← 2015 Paper 2

UPSC 2015 Maths Optional Paper 2 Q2b — Step-by-Step Solution

15 marks · Section A

Riemann integral · Real Analysis · asked 11× in 14 yrs · Read the full method →

Question

Is the function

Riemann integrable? If yes, obtain the value of $\int_0^1 f(x)\,dx$.

Technique

Lebesgue’s criterion (bounded + measure-zero discontinuities) for integrability; partial fractions and telescoping + Basel’s identity for evaluation.

Solution

Step 1 — Understand $f$

$f$ is a step function on $[0,1]$:

• On $(1/2,1]$, $f=1$.
• On $(1/3,1/2]$, $f=1/2$.
• On $(1/4,1/3]$, $f=1/3$.
• …
• On $(1/(n+1),1/n]$, $f=1/n$.
• $f(0)=0$.

$f$ is bounded ($0\le f\le 1$) and has countably many discontinuities — one at each point $1/n$ for $n\ge 2$ (jump from $1/n$ to $1/(n-1)$).

Step 2 — Riemann integrability

A bounded function on $[a,b]$ is Riemann integrable iff the set of its discontinuities has Lebesgue measure zero (Lebesgue’s criterion).

Discontinuities of $f$: $\{1/n:n\ge 2\}\cup\{0\}$ — countable set, measure zero ✓.

So $f$ is Riemann integrable on $[0,1]$.

Step 3 — Compute the integral

Because $f$ is constant $1/n$ on $(1/(n+1),1/n]$, an interval of length $1/n-1/(n+1)=1/(n(n+1))$:

Step 4 — Evaluate $\sum_{n=1}^\infty\dfrac{1}{n^2(n+1)}$

Partial fractions:

$1=An(n+1)+B(n+1)+Cn^2$.

• $n=0$: $1=B$. So $B=1$.
• $n=-1$: $1=C$. So $C=1$.
• Coefficient of $n^2$: $0=A+C\Rightarrow A=-1$.

Check $n=1$: $-1\cdot 1\cdot 2+1\cdot 2+1\cdot 1=-2+2+1=1$ ✓.

So $\dfrac{1}{n^2(n+1)}=-\dfrac{1}{n}+\dfrac{1}{n^2}+\dfrac{1}{n+1}$.

Sum:

The first sum telescopes: $\sum_{n=1}^\infty[\tfrac{1}{n+1}-\tfrac{1}{n}]=-1$ (partial sum to $N$: $\tfrac{1}{N+1}-1\to -1$).

The second is the Basel problem: $\sum_{n=1}^\infty 1/n^2=\pi^2/6$.

So

Answer