Convergence tests for improper integrals — comparison: Worked examples

Before anything, a tiny reminder of the alphabet so no symbol arrives unannounced:

The scenario matrix

Every improper-integral-comparison problem is one (or a blend) of these cells. Read the table first; each example below is tagged with the cell it kills.

Cell	What makes it that case	Weapon of choice	Example
A. Clean "+constant" tail	$f$ is $1/ (polynomial)$ , dominant power obvious	DCT, bound above	Ex 1
B. Messy denominator (no clean inequality)	roots + powers mixed, e.g. $x + x$	LCT	Ex 2
C. Bounded wiggle in numerator	$sin, cos$ trapped in $[- 1, 1]$	DCT after trapping numerator	Ex 3
D. Slow-growing helper ( $ln$ )	$ln x$ beats/loses to powers	DCT, direction from $ln x < x$	Ex 4
E. Exponential benchmark	$e^{- x}$ crushes any power	DCT with $g = e^{- x}$	Ex 5
F. Sign-changing $f$	$f$ not $\geq 0$ ; DCT proof breaks	absolute comparison $	f
G. Degenerate LCT limit $L = 0$ or $L = \infty$	benchmark mismatched	one-way LCT, then fix	Ex 7
H. Type-2 (blow-up at a finite point)	singularity, not $\infty$	DCT near the bad point	Ex 8
I. Word problem / real world	probability, physics tail	translate → compare	Ex 9
J. Exam twist (looks convergent, isn't — or vice versa)	first instinct wrong	forecast then verify	Ex 10

The figure below is the mental picture behind every cell: your curve trapped between benchmarks.

Figure — Convergence tests for improper integrals — comparison

Cell A — clean "+constant" tail

Worked example

Ex 1. Does $\int_{1}^{\infty} \frac{d x}{x ^{3} + 2 x + 7}$ converge? Forecast (guess first!): the denominator grows like $x^{3}$ , and $\int x^{- 3}$ has $p = 3 > 1$ … so guess convergent. Let's prove it.

Step 1 — spot the dominant power. Why this step? For huge $x$ , the $+ 2 x + 7$ is dust next to $x^{3}$ ; the tail behaves like $1/ x^{3}$ . This tells us the benchmark $g = 1/ x^{3}$ .

Step 2 — get the inequality in the right direction. Since $x^{3} + 2 x + 7 > x^{3}$ for all $x \geq 1$ (we only added positive stuff), taking reciprocals flips the inequality: $0 \leq \frac{1}{x ^{3} + 2 x + 7} < \frac{1}{x ^{3}} .$ Why bound above? We want to prove convergence, and "small $\leq$ convergent $\Rightarrow$ convergent" is the useful direction (mnemonic: squeeze to please).

Step 3 — invoke the benchmark. $\int_{1}^{\infty} x^{- 3} d x$ converges because $p = 3 > 1$ . By DCT, our integral converges. ✅

Verify: sanity-check the benchmark value: $\int_{1}^{\infty} x^{- 3} d x = [\frac{x ^{- 2}}{- 2}]_{1}^{\infty} = \frac{1}{2}$ , a finite number. Our area is below $\frac{1}{2}$ , so certainly finite. Consistent.

Cell B — messy denominator, LCT

Worked example

Ex 2. Does $\int_{1}^{\infty} \frac{d x}{2 x + 3 x}$ converge? Forecast: dominant term downstairs is $3 x$ (grows faster than $x$ ), so $f \sim \frac{1}{3 x}$ , which behaves like $1/ x$ ( $p = 1$ ) — guess divergent.

Step 1 — pick $g = 1/ x$ . Why LCT and not DCT? Because $2 x + 3 x$ isn't a single clean power; forcing a DCT inequality is fiddly. LCT lets us just match dominant terms.

Step 2 — compute the ratio limit. $L = lim_{x \to \infty} \frac{1/ ( 2 x + 3 x )}{1/ x} = lim_{x \to \infty} \frac{x}{2 x + 3 x} .$ Step 3 — divide top and bottom by $x$ . Why this step? It exposes the limit by killing the growing terms: $L = lim_{x \to \infty} \frac{1}{2 x ^{- 1/2} + 3} = \frac{1}{0 + 3} = \frac{1}{3} .$ Step 4 — read the verdict. $0 < L = \frac{1}{3} < \infty$ , so $f$ and $g = 1/ x$ share the same fate. Since $\int_{1}^{\infty} x^{- 1} d x$ diverges ( $p = 1$ ), our integral diverges. ❌

Verify: the constant $\frac{1}{3}$ is finite and positive — exactly the regime where LCT's clean two-way conclusion is valid. Good.

Cell C — bounded wiggle

Cell D — slow-growing $ln$

Worked example

Ex 4. Does $\int_{2}^{\infty} \frac{ln x}{x} d x$ converge? Forecast: careful — $ln x$ grows, so $\frac{l n x}{x}$ is bigger than $\frac{1}{x}$ . Since $\int 1/ x$ diverges, and we're above it… guess divergent.

Step 1 — get the inequality. For $x \geq e$ (certainly for $x \geq 3$ ), $ln x \geq 1$ , so $\frac{l n x}{x} \geq \frac{1}{x} .$ Why bound below? We suspect divergence, and "big $\geq$ divergent $\Rightarrow$ divergent" is the useful direction (push to die).

Step 2 — handle the small window $[2, 3]$ . Why? On $[2, 3]$ the function is continuous and finite, contributing a finite bounded chunk; convergence is decided only by the infinite tail, so it's safe to start comparing from $x = 3$ .

Step 3 — benchmark. $\int_{3}^{\infty} x^{- 1} d x$ diverges, and our function lies above it, so by DCT diverges. ❌

Verify: compute directly, $\int_{2}^{t} \frac{l n x}{x} d x = [\frac{( l n x ) ^{2}}{2}]_{2}^{t} = \frac{( l n t ) ^{2} - ( l n 2 ) ^{2}}{2} \to \infty$ . Divergence confirmed exactly.

Cell E — exponential benchmark

Worked example

Ex 5. Does $\int_{0}^{\infty} x^{2} e^{- x} d x$ converge? Forecast: exponentials crush polynomials, so $x^{2} e^{- x} \to 0$ fast — guess convergent.

Step 1 — pick $g = e^{- x /2}$ . Why this benchmark and not a power? A power $x^{- p}$ can't dominate $x^{2} e^{- x}$ — but $e^{- x}$ decays faster than any power, so a slightly weaker exponential $e^{- x /2}$ still swallows the polynomial factor and is trivially integrable.

Step 2 — build the inequality. For large $x$ , $x^{2} e^{- x} = x^{2} e^{- x /2} \cdot e^{- x /2}$ , and $x^{2} e^{- x /2} \to 0$ , so it is $\leq 1$ once $x \geq X_{0}$ . Then $0 \leq x^{2} e^{- x} \leq e^{- x /2} (x \geq X_{0}) .$ Step 3 — benchmark. $\int_{0}^{\infty} e^{- x /2} d x = 2$ converges. The finite window $[0, X_{0}]$ contributes a bounded, continuous chunk. By DCT, converges. ✅

Verify: this is the Gamma integral $\int_{0}^{\infty} x^{2} e^{- x} d x = Γ (3) = 2! = 2$ , a clean finite number. Convergence confirmed with exact value $2$ .

Cell F — sign-changing $f$ (absolute comparison)

Worked example

Ex 6. Does $\int_{1}^{\infty} \frac{sin x}{x ^{2}} d x$ converge? Forecast: $sin x$ is negative half the time, so DCT's " $f \geq 0$ " rule is broken. But $∣ sin x ∣ \leq 1$ , so the size is $\leq 1/ x^{2}$ — guess (absolutely) convergent.

Step 1 — switch to absolute value. Why? DCT's proof needs an increasing area function, which requires $f \geq 0$ . With sign changes, compare $∣ f ∣$ instead (see Absolute vs conditional convergence): $\frac{s i n x}{x ^{2}} = \frac{∣ s i n x ∣}{x ^{2}} \leq \frac{1}{x ^{2}} .$ Step 2 — benchmark on $∣ f ∣$ . $\int_{1}^{\infty} x^{- 2} d x$ converges, so by DCT $\int_{1}^{\infty} \frac{s i n x}{x ^{2}} d x$ converges.

Step 3 — upgrade to the original. Why does absolute convergence give ordinary convergence? Because "absolutely convergent $\Rightarrow$ convergent" — the positive and negative pieces are each finite, so their difference is finite. Thus $\int_{1}^{\infty} \frac{s i n x}{x ^{2}} d x$ converges. ✅

Verify: the tail is bounded: $\int_{1}^{\infty} \frac{s i n x}{x ^{2}} d x \leq \int_{1}^{\infty} \frac{1}{x ^{2}} d x = 1$ . Finite bound confirms convergence.

Cell G — degenerate LCT limit

Worked example

Ex 7. Does $\int_{1}^{\infty} \frac{ln x}{x ^{3}} d x$ converge, tested via a naive LCT? Forecast: $ln x$ grows so slowly it barely matters against $x^{3}$ — guess convergent, but watch the LCT trap.

Step 1 — try the naive benchmark $g = 1/ x^{3}$ . $L = lim_{x \to \infty} \frac{( l n x ) / x ^{3}}{1/ x ^{3}} = lim_{x \to \infty} ln x = \infty.$ Why is this a problem? LCT's clean two-way rule needs $0 < L < \infty$ . Here $L = \infty$ , so we only get the one-way implication: if $g$ diverged, we'd learn nothing useful. Since $g$ converges, $L = \infty$ gives no conclusion — dead end.

Step 2 — fix by choosing a slightly weaker benchmark $g = 1/ x^{2}$ . Why $x^{- 2}$ ? We want to absorb the slow $ln x$ into spare decay. Since $ln x \leq x^{1/2}$ for large $x$ (any power of $x$ beats $ln x$ ): $\frac{l n x}{x ^{3}} \leq \frac{x ^{1/2}}{x ^{3}} = \frac{1}{x ^{5/2}} .$ Step 3 — benchmark. $\int_{1}^{\infty} x^{- 5/2} d x$ converges ( $p = \frac{5}{2} > 1$ ). By DCT, converges. ✅

Verify: direct integration by parts gives $\int_{1}^{\infty} \frac{l n x}{x ^{3}} d x = \frac{1}{4}$ (finite). Convergence confirmed; the exact value is $\frac{1}{4}$ .

Cell H — Type-2 blow-up at a finite point

Worked example

Ex 8. Does $\int_{0}^{1} \frac{d x}{x + x ^{2}}$ converge? Forecast: the trouble is at $x = 0$ (see Improper integrals — infinite discontinuities (type 2)), where $x \to 0$ makes the integrand blow up. Near $0$ , $x$ dominates $x^{2}$ , so $f \sim 1/ x = x^{- 1/2}$ . The p-test near a singularity says $\int_{0}^{1} x^{- p} d x$ converges iff $p < 1$ . Here $p = \frac{1}{2} < 1$ — guess convergent.

Step 1 — locate the singularity. Why? Comparison must happen where the integrand misbehaves — here $x \to 0^{+}$ , not $\infty$ .

Step 2 — inequality near $0$ . For $0 < x \leq 1$ , $x + x^{2} > x$ , so $0 \leq \frac{1}{x + x ^{2}} < \frac{1}{x} .$ Why above? Proving convergence → squeeze under a convergent benchmark.

Step 3 — benchmark for a type-2 endpoint. $\int_{0}^{1} x^{- 1/2} d x = [2 x]_{0}^{1} = 2$ , finite (converges since $p = \frac{1}{2} < 1$ ). By DCT, converges. ✅

Verify: benchmark value $2$ is finite; our integrand is smaller, so its integral is below $2$ . Consistent.

Cell I — real-world word problem

Cell J — exam twist (instinct is wrong)

Worked example

Ex 10. Does $\int_{2}^{\infty} \frac{d x}{x ln x}$ converge? Forecast (the trap): it looks like $1/ x^{something > 1}$ since there's an extra $ln x$ downstairs pushing it smaller — many students guess convergent. Be suspicious.

Step 1 — why comparison to a power fails cleanly. Try $g = 1/ x$ : $L = lim \frac{1/ ( x l n x )}{1/ x} = lim \frac{1}{l n x} = 0$ . LCT with $L = 0$ only gives one-way info and $g$ diverges, so no conclusion — the twist is real.

Step 2 — integrate directly with substitution. Why? When comparison stalls, evaluate. Let $u = ln x$ , so $d u = \frac{d x}{x}$ : $\int \frac{d x}{x l n x} = \int \frac{d u}{u} = ln ∣ u ∣ = ln (ln x) .$ Step 3 — take the limit. $\int_{2}^{t} \frac{d x}{x l n x} = ln (ln t) - ln (ln 2) t \to \infty \infty.$ Because $ln (ln t) \to \infty$ (slowly, but surely), the integral diverges. ❌ The extra $ln x$ was not enough to tame it — instinct busted.

Verify: $ln (ln t)$ is unbounded as $t \to \infty$ , so the antiderivative has no finite limit. Divergence confirmed exactly.

Flashcards

For

\int_{1}^{\infty} \frac{d x}{x ^{3} + 2 x + 7}

, which benchmark and verdict?

g = 1/ x^{3}

, bound above, converges (

p = 3 > 1

LCT on

\int_{1}^{\infty} \frac{d x}{2 x + 3 x}

gives what

L

and verdict?

L = 1/3

, same fate as

1/ x

, diverges.

Why can't

\int_{2}^{\infty} \frac{d x}{x l n x}

be settled by LCT with

g = 1/ x

L = 0

(one-way only); integrate: antiderivative

ln (ln x) \to \infty

, diverges.

Near a type-2 singularity at

0

, when does

\int_{0}^{1} x^{- p} d x

converge?

When

p < 1

(opposite of the

\infty

tail rule).

How do you handle

\int_{1}^{\infty} \frac{s i n x}{x ^{2}} d x

Compare

∣ f ∣ \leq 1/ x^{2}

; absolutely convergent ⇒ convergent.

Why use

e^{- x /2}

to dominate

x^{2} e^{- x}

x^{2} e^{- x /2} \to 0

, so the polynomial gets absorbed and

e^{- x /2}

is integrable.

Value of

\int_{0}^{\infty} x^{2} e^{- x} d x

Γ (3) = 2! = 2

Value of

\int_{1}^{\infty} \frac{d x}{x ^{2} + x}

ln 2 \approx 0.693

(converges, finite energy).

Connections

4.2.12 Convergence tests for improper integrals — comparison (Hinglish)
The p-integral and p-series
Improper integrals — infinite discontinuities (type 2)
Comparison test for infinite series
Absolute vs conditional convergence
Monotone Convergence Theorem
Integral Test for series