Convergence tests for improper integrals — comparison: Worked examples

4.2.12 · D3 · Maths › Calculus II — Integration › Convergence tests for improper integrals — comparison

Shuru karne se pehle, symbols ka ek chhota sa reminder taaki koi symbol achanak aa ke surprise na kare:

Scenario matrix

Har improper-integral-comparison problem in cells mein se ek (ya in ka blend) hai. Pehle table padho; neeche har example us cell ke saath tagged hai jo woh handle karta hai.

Cell	Ise us case ki kya cheez banati hai	Tool ka choice	Example
A. Clean "+constant" tail	$f$ ek $1/ (polynomial)$ hai, dominant power obvious hai	DCT, upar se bound karo	Ex 1
B. Messy denominator (clean inequality nahi)	roots + powers mixed, e.g. $x + x$	LCT	Ex 2
C. Numerator mein bounded wiggle	$sin, cos$ $[- 1, 1]$ mein trapped hain	Numerator ko trap karke DCT	Ex 3
D. Slow-growing helper ( $ln$ )	$ln x$ powers se aage/peeche rehta hai	DCT, direction $ln x < x$ se	Ex 4
E. Exponential benchmark	$e^{- x}$ kisi bhi power ko crush kar deta hai	$g = e^{- x}$ ke saath DCT	Ex 5
F. Sign-changing $f$	$f$ $\geq 0$ nahi hai; DCT proof toot jaata hai	absolute comparison $	f
G. Degenerate LCT limit $L = 0$ ya $L = \infty$	benchmark mismatched hai	one-way LCT, phir fix karo	Ex 7
H. Type-2 (finite point par blow-up)	singularity hai, $\infty$ nahi	bad point ke paas DCT	Ex 8
I. Word problem / real world	probability, physics tail	translate karo → compare karo	Ex 9
J. Exam twist (convergent lagta hai, nahi hota — ya ulta)	pehli instinct galat	forecast karo phir verify karo	Ex 10

Neeche di gayi figure har cell ke peeche ki mental picture hai: tumhara curve benchmarks ke beech trapped hai.

Figure — Convergence tests for improper integrals — comparison

Cell A — clean "+constant" tail

Worked example

Ex 1. Kya $\int_{1}^{\infty} \frac{d x}{x ^{3} + 2 x + 7}$ converge karta hai? Forecast (pehle guess karo!): denominator $x^{3}$ ki tarah badhta hai, aur $\int x^{- 3}$ mein $p = 3 > 1$ hai… toh guess convergent. Isko prove karte hain.

Step 1 — dominant power pehchano. Yeh step kyun? Bade $x$ ke liye, $+ 2 x + 7$ ka $x^{3}$ ke saamne koi wajood nahi; tail $1/ x^{3}$ ki tarah behave karta hai. Isse hume benchmark $g = 1/ x^{3}$ milta hai.

Step 2 — inequality sahi direction mein laao. Kyunki $x^{3} + 2 x + 7 > x^{3}$ sabhi $x \geq 1$ ke liye (humne sirf positive cheezein joodi hain), reciprocal lene se inequality flip hoti hai: $0 \leq \frac{1}{x ^{3} + 2 x + 7} < \frac{1}{x ^{3}} .$ Upar se bound kyun karte hain? Hum convergence prove karna chahte hain, aur "chhota $\leq$ convergent $\Rightarrow$ convergent" woh useful direction hai (mnemonic: squeeze to please).

Step 3 — benchmark invoke karo. $\int_{1}^{\infty} x^{- 3} d x$ converge karta hai kyunki $p = 3 > 1$ . DCT se, hamara integral converges. ✅

Verify: benchmark value sanity-check karo: $\int_{1}^{\infty} x^{- 3} d x = [\frac{x ^{- 2}}{- 2}]_{1}^{\infty} = \frac{1}{2}$ , ek finite number. Hamara area $\frac{1}{2}$ se neeche hai, toh definitely finite hai. Consistent hai.

Cell B — messy denominator, LCT

Worked example

Ex 2. Kya $\int_{1}^{\infty} \frac{d x}{2 x + 3 x}$ converge karta hai? Forecast: neeche dominant term $3 x$ hai ( $x$ se tez badhta hai), toh $f \sim \frac{1}{3 x}$ , jo $1/ x$ ( $p = 1$ ) ki tarah behave karta hai — guess divergent.

Step 1 — $g = 1/ x$ choose karo. DCT nahi, LCT kyun? Kyunki $2 x + 3 x$ ek clean single power nahi hai; DCT inequality force karna jhanjhat bhari baat hai. LCT humein sirf dominant terms match karne deta hai.

Step 2 — ratio limit compute karo. $L = lim_{x \to \infty} \frac{1/ ( 2 x + 3 x )}{1/ x} = lim_{x \to \infty} \frac{x}{2 x + 3 x} .$ Step 3 — upar aur neeche $x$ se divide karo. Yeh step kyun? Yeh growing terms ko khatam karke limit expose karta hai: $L = lim_{x \to \infty} \frac{1}{2 x ^{- 1/2} + 3} = \frac{1}{0 + 3} = \frac{1}{3} .$ Step 4 — verdict padho. $0 < L = \frac{1}{3} < \infty$ , toh $f$ aur $g = 1/ x$ ka same fate hai. Kyunki $\int_{1}^{\infty} x^{- 1} d x$ diverge karta hai ( $p = 1$ ), hamara integral diverges. ❌

Verify: constant $\frac{1}{3}$ finite aur positive hai — exactly woh regime jahan LCT ka clean two-way conclusion valid hai. Achha hai.

Cell C — bounded wiggle

Worked example

Ex 3. Kya $\int_{1}^{\infty} \frac{3 + cos x}{x ^{3/2}} d x$ converge karta hai? Forecast: $cos x$ sirf $- 1$ aur $1$ ke beech wiggle karta hai; "size" $1/ x^{3/2}$ se set hoti hai, aur $p = \frac{3}{2} > 1$ hai — guess convergent.

Step 1 — numerator ko trap karo. Kyun? $cos x$ ki koi limit nahi hai, toh hum ratio nahi le sakte; balki hum ise bound karte hain. Kyunki $- 1 \leq cos x \leq 1$ : $2 \leq 3 + cos x \leq 4.$ Step 2 — positive $x^{3/2}$ se divide karo (inequality direction preserve hoti hai): $0 \leq \frac{3 + c o s x}{x ^{3/2}} \leq \frac{4}{x ^{3/2}} .$ Upar se bound kyun? Convergence prove karna → ek convergent benchmark ke neeche squeeze karo.

Step 3 — benchmark. $\int_{1}^{\infty} 4 x^{- 3/2} d x$ converge karta hai ( $p = \frac{3}{2} > 1$ ; constant $4$ koi fark nahi padta). DCT se, converges. ✅

Verify: $\int_{1}^{\infty} 4 x^{- 3/2} d x = 4 \cdot [\frac{x ^{- 1/2}}{- 1/2}]_{1}^{\infty} = 4 \cdot 2 = 8$ , finite. Hamara area $8$ se neeche baithta hai. Consistent hai.

Cell D — slow-growing $ln$

Worked example

Ex 4. Kya $\int_{2}^{\infty} \frac{ln x}{x} d x$ converge karta hai? Forecast: dhyan se — $ln x$ badhta hai, toh $\frac{l n x}{x}$ , $\frac{1}{x}$ se bada hai. Kyunki $\int 1/ x$ diverge karta hai, aur hum usse upar hain… guess divergent.

Step 1 — inequality laao. $x \geq e$ ke liye (definitely $x \geq 3$ ke liye), $ln x \geq 1$ , toh $\frac{l n x}{x} \geq \frac{1}{x} .$ Neeche se bound kyun? Hume divergence ka shakk hai, aur "bada $\geq$ divergent $\Rightarrow$ divergent" woh useful direction hai (push to die).

Step 2 — chhoti window $[2, 3]$ handle karo. Kyun? $[2, 3]$ par function continuous aur finite hai, jo ek finite bounded chunk contribute karta hai; convergence sirf infinite tail se decide hoti hai, toh $x = 3$ se compare karna shuru karna safe hai.

Step 3 — benchmark. $\int_{3}^{\infty} x^{- 1} d x$ diverge karta hai, aur hamara function uske upar lie karta hai, toh DCT se diverges. ❌

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

Cell E — exponential benchmark

Worked example

Ex 5. Kya $\int_{0}^{\infty} x^{2} e^{- x} d x$ converge karta hai? Forecast: exponentials polynomials ko crush kar dete hain, toh $x^{2} e^{- x} \to 0$ tez — guess convergent.

Step 1 — $g = e^{- x /2}$ choose karo. Yeh benchmark kyun, power kyun nahi? Ek power $x^{- p}$ $x^{2} e^{- x}$ ko dominate nahi kar sakta — lekin $e^{- x}$ kisi bhi power se tez decay karta hai, toh thoda weakly exponential $e^{- x /2}$ polynomial factor ko bhi swallow kar leta hai aur trivially integrable hai.

Step 2 — inequality banao. Bade $x$ ke liye, $x^{2} e^{- x} = x^{2} e^{- x /2} \cdot e^{- x /2}$ , aur $x^{2} e^{- x /2} \to 0$ , toh yeh $\leq 1$ hoga jab $x \geq X_{0}$ . Phir $0 \leq x^{2} e^{- x} \leq e^{- x /2} (x \geq X_{0}) .$ Step 3 — benchmark. $\int_{0}^{\infty} e^{- x /2} d x = 2$ converge karta hai. Finite window $[0, X_{0}]$ ek bounded, continuous chunk contribute karta hai. DCT se, converges. ✅

Verify: yeh Gamma integral hai $\int_{0}^{\infty} x^{2} e^{- x} d x = Γ (3) = 2! = 2$ , ek clean finite number. Convergence exact value $2$ ke saath confirm ho gayi.

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

Worked example

Ex 6. Kya $\int_{1}^{\infty} \frac{sin x}{x ^{2}} d x$ converge karta hai? Forecast: $sin x$ aadhe time negative hai, toh DCT ka " $f \geq 0$ " rule toot jaata hai. Lekin $∣ sin x ∣ \leq 1$ , toh size $\leq 1/ x^{2}$ hai — guess (absolutely) convergent.

Step 1 — absolute value par switch karo. Kyun? DCT ke proof ko ek increasing area function chahiye, jiske liye $f \geq 0$ zaroori hai. Sign changes ke saath, $∣ f ∣$ compare karo (dekho 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 — $∣ f ∣$ par benchmark. $\int_{1}^{\infty} x^{- 2} d x$ converge karta hai, toh DCT se $\int_{1}^{\infty} \frac{s i n x}{x ^{2}} d x$ converges.

Step 3 — original par upgrade karo. Absolute convergence ordinary convergence kyun deta hai? Kyunki "absolutely convergent $\Rightarrow$ convergent" — positive aur negative pieces dono finite hain, toh unka difference bhi finite hai. Iss tarah $\int_{1}^{\infty} \frac{s i n x}{x ^{2}} d x$ converges. ✅

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

Cell G — degenerate LCT limit

Worked example

Ex 7. Kya $\int_{1}^{\infty} \frac{ln x}{x ^{3}} d x$ converge karta hai, naive LCT se test karke? Forecast: $ln x$ itna dhire badhta hai ki $x^{3}$ ke saamne barely matter karta hai — guess convergent, lekin LCT ka trap dekho.

Step 1 — naive benchmark $g = 1/ x^{3}$ try karo. $L = lim_{x \to \infty} \frac{( l n x ) / x ^{3}}{1/ x ^{3}} = lim_{x \to \infty} ln x = \infty.$ Yeh problem kyun hai? LCT ka clean two-way rule $0 < L < \infty$ chahiye. Yahan $L = \infty$ hai, toh hume sirf one-way implication milti hai: agar $g$ diverge karta, toh hum kuch useful nahi seekh paate. Kyunki $g$ converge karta hai, $L = \infty$ koi conclusion nahi deta — dead end.

Step 2 — thoda weaker benchmark $g = 1/ x^{2}$ choose karke fix karo. $x^{- 2}$ kyun? Hum slow $ln x$ ko extra decay mein absorb karna chahte hain. Kyunki $ln x \leq x^{1/2}$ bade $x$ ke liye ( $x$ ki koi bhi power $ln x$ ko beat karti hai): $\frac{l n x}{x ^{3}} \leq \frac{x ^{1/2}}{x ^{3}} = \frac{1}{x ^{5/2}} .$ Step 3 — benchmark. $\int_{1}^{\infty} x^{- 5/2} d x$ converge karta hai ( $p = \frac{5}{2} > 1$ ). DCT se, converges. ✅

Verify: integration by parts se directly $\int_{1}^{\infty} \frac{l n x}{x ^{3}} d x = \frac{1}{4}$ milta hai (finite). Convergence confirm ho gayi; exact value $\frac{1}{4}$ hai.

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

Worked example

Ex 8. Kya $\int_{0}^{1} \frac{d x}{x + x ^{2}}$ converge karta hai? Forecast: problem $x = 0$ par hai (dekho Improper integrals — infinite discontinuities (type 2)), jahan $x \to 0$ integrand ko blow up karta hai. $0$ ke paas, $x$ , $x^{2}$ ko dominate karta hai, toh $f \sim 1/ x = x^{- 1/2}$ . Singularity ke paas p-test kehta hai $\int_{0}^{1} x^{- p} d x$ converge karta hai iff $p < 1$ . Yahan $p = \frac{1}{2} < 1$ — guess convergent.

Step 1 — singularity locate karo. Kyun? Comparison wahan honi chahiye jahan integrand misbehave kare — yahan $x \to 0^{+}$ par, $\infty$ par nahi.

Step 2 — $0$ ke paas inequality. $0 < x \leq 1$ ke liye, $x + x^{2} > x$ , toh $0 \leq \frac{1}{x + x ^{2}} < \frac{1}{x} .$ Upar se bound kyun? Convergence prove karna → ek convergent benchmark ke neeche squeeze karo.

Step 3 — type-2 endpoint ke liye benchmark. $\int_{0}^{1} x^{- 1/2} d x = [2 x]_{0}^{1} = 2$ , finite (converge karta hai kyunki $p = \frac{1}{2} < 1$ ). DCT se, converges. ✅

Verify: benchmark value $2$ finite hai; hamara integrand chhota hai, toh uska integral $2$ se neeche hai. Consistent hai.

Cell I — real-world word problem

Worked example

Ex 9. Ek radioactive sensor ek signal emit karta hai jiska power tail time $x \geq 1$ seconds ke baad $P (x) = \frac{1}{x ^{2} + x}$ watts hai. Kya total emitted energy $\int_{1}^{\infty} P (x) d x$ (joules mein) finite rehti hai? Forecast: $1/ x^{2}$ -jaisi tail tez decay karti hai; finite energy physically sensible hai — guess convergent (finite energy).

Step 1 — dominant behaviour identify karo. Bade $x$ ke liye, $x^{2} + x \sim x^{2}$ , toh $P \sim 1/ x^{2}$ ; benchmark $g = 1/ x^{2}$ .

Step 2 — inequality. Kyunki $x^{2} + x > x^{2}$ , $0 \leq \frac{1}{x ^{2} + x} < \frac{1}{x ^{2}} .$ Step 3 — benchmark. $\int_{1}^{\infty} x^{- 2} d x = 1$ converge karta hai, toh DCT se energy converges (finite). ✅

Verify (exact, kyunki yahan kar sakte hain): partial fractions $\frac{1}{x ^{2} + x} = \frac{1}{x} - \frac{1}{x + 1}$ se $\int_{1}^{\infty} = [ln \frac{x}{x + 1}]_{1}^{\infty} = 0 - ln \frac{1}{2} = ln 2 \approx 0.693$ joules milte hain. Finite — hamari forecast se match karta hai, aur wakai benchmark $1$ se neeche hai.

Cell J — exam twist (instinct galat hai)

Worked example

Ex 10. Kya $\int_{2}^{\infty} \frac{d x}{x ln x}$ converge karta hai? Forecast (the trap): yeh $1/ x^{kuch > 1}$ jaisa lagta hai kyunki neeche extra $ln x$ hai jo ise aur chhota banata hai — bahut se students guess convergent karte hain. Suspicious raho.

Step 1 — power ke saath comparison clean kyun nahi chalta. $g = 1/ x$ try karo: $L = lim \frac{1/ ( x l n x )}{1/ x} = lim \frac{1}{l n x} = 0$ . $L = 0$ ke saath LCT sirf one-way info deta hai aur $g$ diverge karta hai, toh koi conclusion nahi — twist real hai.

Step 2 — substitution se directly integrate karo. Kyun? Jab comparison ruk jaaye, evaluate karo. $u = ln x$ lao, toh $d u = \frac{d x}{x}$ : $\int \frac{d x}{x l n x} = \int \frac{d u}{u} = ln ∣ u ∣ = ln (ln x) .$ Step 3 — limit lo. $\int_{2}^{t} \frac{d x}{x l n x} = ln (ln t) - ln (ln 2) t \to \infty \infty.$ Kyunki $ln (ln t) \to \infty$ (dhire, lekin zaroor), integral diverges. ❌ Extra $ln x$ ise tame karne ke liye kaafi nahi tha — instinct busted.

Verify: $ln (ln t)$ $t \to \infty$ ke saath unbounded hai, toh antiderivative ki koi finite limit nahi. Divergence exactly confirm ho gayi.

Flashcards

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

ke liye kaunsa benchmark aur kya verdict?

g = 1/ x^{3}

, upar se bound, converges (

p = 3 > 1

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

par LCT kya

L

aur kya verdict deta hai?

L = 1/3

1/ x

ke saath same fate, diverges.

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

g = 1/ x

ke saath LCT se kyun settle nahi kiya ja sakta?

L = 0

(sirf one-way); integrate karo: antiderivative

ln (ln x) \to \infty

, diverges.

0

par type-2 singularity ke paas,

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

kab converge karta hai?

Jab

p < 1

ho (

\infty

tail rule ka ulta).

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

ko kaise handle karte hain?

∣ f ∣ \leq 1/ x^{2}

compare karo; absolutely convergent ⇒ convergent.

x^{2} e^{- x}

ko dominate karne ke liye

e^{- x /2}

kyun use karte hain?

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

, toh polynomial absorb ho jaata hai aur

e^{- x /2}

integrable hai.

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

ki value?

Γ (3) = 2! = 2

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

ki value?

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