x\to 0^+}(x\log x-x)=-1-(0-0)=-1$ (finite, using $\lim_{x\to 0^+}x\log x=0$).

Since $\log x\le 0$ on $(0,1)$, we have $\dfrac{\log x}{1+x}\ge\log x$ (dividing a negative by something $\ge 1$ makes it less negative). Also $\dfrac{\log x}{1+x}\ge\dfrac{\log x}{1}=\log x$ which is integrable. So by comparison, the original integrand is integrable near $0$.

Step 3 — Behaviour at $x=1$.

At $x=1$: $\log 1=0,\,1+1=2$. Integrand is $0$, no singularity.

Conclusion

Answer