Improper integrals (analysis perspective)

At a Glance

• Frequency: 3 sub-parts across 3 of 13 years (2014, 2022, 2024)
• Priority tier: T3
• Marks (count): 10 (3)
• Average solve time: ~5 min
• Difficulty mix: easy 3
• Section: A | Dominant type: proof

Why This Chapter Matters

Improper integrals appear in every UPSC Paper 2 Section A diet — three times in the last six years alone — and the question is always the same shape: identify the singularities, bound the integrand, invoke the comparison test, and state convergence. Mastering this archetype is essentially free marks once the two-step pattern is internalised. The comparison and limit-comparison tests are also the gateway to the harder series-convergence problems in RA-17.

Minimum Theory

Definition (Type I and Type II). An improper integral of Type I is $\int_a^\infty f(x)\,dx := \lim_{t\to\infty}\int_a^t f(x)\,dx$ when the limit exists and is finite. An improper integral of Type II arises when $f$ is unbounded near an endpoint: $\int_a^b f(x)\,dx := \lim_{t\to b^-}\int_a^t f(x)\,dx$ (or $\lim_{t\to a^+}$). If the integral has singularities at both endpoints, split the interval and require both parts to converge independently.

Comparison Test. Suppose $0 \le f(x) \le g(x)$ for all $x \ge a$. Then: $\int_a^\infty g$ converges $\Rightarrow$ $\int_a^\infty f$ converges; $\int_a^\infty f$ diverges $\Rightarrow$ $\int_a^\infty g$ diverges. The same test applies for Type II integrals near a singular endpoint.

Limit Comparison Test. If $f, g \ge 0$ and $\lim_{x\to c} f(x)/g(x) = L$ with $0 < L < \infty$, then $\int f$ and $\int g$ converge or diverge together. p-test benchmarks: $\int_1^\infty x^{-p}\,dx$ converges iff $p > 1$; $\int_0^1 x^{-p}\,dx$ converges iff $p < 1$.

Absolute convergence. If $\int_a^\infty |f(x)|\,dx$ converges, so does $\int_a^\infty f(x)\,dx$. This lets you handle oscillating integrands (e.g.\ $\cos x$) by bounding $|\cos x| \le 1$ and comparing the absolute value.

Question Archetypes

improper-convergence (3 question(s); 2014, 2022, 2024)

Recognition Cues — The question says “test convergence” or “examine convergence”. Either the upper limit is $\infty$ (Type I) or the integrand blows up at an endpoint (Type II: look for $\log x$ near $0$, $1/\sqrt{b-x}$ near $x=b$, etc.). Never just substitute limits blindly — first locate every singularity.

Solution Template

1. Locate singularities. Identify whether the problem is Type I ($\infty$ limit) or Type II (unbounded integrand), or both. For a bounded interval, check both endpoints.
2. Split if necessary. If two singularities exist (e.g. 2024’s integral over $[0,2]$), split and treat each piece independently.
3. Choose a comparison or benchmark. Use $1/x^p$ as the standard comparison function. For oscillating integrands, bound $|\cos x| \le 1$ or $|\sin x| \le 1$ and work with absolute values.
4. State and apply the test. Write $0 \le f(x) \le g(x)$ (or compute the limit ratio), cite the comparison test, and conclude for $g$ first (using a known integral or p-test).
5. Conclude. State whether the integral converges or diverges. A brief bound on the value (if easy) earns bonus marks.

Worked Example

2014 Paper 2, 2014-P2-Q1b (10 marks)

Test the convergence of the improper integral $\displaystyle\int_1^\infty\frac{dx}{x^2(1+e^{-x})}$.

Locate singularity. The upper limit is $\infty$ (Type I). At $x=1$ the integrand is finite and positive, so no issue at the lower endpoint.

Bound the integrand. For $x \ge 1$, we have $0 < e^{-x} \le e^{-1} < 1$, so $1 < 1+e^{-x} \le 1+e^{-1}$. Taking reciprocals:

$\frac{1}{1+e^{-1}} \le \frac{1}{1+e^{-x}} < 1 \quad (x \ge 1).$

Therefore

$0 < \frac{1}{x^2(1+e^{-x})} < \frac{1}{x^2}.$

Apply comparison. The integral $\int_1^\infty x^{-2}\,dx = 1$ converges (p-test with $p=2>1$). By the comparison test, the smaller positive integrand also gives a convergent integral.

Bonus bound. The lower bound gives $\int_1^\infty f\,dx \ge e/(e+1) \approx 0.731$, so the integral lies in $(e/(e+1),\,1)$.

$\boxed{\text{The integral converges.}}$

2022 Paper 2, 2022-P2-Q1c (10 marks)

Test the convergence of $\displaystyle\int_0^\infty\dfrac{\cos x}{1+x^2}\,dx$.

Locate singularity. Type I ($\infty$ limit); the integrand is bounded and continuous on $[0,\infty)$.

Use absolute convergence. Since $|\cos x| \le 1$:

$\left|\frac{\cos x}{1+x^2}\right| \le \frac{1}{1+x^2}.$

Evaluate the bounding integral.

$\int_0^\infty \frac{dx}{1+x^2} = [\arctan x]_0^\infty = \frac{\pi}{2} < \infty.$

Conclude. By comparison, $\int_0^\infty |\cos x|/(1+x^2)\,dx \le \pi/2$, so the integral converges absolutely, hence converges.

$\boxed{\int_0^\infty \frac{\cos x}{1+x^2}\,dx \text{ converges (absolutely; exact value } \pi/(2e) \text{).}}$

2024 Paper 2, 2024-P2-Q1c (10 marks)

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

Locate singularities. The integrand is potentially singular at both endpoints: $\log x \to -\infty$ as $x \to 0^+$, and $1/\sqrt{2-x} \to \infty$ as $x \to 2^-$. Split:

$\int_0^2 = \int_0^1 + \int_1^2.$

Near $x = 0$. On $(0,1]$, the factor $1/\sqrt{2-x}$ is bounded (between $1/\sqrt{2}$ and $1$). So the behaviour is governed by $|\log x|$ alone:

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

Since $\log x$ is integrable near $0$ and $1/\sqrt{2-x}$ is bounded there, the product is integrable. No issue at $x=0$.

Near $x = 2$. As $x \to 2^-$, $\log x \to \log 2$ (finite and nonzero). By limit comparison with $g(x) = 1/\sqrt{2-x}$:

$\lim_{x\to 2^-} \frac{|\log x|/\sqrt{2-x}}{1/\sqrt{2-x}} = \log 2 \in (0,\infty).$

Since $\int_1^2 1/\sqrt{2-x}\,dx = [-2\sqrt{2-x}]_1^2 = 2$ converges, the limit comparison gives convergence of $\int_1^2 |\log x|/\sqrt{2-x}\,dx$ as well.

Conclude. Both pieces converge, so

$\boxed{\int_0^2 \frac{\log x}{\sqrt{2-x}}\,dx \text{ converges (value} \approx -2.55\text{).}}$

Common Traps

• Missing the lower endpoint. In the 2024 integral, $\log x$ is the “obvious” singularity but $1/\sqrt{2-x}$ is bounded near $0$ — the piece near $x=0$ is actually fine. Students who only check one endpoint lose half the marks.
• Wrong direction of comparison. The 2014 integrand is smaller than $1/x^2$ (because $1+e^{-x} > 1$). Convergence of the larger function implies convergence of the smaller, not the other way.
• Forgetting absolute convergence for oscillating integrands. In the 2022 integral, $\cos x$ oscillates. The correct route is to bound $|\cos x| \le 1$, prove absolute convergence, and deduce convergence. Do not try to integrate $\cos x/(1+x^2)$ directly.
• Confusing p-test ranges. $\int_1^\infty x^{-p}$ converges iff $p > 1$; $\int_0^1 x^{-p}$ converges iff $p < 1$. The exponent threshold flips at the two standard limits.

Marks-Aware Writing

A 10-mark answer requires: (1) identify singularity type, (2) state and establish the bound $f(x) \le g(x)$ with explicit inequalities, (3) compute or cite the comparison integral, (4) invoke the comparison test by name, (5) state the conclusion. Writing down the exact inequality and referencing the p-test or known antiderivative are the two steps students most often skip.

Practice Set

• 2019-P2-Q4c (15 m) — — Hint: likely a Type II integral; check both endpoints and split if needed.