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

False — width infinite hai lekin height itni tezi se shrink ho sakti hai ki total area finite ho jaaye; e.g. ${\int }_1^{\infty }{x}^{-2}dx=1$. Convergence growing width aur shrinking height ke beech ek race hai.

False — zero tak decay necessary-ish hai lekin kaafi nahi hai; $1/x\to 0$ phir bhi ${\int }_1^{\infty }\frac{dx}{x}$ diverge karta hai kyunki yeh bahut slowly decay karta hai ($p=1$, the knife-edge).

False — exponent $p=0.999<1$ hai, isliye yeh diverge karta hai. Boundary $p=1$ se thoda sa neeche hona bhi "infinity par diverges" side mein hi aata hai.

True — zero ke paas rule flip ho jaata hai, aur $p=0.999<1$ bilkul wahi condition hai convergence ke liye, jo finite $\frac{1}{1-p}=1000$ deta hai.

False — positive curve ke neeche area sirf positive (ya $+\infty$) ho sakta hai; ek negative "answer" red flag hai ki tumne Fundamental Theorem of Calculus ko illegally ek blow-up ke across apply kiya.

True — jab ek baar har piece ek finite number ho, unka sum finite hai; aur proper splitting point ke liye total us specific legal $c$ se independent hai jise tum choose karte ho. Subtlety sirf yeh hai ki tumhe $c$ actually is jagah rakhna chahiye taaki har piece apna bad endpoint isolate kare.

False — har half ${\int }_0^{\infty }xdx$ diverge karta hai, isliye two-sided integral diverge karta hai; "$0$" sirf Cauchy principal value hai, convergence nahi.

True — agar $0\le g\le f$ aur chota $\int g$ diverge karta hai, to $\int f$ bhi diverge karta hai; comparison dono directions mein kaam karta hai depending on which side you bound.

True — chahe $\sin x$ sign mein oscillate karta ho, $\left|\frac{\sin x}{x^2}\right|\le \frac{1}{x^2}$, aur ${\int }_1^{\infty }{x}^{-2}$ converge karta hai; absolute convergence convergence force karta hai.

True — lekin sirf conditionally: yeh Dirichlet's test se converge karta hai (bounded oscillating $\sin$ against the monotone-to-zero factor $1/x$), jabki ${\int }_1^{\infty }\frac{|\sin x|}{x}dx$ diverge karta hai. Isliye yeh absolutely converge kiye bina converge karta hai.

Integrand $x=0$ par blow up karta hai, jo $[-1,1]$ ke andar hai, isliye Fundamental Theorem of Calculus iske across apply nahi hota. Tumhe $0$ par split karna hoga aur limits lene honge — sirf right half hai ${\lim }_{t\to 0^{+}}{\int }_t^1{x}^{-2}dx={\lim }_{t\to 0^{+}}\left(\frac{1}{t}-1\right)=+\infty$, isliye yeh diverge karta hai; true value $+\infty$ hai, aur ek positive function ke liye negative answer impossible hai.

Kyunki tum $\infty$ ko $-e^{-x}$ mein aise substitute nahi kar sakte jaise woh koi number ho; tumhe ${\lim }_{t\to \infty }\left(-e^{-t}+1\right)$ likhna hoga aur observe karna hoga ki $e^{-t}\to 0$. Infinity plug in karne ki yahi aadat logon ko divergent cases mein galat handle karne par majboor karti hai.

Antiderivative hona kaafi nahi hai — tumhe uski limit evaluate karni hogi: ${\lim }_{t\to \infty }\ln t=\infty$, isliye yeh diverge karta hai. $\ln t$ ki unbounded growth exactly woh divergence hai, aur isliye "$\ln \infty$" ko value ke roop mein likhna meaningless hai.

Comparison direction useless hai: apni function ko ek convergent se neeche bound karna kuch prove nahi karta. Convergence conclude karne ke liye tumhe $\frac{1}{x^3+1}\le \frac{1}{x^3}$ chahiye (ek convergent se upar bound karo).

Finite interval ise proper nahi banata — integrand endpoint $0$ par blow up karta hai, jo ise Type II banata hai; limit ${\lim }_{t\to 0^{+}}(-\ln t)=+\infty$ diverge karta hai.

Nahi — woh symmetric limit principal value hai; true definition kisi $c$ par split karti hai aur require karti hai ki har one-sided piece independently converge kare.

Kyunki Fundamental Theorem of Calculus sirf ek closed, bounded interval par continuous integrand ke saath kaam karta hai; $t$ hume legal ground par rakhta hai, aur limit phir poochti hai ki area kisi number par settle hota hai ya nahi.

Kyunki $\int x^{-1}dx=\ln x$, aur $\ln$ dono ends par bahut slowly-phir-bhi-surely grow/shrink karta hai: $\ln t\to \infty$ as $t\to \infty$ aur $\ln t\to -\infty$ as $t\to 0^{+}$. Yeh woh knife-edge hai jahan power rule toot jaata hai.

Alag alag enemies: infinity par tum slow decay se ladte ho, isliye tumhe tail ko khatam karne ke liye ek bada exponent chahiye; zero ke paas tum ek tall spike se ladte ho, isliye tumhe spike ko gentle rakhne ke liye ek chota exponent chahiye.

Kyunki "convergence" ka matlab hai ki total area ek well-defined finite number hai chahe dono limits apne bad points ke kaafi paas kaise bhi jaayein; agar ek half $+\infty$ hai, to ek alag $-\infty$ half ke saath koi legitimate cancellation allowed nahi hai.

Logic "bade finite area ke andar chota area finite hai" sirf tabhi hold karta hai jab koi sign cancellation na ho; mixed signs ke saath ek chhota-dikhne-wala function phir bhi oscillate kar sakta hai aur alag behave kar sakta hai, isliye tum magnitudes compare karte ho ya Dirichlet's jaisa oscillation test use karte ho (dekho Comparison Test for Integrals).

Kyunki convergence yahan cancellation se aati hai: consecutive positive aur negative humps almost cancel ho jaate hain, aur unka leftover $1/x$ ki tarah shrink karta hai. Dirichlet's test ise capture karta hai — ek bounded oscillation times ek monotone factor jo $0$ ki taraf decrease karta hai converge karta hai — ek mechanism jo comparison, jo sign ko ignore karta hai, simply nahi dekh sakta.

Woh finite area exactly wahi hai jo $e^{-x}$ ko $[0,\infty )$ par ek valid probability density banata hai (yeh $1$ tak integrate hota hai), aur wahi convergence Laplace Transform aur Gamma Function ko power karti hai.

Yeh top par diverge karta hai. $u=\ln x$ substitute karo, to $du=\frac{dx}{x}$ milta hai, jo integral ko $\int \frac{du}{u}$ mein badal deta hai jo $\infty$ tak jaata hai — disguise mein $p=1$ divergent integral; extra $\ln x$ ise bachane ke liye bahut weak hai. (Yeh $x=1$ par bhi improper hai jahan $\ln x\to 0$, isliye careful treatment integral ko kisi $x=c>1$ se shuru karti hai.)

Yeh har $p$ ke liye diverge karta hai — koi bhi $p$ dono "$p>1$ at infinity" aur "$p<1$ at zero" simultaneously satisfy nahi kar sakta, isliye ek end hamesha ise khatam kar deta hai. Tumhe $1$ par split karna hoga aur har piece ko alag analyze karna hoga.

Haan — e.g. ${\int }_0^{\infty }\frac{dx}{\sqrt{x}(1+x)}$ mein ek infinite upper limit hai (Type I) aur $0$ par blow-up bhi hai (Type II); kisi convenient point jaise $1$ par split karo aur har end alag check karo.

Haan aur yahi punchline hai — $1/x$ dono ends par diverge karta hai, near-zero aur near-infinity, isliye $p=1$ ko lonely loser kaha jaata hai: yeh har front par fail karta hai. Discrete mirror image ke liye p-series and Integral Test dekho.

Haan — $u=\ln x$, $du=\frac{dx}{x}$ ke saath, aur lower limit $u=\ln e=1$ ban jaati hai, isliye yeh ${\int }_1^{\infty }{u}^{-2}du=1$ mein badal jaata hai. Squared log just enough hai converge karne ke liye, jahan single log fail ho gaya; hum deliberately $e$ se shuru karte hain taaki $u\ge 1$ ho aur sirf ek hi issue infinite upper limit ka ho.

Kyunki $x=1$ se shuru karne par $\ln x\to 0^{+}$ hota hai, isliye $\frac{1}{x(\ln x{)}^2}\to +\infty$ wahan — integral lower end par bhi Type II hai. Wahi substitution $u=\ln x$, $du=\frac{dx}{x}$ ise ${\int }_0^{\infty }{u}^{-2}du$ mein badal deta hai, jiska piece ${\int }_0^1{u}^{-2}du$ diverge karta hai. Tumhe $x=1$ par singularity ko isolate aur check karna hoga celebrate karne se pehle.

Zaruri nahi. Concretely, maano $f$ har jagah $0$ hai sivaay ek triangular spike ke jo height $1$ aur base-width $2^{-n}$ ka hai jo har integer $x=n$ par centred hai; har spike ka area $\frac{1}{2}\cdot 2^{-n}$ hai, isliye total area $\sum \frac{1}{2}{2}^{-n}<\infty$ hai (converge karta hai), phir bhi $f(n)=1$ har $n$ ke liye hai, isliye $f$ nahi tend karta $0$ ki taraf. Monotone ya otherwise "nice" $f$ ke liye, lekin, convergence zarur $f\to 0$ force karti hai.

$u=\ln x$, $du=\frac{dx}{x}$ substitute karo: yeh ${\int }_0^{\infty }{u}^{-1/2}du$ ban jaata hai, jiska tail ${\int }_1^{\infty }{u}^{-1/2}du$ mein $p=\frac{1}{2}<1$ hai aur isliye diverge karta hai. Infinity tak grow karne wala ek factor kaafi nahi hai — use $\ln x$ ki $>1$ power ki tarah grow karna chahiye (jaise $(\ln x{)}^2$ ne kiya tha).