Improper integrals — Type I (infinite limits), Type II (discontinuous integrand): Worked examples

4.2.11 · D3 · Maths › Calculus II — Integration › Improper integrals — Type I (infinite limits), Type II (disc

The scenario matrix

Har improper integral jo tum dekho ge woh in cells mein se kisi ek mein fit hogi. Do "bad-point types" columns hain; "behaviour flavour" rows hain.

Cell	Bad point	Flavour	Example jo isse hit karta hai
A	Type I ( $\infty$ )	power decay, converges ( $p > 1$ )	Ex 1
B	Type I ( $\infty$ )	knife-edge, diverges ( $p = 1$ )	Ex 2
C	Type I ( $\infty$ )	exponential decay, converges	Ex 3
D	Type I ( $- \infty$ AND $+ \infty$ )	dono halves, split zaroori	Ex 4
E	Type I	signed integrand — principal-value trap	Ex 5
F	Type II (left-end $0$ )	mild spike, converges ( $p < 1$ )	Ex 6
G	Type II (interior blow-up)	hidden singularity, diverges	Ex 7
H	Type I ( $\infty$ ), koi antiderivative nahi	comparison decide karta hai	Ex 8
I	real-world word problem	probability / physics	Ex 9
J	exam twist	ek $\infty$ end bhi aur $0$ pe spike bhi	Ex 10

Recall Do decision rules jo apni pocket mein rakhna

Infinity pe tumhe fast decay chahiye ( $p > 1$ ); zero pe tumhe mild spike chahiye ( $p < 1$ ). $\infty$ pe, kaunsa $p$ converge karta hai? ::: $p > 1$ $0$ pe, kaunsa $p$ converge karta hai? ::: $p < 1$ $p = 1$ pe kya hota hai? ::: DONO cases mein diverges karta hai

Figure — Improper integrals — Type I (infinite limits), Type II (discontinuous integrand)

Cell A — Type I, power decay, converges

Worked example Example 1 —

$\int_{1}^{\infty} \frac{d x}{x ^{3/2}}$ Forecast: Yahan $p = \frac{3}{2}$ hai. Kya $\frac{3}{2} > 1$ hai? Aage padhne se pehle convergence ya divergence guess karo.

Step 1. Forbidden $\infty$ ko ek real number $t$ se replace karo: $\int_{1}^{\infty} x^{- 3/2} d x = lim_{t \to \infty} \int_{1}^{t} x^{- 3/2} d x .$ Yeh step kyun? Hum Fundamental Theorem of Calculus mein $\infty$ feed nahi kar sakte — woh sirf finite, closed interval $[1, t]$ pe kaam karta hai. Isliye pehle ek movable wall $t$ tak integrate karte hain.

Step 2. Power rule, $\int x^{- 3/2} d x = \frac{x ^{- 1/2}}{- 1/2} = - \frac{2}{x}$ : $\int_{1}^{t} x^{- 3/2} d x = [- \frac{2}{x}]_{1}^{t} = - \frac{2}{t} + 2.$ Yeh step kyun? Ab interval $[1, t]$ finite hai aur integrand wahan continuous hai, isliye FTC legal hai.

Step 3. Wall ko infinity tak slide karo. Jab $t \to \infty$ , $\frac{2}{t} \to 0$ (dekho Limits at Infinity): $lim_{t \to \infty} (2 - \frac{2}{t}) = 2.$ Yeh step kyun? Limit lena woh final move hai jo movable-wall answer ko true infinite-area answer mein badalta hai — hum dekhte hain ki leftover term $2/ t$ shrink hokar zero ho jaata hai.

Converges to $2$ . Cell A confirmed: $p = \frac{3}{2} > 1$ .

Verify: $p$ -integral formula se answer hona chahiye $\frac{1}{p - 1} = \frac{1}{3/2 - 1} = \frac{1}{1/2} = 2$ . ✓ Match karta hai.

Cell B — Type I, knife-edge, diverges

Worked example Example 2 —

$\int_{2}^{\infty} \frac{d x}{x}$ Forecast: $p = 1$ exactly — akela loser. Verdict predict karo.

Step 1. Movable wall: $\int_{2}^{\infty} \frac{d x}{x} = t \to \infty lim \int_{2}^{t} \frac{d x}{x}$ . Yeh step kyun? Same forbidden-infinity reason jaise Example 1 mein — $\infty$ koi number nahi hai jise hum FTC mein feed kar sakein, isliye ek real wall $t$ tak integrate karte hain.

Step 2. Power rule fail ho jaata hai $p = 1$ pe (woh $1 - p = 0$ se divide kar deta). $1/ x$ ka sahi antiderivative $ln x$ hai: $\int_{2}^{t} \frac{d x}{x} = [ln x]_{2}^{t} = ln t - ln 2.$ Yeh step kyun? $ln x$ woh ek function hai jiska derivative $1/ x$ hai — yahi woh case hai jise power rule touch nahi kar sakta.

Step 3. Jab $t \to \infty$ , $ln t \to \infty$ : $lim_{t \to \infty} (ln t - ln 2) = \infty.$ Yeh step kyun? Hum wall ko door slide karte hain aur area total dekhte hain; kyunki $ln t$ ki koi ceiling nahi hai, area kabhi kisi number pe settle nahi karta.

Diverges. Cell B confirmed. Logarithm bina ruke badhta hai — dheere, lekin hamesha ke liye.

Verify: $t = e^{10}$ pe partial area hai $ln (e^{10}) - ln 2 = 10 - ln 2 \approx 9.307$ , aur $t$ ke saath badhta rehta hai — koi finite ceiling nahi. ✓ Diverges.

Cell C — Type I, exponential decay

Worked example Example 3 —

$\int_{0}^{\infty} x e^{- x} d x$ Forecast: Exponential decay kisi bhi polynomial ko beat kar deta hai. Guess: kya $x e^{- x}$ finite area enclose karta hai?

Step 1. $\int_{0}^{\infty} x e^{- x} d x = t \to \infty lim \int_{0}^{t} x e^{- x} d x .$ Yeh step kyun? Infinite upper limit → $t$ se replace karo taaki FTC finite interval $[0, t]$ pe act kar sake.

Step 2. Integration by parts karo $u = x$ , $d v = e^{- x} d x$ ke saath, toh $d u = d x$ , $v = - e^{- x}$ : $\int_{0}^{t} x e^{- x} d x = [- x e^{- x}]_{0}^{t} + \int_{0}^{t} e^{- x} d x = - t e^{- t} + [- e^{- x}]_{0}^{t} = - t e^{- t} - e^{- t} + 1.$ Yeh step kyun? Hum $x e^{- x}$ ko ek shot mein antidifferentiate nahi kar sakte; integration by parts $x$ factor ko peel off karta hai aur ek plain $e^{- x}$ chhod deta hai jise hum integrate karna jaante hain.

Step 3. Jab $t \to \infty$ : $e^{- t} \to 0$ , aur $t e^{- t} \to 0$ bhi kyunki exponential linear factor ko crush kar deta hai (dekho Limits at Infinity): $lim_{t \to \infty} (1 - t e^{- t} - e^{- t}) = 1 - 0 - 0 = 1.$ Yeh step kyun? Yahi woh jagah hai jahan exponential decay apni reputation earn karta hai — woh badhte hue $t$ ko zero pe gira deta hai, toh leftover terms gayab ho jaate hain aur sirf constant $1$ bachta hai.

Converges to $1$ . Cell C confirmed.

Verify: Yeh Gamma Function ki value $Γ (2) = 1! = 1$ hai. ✓ Match karta hai.

Cell D — Type I DONO ends pe, split zaroori

Worked example Example 4 —

$\int_{- \infty}^{\infty} \frac{d x}{1 + x ^{2}}$ Forecast: Dono ends infinity tak jaate hain. Rule kehta hai split karo aur dono halves check karo. Total guess karo.

Step 1. Convenient point $c = 0$ pe split karo: $\int_{- \infty}^{\infty} \frac{d x}{1 + x ^{2}} = \int_{- \infty}^{0} \frac{d x}{1 + x ^{2}} + \int_{0}^{\infty} \frac{d x}{1 + x ^{2}} .$ Yeh step kyun? Definition maangti hai ki har half independently converge kare — ek single symmetric limit sirf principal value dega, true convergence nahi.

Step 2. $\frac{1}{1 + x ^{2}}$ ka antiderivative $arctan x$ hai. Right half: $\int_{0}^{\infty} = lim_{t \to \infty} [arctan x]_{0}^{t} = lim_{t \to \infty} (arctan t - 0) = \frac{π}{2} .$ Yeh step kyun? $arctan$ ka matlab hai "kaunsa angle is tangent ko rakhta hai?", aur jab $t \to \infty$ woh angle $9 0^{\circ} = \frac{π}{2}$ radians ke paas aata hai.

Step 3. Ab left half. Hum symmetry blindly assume nahi karte — hum ise directly apni movable wall $s \to - \infty$ ke saath compute karte hain: $\int_{- \infty}^{0} = lim_{s \to - \infty} [arctan x]_{s}^{0} = 0 - (- \frac{π}{2}) = \frac{π}{2} .$ Yeh step kyun? Definition require karti hai ki har half ko apni limit pe khud test kiya jaaye; yahan $arctan s \to - \frac{π}{2}$ jab $s \to - \infty$ , aur us negative ko subtract karne se $+ \frac{π}{2}$ milta hai. (Yeh right half se match karta hai kyunki $\frac{1}{1 + x ^{2}}$ even hai, lekin yeh result humne compute karke earn kiya, hand-waving se nahi.)

Dono halves converge karte hain, isliye total $= \frac{π}{2} + \frac{π}{2} = π$ .

Verify: $arctan (\infty) - arctan (- \infty) = \frac{π}{2} - (- \frac{π}{2}) = π \approx 3.14159$ . ✓ $π$ pe converge karta hai.

Cell E — signed integrand, principal-value trap

Worked example Example 5 —

$\int_{- \infty}^{\infty} x d x$ Forecast: Graph origin ke baare mein symmetric hai, toh "obviously" positive aur negative areas cancel hokar $0$ ho jaate hain — hai na? Dhyan rakho.

Step 1. Split karo: $\int_{- \infty}^{\infty} x d x = \int_{- \infty}^{0} x d x + \int_{0}^{\infty} x d x .$ Yeh step kyun? Kabhi bhi ek symmetric limit mein merge mat karo — woh sirf Cauchy principal value deta hai, true convergence nahi.

Step 2. Right half akele test karo: $\int_{0}^{\infty} x d x = lim_{t \to \infty} [\frac{x ^{2}}{2}]_{0}^{t} = lim_{t \to \infty} \frac{t ^{2}}{2} = \infty.$ Yeh step kyun? Agar ek bhi piece diverge kare, poora integral diverge karta hai — doosra compute karne ki zaroorat nahi, isliye pehle sabse simple half check karo.

Step 3. Isliye $\int_{- \infty}^{\infty} x d x$ diverges, bhale hi symmetric limit $lim_{t \to \infty} \int_{- t}^{t} x d x = 0$ ho. Yeh step kyun? Ek bad half ke baad turant diverging conclusion state karna exactly "one divergence kills the whole" rule ko action mein dikhana hai — symmetric $0$ ek mirage hai.

Trap: $0$ principal value hai, true value nahi. Cell E confirmed.

Verify: $t = 1000$ pe right-half partial area $\frac{100 0 ^{2}}{2} = 500000$ hai aur bina ruke badhta hai. ✓ Tempting symmetric $0$ ke bawajood diverges karta hai.

Cell F — Type II, left end pe mild spike

Worked example Example 6 —

$\int_{0}^{1} \frac{d x}{x ^{2/3}}$ Forecast: Integrand $x = 0$ pe blow up karta hai. Yahan $p = \frac{2}{3} < 1$ hai — ek mild spike. Guess: finite area, ya nahi?

Step 1. Bad point left end $x = 0$ hai, isliye $t > 0$ se creep up karo: $\int_{0}^{1} x^{- 2/3} d x = lim_{t \to 0^{+}} \int_{t}^{1} x^{- 2/3} d x .$ Yeh step kyun? $1/ x^{2/3} \to \infty$ jab $x \to 0^{+}$ ; FTC $x = 0$ ko touch nahi kar sakta, isliye hum right se uski taraf tiptoe karte hain.

Step 2. Power rule, $\int x^{- 2/3} d x = \frac{x ^{1/3}}{1/3} = 3 x^{1/3}$ : $\int_{t}^{1} x^{- 2/3} d x = [3 x^{1/3}]_{t}^{1} = 3 - 3 t^{1/3} .$ Yeh step kyun? $[t, 1]$ pe integrand continuous hai, isliye FTC legally apply hota hai.

Step 3. Jab $t \to 0^{+}$ , $t^{1/3} \to 0$ : $lim_{t \to 0^{+}} (3 - 3 t^{1/3}) = 3.$ Yeh step kyun? $t$ ko spike tak slide karte hue, correction term $3 t^{1/3}$ gayab ho jaata hai — tall-but-skinny sliver sirf finite area rakhta hai.

Converges to $3$ . Cell F confirmed: $p = \frac{2}{3} < 1$ .

Verify: Type II $p$ -integral deta hai $\frac{1}{1 - p} = \frac{1}{1 - 2/3} = \frac{1}{1/3} = 3$ . ✓ Match karta hai.

Cell G — Type II, hidden interior singularity

Worked example Example 7 —

$\int_{- 1}^{2} \frac{d x}{x ^{2}}$ Forecast: $[- 1/ x]_{- 1}^{2}$ seedha likhne ka mann karta hai. Lekin kya $x^{2}$ poore $[- 1, 2]$ pe well-behaved hai? Blow-up dhundo.

Step 1. Singularity spot karo: $\frac{1}{x ^{2}} \to + \infty$ at $x = 0$ , jo $[- 1, 2]$ ke andar hai. Yeh step kyun? Koi bhi antiderivative pakadne se pehle, interval ko scan karo un points ke liye jahan integrand blast karta hai. Yeh wala origin pe chhupa hai.

Step 2. $c = 0$ pe split karo aur require karo ki dono halves converge karein: $\int_{- 1}^{2} \frac{d x}{x ^{2}} = \int_{- 1}^{0} \frac{d x}{x ^{2}} + \int_{0}^{2} \frac{d x}{x ^{2}} .$ Right half test karo: $p = 2 \geq 1$ ke saath endpoint $0$ pe, $\int_{0}^{2} x^{- 2} d x = lim_{t \to 0^{+}} [- \frac{1}{x}]_{t}^{2} = lim_{t \to 0^{+}} (- \frac{1}{2} + \frac{1}{t}) = + \infty.$ Yeh step kyun? Ek interior blow-up poore interval mein FTC ko tod deta hai, isliye hum wahan split karte hain; phir $\frac{1}{t} \to + \infty$ dikhata hai ki spike $1/ x^{2}$ itna tall hai ki finite area fit nahi hoti.

Step 3. Ek half diverge karta hai → poora integral diverges. Cell G confirmed. Yeh step kyun? "One divergence kills the whole" rule se, jaise hi ek piece fail ho jaata hai hum rok sakte hain.

Trap: Naive plug-in $[- 1/ x]_{- 1}^{2} = - \frac{1}{2} - 1 = - \frac{3}{2}$ ek strictly positive integrand ke liye negative number deta hai — ek impossible negative area, red flag ki FTC ek discontinuity ke across illegally apply hua.

Verify: $t = 1 0^{- 6}$ pe right-half partial $- \frac{1}{2} + 1 0^{6} = 999999.5$ hai, blast hota hua jab $t \to 0$ . ✓ Diverges (aur naive $- \frac{3}{2}$ bakwaas hai).

Cell H — Type I, koi antiderivative nahi, comparison use karo

Worked example Example 8 — Kya

$\int_{1}^{\infty} \frac{d x}{x ^{4} + 1}$ converge karta hai? Forecast: Koi elementary antiderivative nahi, lekin interval infinity tak jaata hai isliye yeh Type I hai. Kyunki large $x$ ke liye $x^{4} + 1 \approx x^{2}$ , verdict guess karo, phir saaf justify karo.

Step 1. Integrand ko kisi aise cheez se bound karo jise hum integrate KAR SAKEIN. $x \geq 1$ ke liye, kyunki $x^{4} + 1 > x^{4}$ : $x^{4} + 1 > x^{4} = x^{2} ⟹ \frac{1}{x ^{4} + 1} < \frac{1}{x ^{2}} .$ Yeh step kyun? Comparison Test for Integrals kehta hai: agar ek badi non-negative function ka finite area hai, toh chhoti ka bhi hoga — isliye hum ek simple upper bound dhundte hain.

Step 2. Dominating integral converge karta hai — yeh $p = 2 > 1$ wala $p$ -integral hai (yeh exactly wahi master $\int_{1}^{\infty} x^{- p} d x$ result hai jo parent note mein hai, aur yeh p-series and Integral Test ko bhi underlie karta hai): $\int_{1}^{\infty} \frac{d x}{x ^{2}} = lim_{t \to \infty} [- \frac{1}{x}]_{1}^{t} = lim_{t \to \infty} (1 - \frac{1}{t}) = 1 < \infty.$ Yeh step kyun? Humein compare karne ke liye ek known finite value chahiye; $1/ x^{2}$ sabse simple curve hai jo hamare wale ko dominate karta hai aur $p > 1$ rakhta hai, isliye uski area finite hai.

Step 3. Kyunki $0 < \frac{1}{x ^{4} + 1} < \frac{1}{x ^{2}}$ aur upper integral finite hai, hamaara integral converges (kisi value pe $\leq 1$ ). Cell H confirmed. Yeh step kyun? Ek non-negative area jo strictly ek finite area ke neeche trapped hai, woh khud bhi finite hogi — yahi comparison test ki poori logic hai.

Verify: Hamaara integral $\int_{1}^{\infty} x^{- 2} d x = 1$ ke neeche aur $0$ ke upar squeezed hai, isliye finite hai. (Numerically yeh $\approx 0.927$ hai, aaram se $(0, 1)$ mein.) ✓ Converge karta hai.

Cell I — real-world word problem

Worked example Example 9 — Exponential lifetime (probability)

Ek machine ka failure time $X$ (years mein) ka probability density $f (x) = 2 e^{- 2 x}$ hai $x \geq 0$ ke liye. (a) Confirm karo ki yeh valid density hai (total probability $= 1$ ). (b) Mean lifetime $\int_{0}^{\infty} x f (x) d x$ nikalo. Forecast: Ek valid density ke liye saare $x$ pe $f$ ke neeche area exactly $1$ hona chahiye. Compute karne se pehle mean guess karo.

Step 1 (a). Total probability: $\int_{0}^{\infty} 2 e^{- 2 x} d x = lim_{t \to \infty} [- e^{- 2 x}]_{0}^{t} = lim_{t \to \infty} (1 - e^{- 2 t}) = 1.$ Yeh step kyun? Ek density ka total area $1$ hona chahiye; infinite upper limit is Type I improper integral banata hai, isliye hum movable-wall limit use karte hain. Valid.

Step 2 (b). Mean $= \int_{0}^{\infty} x \cdot 2 e^{- 2 x} d x$ . Integration by parts karo ( $u = x$ , $d v = 2 e^{- 2 x} d x$ , $v = - e^{- 2 x}$ ): $\int_{0}^{t} 2 x e^{- 2 x} d x = [- x e^{- 2 x}]_{0}^{t} + \int_{0}^{t} e^{- 2 x} d x = - t e^{- 2 t} + [- \frac{1}{2} e^{- 2 x}]_{0}^{t} .$ Yeh step kyun? $x$ factor integration by parts demand karta hai, bilkul Cell C jaisa — yeh $x$ ko strip off karta hai aur ek clean exponential chhod deta hai.

Step 3. Jab $t \to \infty$ : $t e^{- 2 t} \to 0$ aur $e^{- 2 t} \to 0$ : $mean = 0 + (\frac{1}{2} - 0) = \frac{1}{2} .$ Yeh step kyun? Exponential decay phir se linear $t$ ko defeat karta hai, isliye saare wall-dependent terms gayab ho jaate hain aur sirf $\frac{1}{2}$ bachta hai.

Mean lifetime $= \frac{1}{2}$ year (units: years — integrand hai $x [yr] \cdot f [yr^{- 1}] \cdot d x [yr]$ , years deta hai). Cell I confirmed. Yeh Laplace Transform se connect hota hai, jo exactly $\int_{0}^{\infty} e^{- s x} f (x) d x$ hai.

Verify: Rate $λ = 2$ ke liye, exponential distribution ka mean $1/ λ = 1/2$ hai. ✓ Match karta hai.

Cell J — exam twist: infinite end AUR spike dono saath

Worked example Example 10 —

$\int_{0}^{\infty} \frac{d x}{x ( 1 + x )}$ Forecast: DO buri cheezein ek saath — $x = 0$ pe spike ( $1/ x$ se) AUR infinite upper limit. Dono dangers mein split karo aur har ek check karo.

Step 1. $x = 1$ pe split karo (koi bhi positive finite point kaam karega): piece $\int_{0}^{1}$ spike ko $0$ pe isolate karta hai, piece $\int_{1}^{\infty}$ infinity ko isolate karta hai. $\int_{0}^{\infty} = \int_{0}^{1} + \int_{1}^{\infty} .$ Yeh step kyun? Har improper feature ko isolate karna zaroori hai taaki har piece mein exactly ek danger ho — yeh ek combined Type I + Type II integral hai aur dono failures ko alag alag test karna hoga.

Step 2. Substitution $u = x$ se antiderivative nikalo, toh $x = u^{2}$ aur $d x = 2 u d u$ : $\int \frac{d x}{x ( 1 + x )} = \int \frac{2 u d u}{u ( 1 + u ^{2} )} = \int \frac{2 d u}{1 + u ^{2}} = 2 arctan u = 2 arctan x .$ Yeh step kyun? $x$ direct integration mushkil banata hai; $u = x$ substitute karne se $x$ cancel ho jaata hai $2 u$ ke saath jo $d x$ se aata hai, aur integral standard $\frac{1}{1 + u ^{2}}$ pattern ban jaata hai jiska antiderivative $arctan$ hai (wohi tool jo Cell D mein use kiya).

Step 3. Is single antiderivative se dono pieces evaluate karo, har danger pe ek limit lete hue: $\int_{0}^{1} = lim_{s \to 0^{+}} [2 arctan x]_{s}^{1} = 2 arctan 1 - lim_{s \to 0^{+}} 2 arctan s = 2 \cdot \frac{π}{4} - 0 = \frac{π}{2},$ $\int_{1}^{\infty} = lim_{t \to \infty} [2 arctan x]_{1}^{t} = lim_{t \to \infty} 2 arctan t - 2 arctan 1 = 2 \cdot \frac{π}{2} - \frac{π}{2} = \frac{π}{2} .$ Yeh step kyun? Har piece ko apni movable wall milti hai (spike ke liye $s \to 0^{+}$ , infinity ke liye $t \to \infty$ ); $0$ ke paas term $2 arctan s \to 0$ (mild spike, finite) aur infinity pe $2 arctan t \to 2 \cdot \frac{π}{2} = π$ (angle $9 0^{\circ}$ pe saturate ho jaata hai).

Dono pieces converge karte hain, isliye total $= \frac{π}{2} + \frac{π}{2} = π$ . Converges to $π$ . Cell J confirmed.

Verify: Overall $2 arctan x$ evaluate kiya $0$ se $\infty$ tak $2 \cdot \frac{π}{2} - 0 = π \approx 3.14159$ hai, aur dono split pieces $\frac{π}{2} + \frac{π}{2}$ agree karte hain. ✓ $π$ pe converge karta hai.

Recall Matrix pe quick self-check

$\int_{- 1}^{1} \frac{d x}{x ^{2}}$ kaunsa cell hai? ::: Cell G — $x = 0$ pe interior blow-up, diverges. $\int_{0}^{\infty} e^{- x} d x$ kaunsa cell hai? ::: Cell C — infinity pe exponential decay, converges. $\int_{- \infty}^{\infty} x d x$ kaunsa cell hai? ::: Cell E — signed, principal-value trap, diverges. $\int_{0}^{1} x^{- 1/2} d x$ kaunsa cell hai? ::: Cell F — $0$ pe mild spike ( $p < 1$ ), converges.