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UPSC 2026 Maths Optional Paper 2 Q4b — Step-by-Step Solution

15 marks · Section A

Improper integrals (analysis perspective) · Real Analysis · asked 4× in 14 yrs · Read the full method →

Question

Prove that the improper integral $\displaystyle\int_{0}^{\infty} \frac{1}{1 + x^2\sin^2 x}\,dx$ is divergent.

Technique

The integrand is continuous and satisfies $0<f(x)\le 1$, so there is no blow-up anywhere and the only question is the behaviour at $\infty$; also $f(k\pi)=1$ for every integer $k$, so $f\not\to0$ and the mass sits in thin spikes at the zeros of $\sin x$. Divergence therefore needs a lower bound, and the natural one is obtained by chopping $[0,\infty)$ at the points $k\pi$: on $[k\pi,(k+1)\pi]$ the substitution $x=k\pi+t$ makes $\sin^2x=\sin^2t$, and replacing $x$ by its largest value on the block decreases the integrand, leaving the standard integral $\int_0^\pi \dfrac{dt}{1+c^2\sin^2t}=\dfrac{\pi}{\sqrt{1+c^2}}$. That produces terms of size $1/k$ — comparison with the harmonic series finishes it.

Solution

Write

Step 1 — What has to be proved, and what kind of improper integral this is.

The denominator satisfies $1+x^{2}\sin^{2}x\ge 1>0$ for all $x$, so $f$ is continuous on the whole of $[0,\infty)$ and

Hence $f$ has no infinite discontinuity: this is an improper integral of the first kind only (infinite range), and on every bounded interval $[0,R]$ the integral $F(R)$ is an ordinary (proper) Riemann integral of a continuous function.

Since $f>0$, the function $R\mapsto F(R)$ is strictly increasing, so

The integral converges iff this limit is finite. To prove divergence it is therefore enough to exhibit a lower bound for $F(R)$ that tends to $+\infty$. (Note that the boundedness $f\le1$ is of no help at all here — an upper bound can never prove divergence. This is the direction in which candidates most often go wrong.)

Observe also that $f(k\pi)=1$ for every integer $k\ge0$: $f$ does not tend to $0$ at infinity. All the “mass” of the integral is in thin spikes of height $1$ sitting at the zeros of $\sin x$, and the whole proof is a bookkeeping of those spikes.

Step 2 — Chop the range at the zeros of $\sin x$.

For $N\in\mathbb{N}$, additivity of the integral gives

Step 3 — Normalise each block by the substitution $x=k\pi+t$.

Put $x=k\pi+t$, $dx=dt$; as $x$ runs over $[k\pi,(k+1)\pi]$, $t$ runs over $[0,\pi]$. This is a $C^{1}$ increasing bijection, so the substitution is legitimate in the Riemann integral. Since

we get, exactly (no inequality yet),

Step 4 — Bound each block from below (mind the direction).

For $t\in[0,\pi]$ we have $k\pi+t\le k\pi+\pi=(k+1)\pi$, hence

Enlarging a positive denominator decreases the fraction, so the integrand in $(2)$ is at least $\bigl(1+(k+1)^{2}\pi^{2}\sin^{2}t\bigr)^{-1}$, and by monotonicity of the integral

Step 5 — The standard integral.

Lemma. For every real $c$, $\displaystyle\int_{0}^{\pi}\frac{dt}{1+c^{2}\sin^{2}t}=\frac{\pi}{\sqrt{1+c^{2}}}$.

Proof. The substitution $t\mapsto \pi-t$ leaves $\sin^{2}t$ unchanged, so the integrand is symmetric about $t=\pi/2$ and

On $[0,\pi/2)$ divide numerator and denominator by $\cos^{2}t>0$:

using $\sec^{2}t=1+\tan^{2}t$. The original integrand is continuous on the closed interval $[0,\pi/2]$, so

On $[0,\pi/2-\varepsilon]$ put $u=\tan t$ (a $C^{1}$ increasing bijection onto $[0,\tan(\pi/2-\varepsilon)]$), $du=\sec^{2}t\,dt$:

Finally

Doubling gives $\pi/\beta=\pi/\sqrt{1+c^{2}}$. $\qquad\blacksquare$

Applying the Lemma with $c=c_{k}=(k+1)\pi$ in $(3)$:

Step 6 — Compare with the harmonic series.

Two elementary estimates turn $(4)$ into something recognisable. Write $m=k+1\ge1$.

(a) $\sqrt{1+m^{2}\pi^{2}}\le 1+m\pi$, since squaring the right side gives $1+2m\pi+m^{2}\pi^{2}\ge 1+m^{2}\pi^{2}$ (both sides positive). Hence

(b) For $m\ge1$ we have $1\le\pi\le m\pi$, so $1+m\pi\le 2m\pi$, and therefore

Combining with $(4)$:

Step 7 — Conclusion: the integral diverges.

Substituting $(5)$ into $(1)$, for every $N\ge1$

where $H_{N}$ is the $N$-th partial sum of the harmonic series, which is divergent ($H_N\to\infty$). Hence $F(N\pi)\to\infty$ as $N\to\infty$.

For an arbitrary $R\ge\pi$ put $N=\lfloor R/\pi\rfloor\ge1$, so $N\pi\le R$; since $f>0$,

Therefore $F(R)\to+\infty$ as $R\to\infty$, i.e.

and the improper integral is divergent. $\qquad\blacksquare$

Step 8 — How fast it diverges (sharpness; one extra line, and it proves you understand the mechanism).

The same block estimate run in the opposite direction gives an upper bound. For $t\in[0,\pi]$, $k\pi+t\ge k\pi$, so by $(2)$ and the Lemma with $c=k\pi$,

while $a_{0}\le\pi$ (as $f\le1$). Hence

and since $H_{N}=\ln N+O(1)$,

The divergence is only logarithmic — the integral does diverge, but as slowly as the harmonic series. This is exactly what $a_{k}\sim 1/k$ predicts.

Step 9 — Alternative elementary route (no standard integral needed).

If the Lemma is not at hand, bound the spikes crudely. For $k\ge1$ let

so the $I_{k}$ are pairwise disjoint subintervals of $(0,\infty)$. For $x\in I_{k}$, using $|\sin x|=|\sin(x-k\pi)|\le|x-k\pi|\le\delta_{k}$ and $x\le k\pi+1$,

so $f(x)>\tfrac13$ on $I_{k}$, and $|I_{k}|=2\delta_{k}=\dfrac{2}{k\pi}$. Therefore

Same conclusion, same comparison series. (It gives a worse constant, which is why Steps 3–6 are the version to write.)

Answer