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

4.2.11 · D4 · Maths › Calculus II — Integration › Improper integrals — Type I (infinite limits), Type II (disc

Intuition Is page ko kaise use karein

Ye 12 problems ek ladder ki tarah chadh'te hain. L1 Recognition — kya aap pehchaan sakte ho kis tarah ka improper integral hai? L2 Application — kya aap machine chala sakte ho (bad point ko $t$ se replace karo → antiderivative → limit lo)? L3 Analysis — kya aap convergence decide kar sakte ho jab integration mushkil ho? L4 Synthesis — kya aap kai saari ideas combine kar sakte ho? L5 Mastery — kya aap un traps ko handle kar sakte ho jo careless students ko todh dete hain?

Har problem clearly likha gaya hai. Har solution ek collapsible [!recall]- box ke andar chhupa hua hai taaki aap pehle try karo, phir check karo. Core machinery parent note mein hai: the parent topic. Jinpe hum rely karte hain prerequisites: Fundamental Theorem of Calculus, Limits at Infinity, Comparison Test for Integrals, aur p-series and Integral Test.

Recall Ek nazar mein do master rules ki yaad-daasht

$\int_{1}^{\infty} \frac{d x}{x ^{p}} converges ⟺ p > 1, \int_{0}^{1} \frac{d x}{x ^{p}} converges ⟺ p < 1.$ "Infinity par bada, zero par chhota." $p = 1$ diverges in both cases.

Level 1 — Recognition

Goal: type ko naam do aur "bad point" locate karo — kuch compute karne se pehle.

Problem 1.1

Har ek ko proper, Type I, Type II, ya both classify karo, aur bad point(s) batao: $(a) \int_{2}^{5} \frac{d x}{x - 1}, (b) \int_{1}^{\infty} \frac{d x}{x ^{2}}, (c) \int_{0}^{2} \frac{d x}{x - 2}, (d) \int_{0}^{\infty} \frac{d x}{x} .$

Recall Solution 1.1

Test yeh hai: kya integrand closed interval par kahin blow up karta hai (Type II), aur/ya kya koi limit infinite hai (Type I)?

(a) $\frac{1}{x - 1}$ blow up karta hai $x = 1$ par, lekin $1 \in / [2, 5]$ . Interval finite hai. → Proper (ordinary integral, kuch bhi improper nahi).
(b) Upper limit $\infty$ hai; integrand $[1, \infty)$ par theek hai. → Type I, bad point $+ \infty$ par.
(c) $\frac{1}{x - 2} \to \pm \infty$ jab $x \to 2$ , aur $2$ right endpoint hai $[0, 2]$ ka. → Type II, bad point $x = 2^{-}$ par.
(d) Upper limit $\infty$ hai aur $\frac{1}{x} \to \infty$ jab $x \to 0^{+}$ (left endpoint). → Both (Type I at $\infty$ , Type II at $0$ ). Split karna zaroori hai, jaise $\int_{0}^{1} + \int_{1}^{\infty}$ .

Problem 1.2

Kya $\int_{- 1}^{1} \frac{d x}{x ^{2/3}}$ improper hai? Agar haan, toh kyun, aur kahan?

Recall Solution 1.2

$x^{2/3} = (3 x)^{2} \geq 0$ aur sirf $x = 0$ par $0$ hota hai. Toh $\frac{1}{x ^{2/3}} \to + \infty$ jab $x \to 0$ . Kyunki $0$ interior mein hai $[- 1, 1]$ ke, yeh ek Type II integral hai jisme interior singularity hai $x = 0$ par. Aapko split karna hi padega $0$ par aur dono halves check karne padenge. (Split skip karna classic error hai — neeche dekho.)

Level 2 — Application

Goal: poori machine chalao — bad point ko $t$ se replace karo, integrate karo, limit lo.

Problem 2.1

$\int_{0}^{\infty} e^{- 2 x} d x$ evaluate karo.

Recall Solution 2.1

Type I. $\infty$ ko $t$ se replace karo: $\int_{0}^{t} e^{- 2 x} d x = [- \frac{1}{2} e^{- 2 x}]_{0}^{t} = - \frac{1}{2} e^{- 2 t} + \frac{1}{2} .$ Jab $t \to \infty$ , $e^{- 2 t} \to 0$ , toh value hai $\frac{1}{2}$ . Converges.

Problem 2.2

$\int_{1}^{\infty} \frac{d x}{x ^{3/2}}$ evaluate karo.

Recall Solution 2.2

Type I, $p = \frac{3}{2} > 1$ → convergence expected. Directly: $\int_{1}^{t} x^{- 3/2} d x = [\frac{x ^{- 1/2}}{- 1/2}]_{1}^{t} = [- \frac{2}{x}]_{1}^{t} = - \frac{2}{t} + 2.$ Jab $t \to \infty$ , $\frac{2}{t} \to 0$ , toh value $2$ milti hai. (Formula $\frac{1}{p - 1} = \frac{1}{3/2 - 1} = 2$ se cross-check ✓.)

Problem 2.3

$\int_{0}^{4} \frac{d x}{4 - x}$ evaluate karo.

Recall Solution 2.3

Integrand $\frac{1}{4 - x} \to \infty$ jab $x \to 4^{-}$ (right endpoint). Type II. $t \to 4^{-}$ se creep karo: $\int_{0}^{t} (4 - x)^{- 1/2} d x .$ $u = 4 - x$ , $d u = - d x$ let karo. Antiderivative: $- (- 2) (4 - x)^{1/2} = ...$ dhyan se — $\int (4 - x)^{- 1/2} d x = - 2 4 - x$ . $[- 2 4 - x]_{0}^{t} = - 2 4 - t + 24 = - 2 4 - t + 4.$ Jab $t \to 4^{-}$ , $4 - t \to 0$ , toh value $= 4$ . Converges.

Problem 2.4

$\int_{0}^{1} ln x d x$ evaluate karo.

Recall Solution 2.4

$ln x \to - \infty$ jab $x \to 0^{+}$ : Type II left end par. $t \to 0^{+}$ se creep karo. Antiderivative (by parts): $\int ln x d x = x ln x - x$ . $\int_{t}^{1} ln x d x = [x ln x - x]_{t}^{1} = (0 - 1) - (t ln t - t) = - 1 - t ln t + t .$ Jab $t \to 0^{+}$ : $t \to 0$ , aur $t ln t \to 0$ ( $t$ slow $ln t \to - \infty$ ko crush kar deta hai; Limits at Infinity dekho). Toh value $= - 1$ . Converges. (Yahan negative answer bilkul theek hai — $ln x < 0$ on $(0, 1)$ , toh "area" genuinely axis ke neeche hai.)

Level 3 — Analysis

Goal: convergence decide karo jab clean antiderivative mushkil ho — comparison use karo.

Problem 3.1

Kya $\int_{1}^{\infty} \frac{d x}{x ^{2} + x}$ converge karta hai? Agar haan toh exact value nikalo.

Recall Solution 3.1

Do routes hain. Route A (comparison, quick verdict): $x \geq 1$ ke liye, $x^{2} + x \geq x^{2}$ , toh $\frac{1}{x ^{2} + x} \leq \frac{1}{x ^{2}}$ , aur $\int_{1}^{\infty} x^{- 2} d x$ converge karta hai ( $p = 2 > 1$ ). Comparison Test for Integrals se, chhota non-negative integral converges. Route B (exact value): partial fractions $\frac{1}{x ^{2} + x} = \frac{1}{x} - \frac{1}{x + 1}$ . $\int_{1}^{t} (\frac{1}{x} - \frac{1}{x + 1}) d x = [ln \frac{x}{x + 1}]_{1}^{t} = ln \frac{t}{t + 1} - ln \frac{1}{2} .$ Jab $t \to \infty$ , $\frac{t}{t + 1} \to 1$ toh $ln \to 0$ . Value $= 0 - ln \frac{1}{2} = ln 2 \approx 0.6931$ . Toh $ln 2$ .

Problem 3.2

Comparison use karke, bina integrate kiye, decide karo: $\int_{1}^{\infty} \frac{2 + sin x}{x} d x$ converge ya diverge karta hai?

Recall Solution 3.2

Kyunki $- 1 \leq sin x \leq 1$ , numerator satisfy karta hai $2 + sin x \geq 1 > 0$ . Toh $\frac{2 + s i n x}{x} \geq \frac{1}{x} (x \geq 1) .$ Lekin $\int_{1}^{\infty} \frac{d x}{x}$ diverges ( $p = 1$ ). Ek non-negative function jo ek divergent function se badi ho, woh bhi diverge karegi (uska area aur bhi bada hai). → Diverges. Direction matters: divergence prove karne ke liye aap neeche se bound karo kisi divergent cheez se; convergence prove karne ke liye upar se bound karo kisi convergent cheez se.

Problem 3.3

Kya $\int_{0}^{1} \frac{d x}{x + x ^{2}}$ converge karta hai?

Recall Solution 3.3

Bad point $x = 0$ par hai (denominator $\to 0$ ): Type II. $0$ ke paas, $x^{2}$ , $x$ ke muqable mein negligible hai, toh $x + x^{2} \geq x$ , jisse milta hai $\frac{1}{x + x ^{2}} \leq \frac{1}{x} = x^{- 1/2} .$ $\int_{0}^{1} x^{- 1/2} d x$ converges ( $p = \frac{1}{2} < 1$ , value $2$ ). Comparison se, chhota non-negative integral converges.

Level 4 — Synthesis

Goal: splitting, multiple bad points, aur dono types ek hi problem mein combine karo.

Problem 4.1

$\int_{- \infty}^{\infty} \frac{d x}{1 + x ^{2}}$ evaluate karo.

Recall Solution 4.1

Dono limits infinite hain → ek convenient $c = 0$ par split karo aur har half check karo. Antiderivative: $arctan x$ . $\int_{0}^{\infty} \frac{d x}{1 + x ^{2}} = lim_{t \to \infty} [arctan x]_{0}^{t} = lim_{t \to \infty} arctan t - 0 = \frac{π}{2} .$ $\int_{- \infty}^{0} \frac{d x}{1 + x ^{2}} = lim_{s \to - \infty} [arctan x]_{s}^{0} = 0 - (- \frac{π}{2}) = \frac{π}{2} .$ Dono halves converge karte hain, toh poora integral converge karta hai $\frac{π}{2} + \frac{π}{2} = π$ par.

Problem 4.2

$\int_{0}^{\infty} \frac{d x}{x ( 1 + x )}$ evaluate karo (dono ends par bad hai).

Recall Solution 4.2

Type II at $0$ ( $\frac{1}{x} \to \infty$ ) aur Type I at $\infty$ . $1$ par split karo: $\int_{0}^{1} + \int_{1}^{\infty}$ . Substitution dono ko unify karta hai. $x = u^{2}$ , $d x = 2 u d u$ , $x = u$ let karo. Phir $\int \frac{2 u d u}{u ( 1 + u ^{2} )} = \int \frac{2 d u}{1 + u ^{2}} = 2 arctan u = 2 arctan x .$ Jab $x$ , $0 \to \infty$ run karta hai, $x$ , $0 \to \infty$ run karta hai, aur $arctan$ , $0 \to \frac{π}{2}$ run karta hai: $\int_{0}^{\infty} \frac{d x}{x ( 1 + x )} = [2 arctan x]_{0}^{\infty} = 2 \cdot \frac{π}{2} - 0 = π .$ Substitution ke under dono endpoint behaviours tame hain → cleanly converges.

Problem 4.3

Kin real $p$ ke liye $\int_{0}^{\infty} \frac{d x}{x ^{p}}$ converge karta hai?

Recall Solution 4.3

Yeh ek disguised trap hai. $1$ par split karo: $\int_{0}^{1} x^{- p} d x + \int_{1}^{\infty} x^{- p} d x$ .

$\int_{0}^{1} x^{- p}$ converges $⟺ p < 1$ (Type II rule).
$\int_{1}^{\infty} x^{- p}$ converges $⟺ p > 1$ (Type I rule). Ye dono conditions mutually exclusive hain — koi ek $p$ dono satisfy nahi kar sakta. Isliye $\int_{0}^{\infty} x^{- p} d x$ har real $p$ ke liye diverge karta hai. Neeche di gayi picture dikhati hai kyun: jis bhi end ko aap tame karo, doosra end blow up ho jaata hai.

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

Level 5 — Mastery

Goal: subtle limits, Cauchy-principal-value trap, aur named functions se connections.

Problem 5.1

Kya $\int_{- \infty}^{\infty} x d x$ converge karta hai? "Symmetric limit" kya deta hai, aur woh misleading kyun hai?

Recall Solution 5.1

$0$ par split karo. Right half $\int_{0}^{\infty} x d x = lim_{t \to \infty} \frac{t ^{2}}{2} = \infty$ diverges. Kyunki ek piece diverge karta hai, poora integral diverges — bas, full stop. Tempting "symmetric limit" $lim_{t \to \infty} \int_{- t}^{t} x d x = lim_{t \to \infty} [\frac{x ^{2}}{2}]_{- t}^{t} = lim_{t \to \infty} (\frac{t ^{2}}{2} - \frac{t ^{2}}{2}) = 0$ . Yeh clean $0$ Cauchy principal value hai, convergence nahi: yeh sirf isliye kaam karta hai kyunki badhte negative aur positive areas ko same rate par cancel hone ke liye force kiya gaya hai. Genuine convergence demand karta hai ki har half apne aap settle ho. Answer: diverges (PV $= 0$ ek alag, weaker notion hai).

Problem 5.2

$\int_{0}^{\infty} x e^{- x} d x$ evaluate karo aur ise Gamma Function se connect karo.

Recall Solution 5.2

Type I. By parts integrate karo, $u = x, d v = e^{- x} d x$ : $\int x e^{- x} d x = - x e^{- x} - e^{- x} = - (x + 1) e^{- x}$ . $\int_{0}^{t} x e^{- x} d x = [- (x + 1) e^{- x}]_{0}^{t} = - (t + 1) e^{- t} + 1.$ Jab $t \to \infty$ , $(t + 1) e^{- t} \to 0$ (exponential linear factor ko beat kar deta hai). Value $= 1$ . Connection: Gamma Function hai $Γ (n) = \int_{0}^{\infty} x^{n - 1} e^{- x} d x$ , aur $Γ (n) = (n - 1)!$ . Yahan $n = 2$ , toh yeh $Γ (2) = 1! = 1$ hai ✓. Yahi integral mean of the exponential distribution bhi deta hai rate $1$ ke saath.

Problem 5.3

Constant $s > 0$ ke liye $\int_{0}^{\infty} e^{- s x} d x$ evaluate karo, aur batao $s \leq 0$ hone par kya hota hai. (Yeh ek miniature Laplace Transform hai.)

Recall Solution 5.3

$\int_{0}^{t} e^{- s x} d x = [- \frac{1}{s} e^{- s x}]_{0}^{t} = - \frac{1}{s} e^{- s t} + \frac{1}{s} .$

Agar $s > 0$ : $e^{- s t} \to 0$ jab $t \to \infty$ , value $= \frac{1}{s}$ . Converges.
Agar $s = 0$ : integrand $1$ hai, $\int_{0}^{t} 1 d x = t \to \infty$ . Diverges.
Agar $s < 0$ : exponent $- s x > 0$ grow karta hai, $e^{- s t} \to \infty$ . Diverges. Toh $\int_{0}^{\infty} e^{- s x} d x = \frac{1}{s}$ sirf $s > 0$ ke liye — yeh exactly $L {1} (s) = \frac{1}{s}$ hai, constant function $1$ ka Laplace Transform, exactly apne region of convergence par valid.

Recall Final self-check (verdict zyaban par laao)

$\int_{1}^{\infty} x^{- 3/2} d x$ ::: $2$ par converge karta hai ( $p = 3/2 > 1$ ) $\int_{0}^{1} ln x d x$ ::: $- 1$ par converge karta hai $\int_{- \infty}^{\infty} \frac{d x}{1 + x ^{2}}$ ::: $π$ par converge karta hai $\int_{0}^{\infty} x^{- p} d x$ ::: sabhi real $p$ ke liye diverge karta hai $\int_{0}^{\infty} e^{- s x} d x$ ::: $s > 0$ ke liye $1/ s$ hai, $s \leq 0$ ke liye diverge karta hai $\int_{- \infty}^{\infty} x d x$ ::: diverge karta hai (PV $= 0$ convergence nahi hai)