Convergence tests for improper integrals — comparison: Question bank

4.2.12 · D5 · Maths › Calculus II — Integration › Convergence tests for improper integrals — comparison

Har prompt padho, apna jawab zor se bolo, phir reveal karo. Agar tumhara justification mujhse alag hai, toh woh gap hi woh misconception hai jise fix karna hai.

True or false — justify

A bound

f < g

with

\int g

divergent proves

\int f

diverges.

False. Kisi infinite area se neeche rehna phir bhi finite ho sakta hai (jaise

1/ x^{2} < 1/ x

, phir bhi

\int 1/ x^{2}

converge karta hai). Ek useful lower bound ko ek divergent

g

ke upar hona chahiye.

0 \leq f \leq g

and

\int f

diverges, then

\int g

diverges.

True. Agar chhoti area already infinite hai, toh badi area (jo usse

\geq

hai) finite nahi ho sakti — yeh DCT ka divergence wala half hai.

The DCT works for any two functions as long as

f \leq g

False. Iske liye

f \geq 0

bhi chahiye (equivalently

0 \leq f \leq g

); proof rely karta hai

F (t) = \int_{a}^{t} f

ke increasing hone par, jo fail ho jaata hai agar

f

ka sign change ho.

\int_{a}^{\infty} ∣ f ∣ d x

converges, then

\int_{a}^{\infty} f d x

converges.

True. Yeh absolute convergence hai; ek convergent

g

∣ f ∣

ko control karna sign-changing

f

ko handle karne deta hai — dekho Absolute vs conditional convergence.

\int_{a}^{\infty} f d x

converges, then

\int_{a}^{\infty} ∣ f ∣ d x

must converge too.

False. Convergence conditional ho sakta hai: positive aur negative area ki cancellation

\int f

ko finite bana sakti hai jabki

\int ∣ f ∣

diverge kare.

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

converges for

p = 1

False. Exactly

p = 1

par antiderivative

ln t \to \infty

hai, isliye yeh diverge karta hai. Boundary divergent side se belong karti hai; convergence ke liye strictly

p > 1

chahiye.

For

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

(a type-2 improper integral at

0

), convergence also needs

p > 1

False.

0

ke paas rule flip ho jaata hai:

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

converge karta hai iff

p < 1

. Chhote powers

0

ke paas integrable hain, bade powers

\infty

ke paas — dekho Improper integrals — infinite discontinuities (type 2).

lim_{x \to \infty} f (x) = 0

, then

\int_{1}^{\infty} f d x

converges.

False. Integrand ka zero hona zaroori lagta hai lekin sufficient nahi hai:

1/ x \to 0

phir bhi

\int_{1}^{\infty} 1/ x d x

diverge karta hai. Decay ki speed matter karti hai.

The Limit Comparison Test requires

f

and

g

to be eventually positive.

True. LCT ko

f, g > 0

for large

x

chahiye; ratio-sandwich argument

\frac{L}{2} g \leq f \leq 2 Lg

sirf positive functions ke saath sense deta hai.

L = lim f / g = 1

, then

\int f

and

\int g

literally have equal value.

False.

0 < L < \infty

sirf same fate guarantee karta hai (dono converge ya dono diverge), kabhi equal numerical values nahi — sirf yeh ki unke tails proportionally scale karte hain.

Spot the error

\frac{1}{ln x} < \frac{1}{x}

for large

x

, so since

\int 1/ x

diverges,

\int 1/ ln x

diverges."

Inequality ulti hai: kyunki

ln x < x

hai, isliye

\frac{1}{l n x} > \frac{1}{x}

hota hai. Corrected bound (ek divergent integral se bada hona) exactly wahi hai jo divergence force karta hai.

\frac{2 + sin x}{x ^{2}} \leq \frac{2}{x ^{2}}

, and

\int 2/ x^{2}

converges, so the integral converges."

Bound wahan false hai jahan

sin x > 0

. Safe envelope hai

2 + sin x \leq 3

, jo

\leq 3/ x^{2}

deta hai; conclusion survive karta hai lekin sirf sahi constant ke saath.

"I want convergence, so I found

g

with

f \geq g

and

\int g

convergent — done."

Wrong direction:

f

ko ek convergent

g

se neeche bound karna

f

ke baare mein kuch nahi batata. Convergence prove karne ke liye

f

ko ek convergent

g

se upar bound karna hoga.

"LCT gave

L = 0

, and

\int g

diverges, so

\int f

diverges."

L = 0

ke saath sirf one-way link milta hai: "

\int g

converges

\Rightarrow \int f

converges."

g

ka diverge hona kuch nahi kehta;

f

g

se faster decay karta hai, isliye woh converge bhi kar sakta hai.

f \leq g

pointwise everywhere except near

x = 3

, but that's fine, comparison still applies globally."

Comparison ko inequality sabhi $x \geq a$ ke liye (ya eventually,

a

adjust karke) hold karna chahiye. Ek bhi interval jahan

f > g

ho, extra area add kar sakta hai, isliye tumhe wahan restrict karna hoga jahan bound truly hold kare.

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

: since

sin x

isn't positive, DCT can't say anything, so it diverges."

Premise sahi hai (DCT ko

f \geq 0

chahiye) lekin leap galat hai.

\frac{s i n x}{x ^{2}} \leq \frac{1}{x ^{2}}

use karo; absolute comparison se convergence milti hai.

"Comparing

\frac{1}{x + x}

\frac{1}{x}

by LCT: dominant term is

x

Dominant term as

x \to \infty

actually

x

hai (yeh

x

se faster grow karta hai), isliye sahi benchmark

1/ x

hai, jo

L = 1

aur divergence deta hai —

1/ x

nahi.

Why questions

Why must

f \geq 0

for the Direct Comparison Test but not obviously for the Limit Comparison Test?

DCT ka proof use karta hai ki

F (t) = \int_{a}^{t} f

increasing hai (iske liye

f \geq 0

chahiye); LCT ko bhi secretly positivity chahiye taaki ratio bound

\frac{L}{2} g \leq f \leq 2 Lg

valid two-sided squeeze ho — dono Monotone Convergence Theorem ke zariye nonnegativity par rely karte hain.

Why do we define

\int_{a}^{\infty} f = lim_{t \to \infty} \int_{a}^{t} f

instead of just "integrating to infinity"?

\infty

koi number nahi hai jise tum antiderivative mein substitute kar sako; tum ek legitimate finite window

[a, t]

par integrate karte ho aur poochho ki kya accumulated area settle hoti hai jaise

t

badhta hai.

Why is

1/ x^{2}

the go-to convergent benchmark and

1/ x

the go-to divergent one?

Yeh dono

p = 1

threshold ke doono taraf hain:

p = 2 > 1

converge karta hai,

p = 1

diverge karta hai. Koi bhi tail jo "

1/ x

se faster" behave kare woh convergent hai, "

1/ x

jaisi ya slower" woh divergent hai — dekho The p-integral and p-series.

Why does the comparison test for integrals mirror the comparison test for series?

Dono nonnegative pieces accumulate karte hain aur "increasing + bounded

\Rightarrow

converges" par rely karte hain; Integral Test for series monotone terms ke liye correspondence exact banata hai.

Why does bounding a wiggling numerator like

2 + sin x

by a constant work?

Kyunki

sin x

[- 1, 1]

mein trapped hai, isliye

2 + sin x \in [1, 3]

; oscillation ko uske worst-case constant

3

se replace karna ek clean comparison deta hai bina correctness khoye.

Why can't "

f (x) \to 0

" alone guarantee convergence, when it is required for series?

Series ke liye bhi

a_{n} \to 0

sirf necessary hai, sufficient nahi (

\sum 1/ n

diverge karta hai); integrals wahi lesson inherit karte hain — decay ka rate decide karta hai, sirf decay ka fact nahi.

Why does LCT free us from choosing the correct inequality direction?

Yeh problem ko ek single limit

L = lim f / g

mein convert kar deta hai; ek baar

0 < L < \infty

ho jaaye, two-sided squeeze automatically dono DCT inequalities supply kar deta hai, isliye tumhe kabhi guess nahi karna hota ki bound kis taraf point kare.

Edge cases

What happens in LCT when

L = \infty

f

g

par dominate karta hai: sirf "

\int f

converges

\Rightarrow \int g

converges" milta hai (equivalently "

\int g

diverges

\Rightarrow \int f

diverges"). Yeh ek one-way tool hai, useful sirf tab jab

g

diverge kare.

What happens in LCT when

L = 0

f

g

ke side mein negligible hai: sirf "

\int g

converges

\Rightarrow \int f

converges" milta hai. Agar

g

diverge kare toh

f

ke baare mein kuch nahi pata.

An integral is improper in two ways at once (e.g.

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

). How do you test it?

Ek convenient interior point par split karo, jaise

\int_{0}^{1} + \int_{1}^{\infty}

; poora tab converge karta hai jab dono pieces converge karein. Yahan

\int_{0}^{1}

diverge karta hai (

0

par

p = 1

) aur

\int_{1}^{\infty}

bhi diverge karta hai, isliye yeh kisi bhi taraf se diverge karta hai.

Does

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

converge at

p

slightly above

1

, like

p = 1.0001

Yes. Criterion hai strictly

p > 1

, koi gap nahi: koi bhi

p > 1

, chahe kitna bhi close ho,

t^{1 - p} \to 0

deta hai aur ek finite value. Convergence razor-thin hai lekin genuine hai.

f \leq g

only holds for

x \geq M

(not from

x = a

), can we still conclude convergence?

Haan — convergence sirf tail par depend karta hai.

\int_{a}^{\infty} = \int_{a}^{M} + \int_{M}^{\infty}

split karo; pehla finite piece harmless hai, aur DCT

\int_{M}^{\infty}

par apply hota hai jahan bound hold karta hai.

What if

f \geq 0

but

f

is unbounded on the finite part while decaying at infinity?

Tab yeh dono types se improper hai; har singular point ko alag test karo.

\infty

par convergence (

x^{- p}

p > 1

se comparison ke zariye) aur blow-up point par integrability independent conditions hain — dono hold karni chahiye.

Can two integrals both diverge yet have a finite ratio limit

L

Yes, yahi LCT ka poora point hai:

\int 1/ x

aur

\int 1/ (x + x)

dono

L = 1

ke saath diverge karte hain. Same fate ka matlab finite nahi hai; matlab hai ki yeh saath mein diverge karte hain.

Connections

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