4.2.12
Ek improper integral basically infinite sum of areas hota hai. Agar tumhari function ki area kisi aisi function ke neeche trapped hai jiska area finite hai, toh tumhari area bhi finite hogi. Agar yeh kisi aisi function ke upar baith rahi hai jiska area infinite hai, toh tumhari area bhi blow up karegi. Comparison ka matlab hai: us mushkil integral ko compute mat karo — usse ek aasaan integral ke beech squeeze karo.
Definition Improper integralEk integral improper hoti hai agar ya toh
integration ki ek limit infinite ho: ∫ a ∞ f d x
integrand kisi point par [ a , b ]
Hum isse ek limit se define karte hain:
∫ a ∞ f ( x ) d x = lim t → ∞ ∫ a t f ( x ) d x . converges hoti hai; warna yeh diverges hoti hai.
Limit kyun? Kyunki ∞ finite window [ a , t ] t → ∞
Yeh kyun important hai: almost har comparison mein x
Yeh kyun kaam karta hai (first principles se): Maano F ( t ) = ∫ a t f G ( t ) = ∫ a t g f ≥ 0 F increasing hai. Kyunki f ≤ g F ( t ) ≤ G ( t ) aur upar se bounded ho, woh finite limit mein converge zaroor karega (Monotone Convergence). Agar ∫ g G ( t ) F ( t ) ∫ f
Isse use kaise karein (strategy):
Forecast karo: jab x → ∞ f x
Ek clean g x − p e − x
Convergence prove karni hai? f g Divergence prove karni hai? f g
Kyun: agar f / g → L x 2 L g ≤ f ≤ 2 L g f g Kaise: bas dominant terms match karo; inequalities se mat lado.
∫ 1 ∞ x 2 + 5 d x g = 1/ x 2 x + 5 f 1/ x 2 x 2 + 5 > x 2 ⇒ x 2 + 5 1 < x 2 1 Yeh direction kyun? Hum convergence prove karna chahte hain, isliye upar se bound karo.
∫ 1 ∞ x − 2 d x p = 2 > 1 converges . ✅
∫ 1 ∞ x + x d x LCT yahan kyun? Denominator ek clean power nahi hai. x → ∞ x g = 1/ x L = lim x → ∞ 1/ x 1/ ( x + x ) = lim x → ∞ x + x x = lim x → ∞ x − 1/2 + 1 1 = 1. Yeh step kyun? Limit expose karne ke liye top aur bottom ko x 0 < L = 1 < ∞ ∫ 1/ x diverge karti hai. Toh hamara integral diverges . ❌
∫ 1 ∞ x 2 2 + sin x d x Numerator ko bound kyun karein? sin x − 1 ≤ sin x ≤ 1 1 ≤ 2 + sin x ≤ 3 0 ≤ x 2 2 + sin x ≤ x 2 3 Upar bound kyun? Convergence prove kar rahe hain.
∫ 1 ∞ 3 x − 2 d x converges . ✅
∫ 2 ∞ ln x d x 1/ x x ≥ 2 x ln x < x ln x 1 > x 1 Neeche bound kyun? Hume divergence ka shak hai; isse ek divergent integral ke upar trap karo.
∫ 2 ∞ x 1 d x diverges . ❌
Common mistake Classic errors ko steel-man karna(a) "Maine f < g ∫ g ∫ f
Yeh sahi kyun lagta hai: bounds transitive lagte hain. Yeh WRONG kyun hai: agar f chhota hai, toh f Fix: useful inequalities yeh hain: (small ≤ ≥
(b) f ≥ 0 DCT ke liye nonnegative functions chahiye, kyunki proof "increasing F f absolute comparison (∣ f ∣ ≤ g
(c) LCT mein L = 0 L = ∞ Clean conclusion ke liye 0 < L < ∞ L = 0 g f
Recall Feynman: ek 12-saal ke bacche ko explain karoSocho ki tum ek tall box mein sand daal rahe ho jo forever upar jaati hai. Agar sand ka ek chhota dher kabhi floor se overflow nahi karta (finite total), toh us pile ke neeche wala pile bhi finite rahega. Aur agar tumse bada pile forever overflow karta hai, toh tumhara bhi overflow karega. Magic trick yeh hai: apni weird pile ko measure karne ki jagah, ise ek aasaan pile se compare karo (1/ x 2 1/ x
Mnemonic Directions yaad rakho"Squeeze to please, push to die."
Converge prove karne ke liye: f g neeche squeeze karo (please = finite area).Diverge prove karne ke liye: f g upar push karo (die = infinite area).
Aur benchmark ke liye: "p bigger than 1, the area is done" (p > 1
∫ 1 ∞ x − p d x Exactly jab p > 1 p ≤ 1
Direct Comparison Test convergence ke liye state karo. Agar 0 ≤ f ≤ g ∫ g ∫ f
Direct Comparison Test divergence ke liye state karo. Agar 0 ≤ f ≤ g ∫ f ∫ g
DCT mein f ≥ 0 Taaki F ( t ) = ∫ a t f
LCT ka criterion aur conclusion kya hai? Agar L = lim f / g 0 < L < ∞ ∫ f ∫ g
∫ 1/ ( x 2 + 5 ) g g = 1/ x 2 f < g
∫ 1/ ( x + x ) LCT with g = 1/ x L = 1
f < g ∫ g Infinite area se chhota hona phir bhi finite ho sakta hai — koi conclusion nahi milta.
Directions ke liye mnemonic kya hai? "Squeeze to please (convergent ke neeche), push to die (divergent ke upar)."
∫ a ∞ f ∞ t t → ∞
converges iff p greater 1
bound above by convergent g
bound below by divergent g
