Exercises — Convergence tests for improper integrals — comparison

Yahan $f(x)=\frac{1}{x^3+7}$ hai. Jab $x\to \infty$ tab $+7$, $x^3$ ke saamne dwarf ho jaata hai, isliye $f$ behave karta hai $==1/x^3==$ ki tarah. Benchmark $g(x)=1/x^3$ lo. Kyunki $x^3+7>x^3$ hai, hume milta hai $\frac{1}{x^3+7}<\frac{1}{x^3}$ — yaani $f$, $g$ ke neeche baitha hai. ${\int }_1^{\infty }{x}^{-3}dx$ converge karta hai ($p=3>1$), aur hum upar se ek convergent benchmark se bound karte hain. Verdict: converges (top figure ka left panel).

Integrand $f(x)=\frac{x}{x^2+1}$ hai. Bade $x$ ke liye, numerator $\approx x$ hai, denominator $\approx x^2$ hai, isliye $f\approx x/x^2=1/x$ hai. Benchmark $g(x)=1/x$ hai. ${\int }_1^{\infty }{x}^{-1}dx$ diverge karta hai ($p=1$). Hum expect karte hain ki hamara integral diverge kare — isse prove karne ke liye hume $f$ ko $1/x$ ke kisi multiple se neeche se bound karna hoga (top figure ka right panel). (Full proof L2 mein hai.)

Integrand $f(x)=\frac{1}{x^{3/2}+\sqrt{x}}$ hai. Denominator mein dono powers compare karo: $x^{3/2}$ vs $x^{1/2}$. Badi power jab $x\to \infty$ tab dominate karti hai, isliye denominator $\approx x^{3/2}$ aur $f\approx x^{-3/2}$ hai. Benchmark $g(x)=x^{-3/2}$ hai, jo converge karta hai ($p=3/2>1$). Expect karo convergence.

Integrand $f(x)=\frac{1}{x^3+7}$, benchmark $g(x)=\frac{1}{x^3}$ hai. $[1,\infty )$ par, $f\ge 0$ ✅ (DCT ke liye nonnegativity zaroori hai). $x^3+7>x^3⟹0\le \frac{1}{x^3+7}<\frac{1}{x^3}.$ ${\int }_1^{\infty }\frac{dx}{x^3}$ converge karta hai ($p=3>1$). Isliye area-so-far $F(t)={\int }_1^{t}f$ increasing hai (kyunki $f\ge 0$) aur $g$ ki finite area se upar bounded hai. Monotone Convergence Theorem ke by (jo upar stated hai), $F(t)$ ek finite limit par settle ho jaata hai. Converges. ✅ (DCT strategy hai; Monotone Convergence woh reason hai ki yeh kaam kyun karta hai.)

Integrand $f(x)=\frac{x}{x^2+1}$, benchmark $g(x)=1/x$ hai (dono $f,g>0$ hain). $L={\lim }_{x\to \infty }\frac{f}{g}={\lim }_{x\to \infty }\frac{x/(x^2+1)}{1/x}={\lim }_{x\to \infty }\frac{x^2}{x^2+1}=1.$ Kyunki $0<L=1<\infty$ hai, $f$ aur $g$ same fate share karte hain. ${\int }_1^{\infty }1/xdx$ diverge karta hai, isliye hamara integral diverge karta hai. ❌ (Cross-check: ${\int }_1^t\frac{x}{x^2+1}dx=\frac{1}{2}\ln (x^2+1){|}_1^t=\frac{1}{2}\ln \frac{t^2+1}{2}\to \infty$. Consistent.)

Integrand $f(x)=\frac{1}{x^{3/2}+\sqrt{x}}$, benchmark $g(x)=x^{-3/2}$ hai. $L={\lim }_{x\to \infty }\frac{1/(x^{3/2}+\sqrt{x})}{x^{-3/2}}={\lim }_{x\to \infty }\frac{x^{3/2}}{x^{3/2}+x^{1/2}}={\lim }_{x\to \infty }\frac{1}{1+x^{-1}}=1.$ (Kyun $x^{3/2}$ se divide kiya? Leading term expose karne ke liye.) $0<L=1<\infty$ hai, $\int x^{-3/2}$ jaisa hi fate, jo converge karta hai ($p=3/2>1$). Converges. ✅

Integrand $f(x)=\frac{2+\cos x}{x^3}$ hai. $\cos x$ wiggle karta hai lekin caged hai: $-1\le \cos x\le 1$, isliye $1\le 2+\cos x\le 3$ hai. Isliye $0\le \frac{2+\cos x}{x^3}\le \frac{3}{x^3}.$ Benchmark $g(x)=3/x^3$ se upar se bound karo (convergence prove karne ke liye); ${\int }_1^{\infty }3x^{-3}dx$ converge karta hai ($p=3>1$). DCT se, converges. ✅

$x$ ki powers yeh settle nahi kareingi — $\ln x$ har positive power se slower grow karta hai, isliye $x^{-p}$ ke against DCT inconclusive hai. Substitution $u=\ln x$, $du=dx/x$ se directly integrate karo: $\int \frac{dx}{x\ln x}=\int \frac{du}{u}=\ln |u|=\ln (\ln x).$ ${\int }_2^t\frac{dx}{x\ln x}=\ln (\ln t)-\ln (\ln 2)\underset{t\to \infty }{\to }\infty .$ Diverges. ❌ (Yeh Integral Test for series mein log series ka integral-test cousin hai.)

Integrand $f(x)=\frac{\ln x}{x^2}$ hai. Dhyan se: $\ln x\to \infty$ hai, toh kya $f$ sach mein chhota hai? Haan — $\ln x$ kisi bhi power se slower grow karta hai, isliye hum $x^2$ ka ek zara sa tukda sacrifice kar sakte hain log ko swallow karne ke liye. Key inequality $\ln x\le \sqrt{x}$ SARE $x\ge 1$ ke liye (sirf asymptotically nahi): define karo $h(x)=\sqrt{x}-\ln x$. Tab $h(1)=1-0=1>0$ hai, aur $h^{\prime }(x)=\frac{1}{2\sqrt{x}}-\frac{1}{x}=\frac{\sqrt{x}-2}{2x}$ hai. Isliye $h^{\prime }(x)<0$ on $[1,4)$ aur $h^{\prime }(x)>0$ on $(4,\infty )$: $h$ ek minimum at $x=4$ par dip karta hai jahan $h(4)=2-\ln 4\approx 2-1.386=0.614>0$ hai. Kyunki $[1,\infty )$ par global minimum positive hai, $h(x)>0$ har jagah hai, yaani $==\ln x<\sqrt{x}\text{for all }x\ge 1==$ hai. Koi hand-waving nahi chahiye. Isliye $0\le \frac{\ln x}{x^2}\le \frac{x^{1/2}}{x^2}=\frac{1}{x^{3/2}}.$ Benchmark $g(x)=x^{-3/2}$ se upar se bound karo; ${\int }_1^{\infty }{x}^{-3/2}dx$ converge karta hai ($p=3/2>1$). DCT se, converges. ✅ Exactly $1$ kyun? (direct check). Convergence ab prove ho gayi, lekin hum integration by parts se value pin kar sakte hain jahan $u=\ln x,dv=x^{-2}dx$ hai, isliye $du=\frac{dx}{x},v=-x^{-1}$ hai: $\int \frac{\ln x}{x^2}dx=-\frac{\ln x}{x}+\int \frac{dx}{x^2}=-\frac{\ln x}{x}-\frac{1}{x}.$ $1$ se $t$ tak evaluate karo: lower limit par ($x=1$) yeh $-0-1=-1$ hai; upper limit par, $-\frac{\ln t}{t}-\frac{1}{t}\to 0$ jab $t\to \infty$ (dono terms $\to 0$ kyunki $x$, $\ln x$ ko beat karta hai). Toh integral $=0-(-1)=\boxed{1}$ hai. ✅ (VERIFY mein checked hai.)

Do dangers hain, isliye $x=1$ par split karo: $x=0^{+}$ ke paas $\sqrt{x}\to 0$ hota hai jo integrand ko blow up karta hai (ek type-2 improper point, Improper integrals — infinite discontinuities (type 2) dekho); $x=\infty$ ke paas yeh type-1 hai. ${\int }_0^{\infty }=\underset{\text{type 2 at }0}{\underbrace{{\int }_0^1}}+\underset{\text{type 1}}{\underbrace{{\int }_1^{\infty }}}.$ 0 ke paas wala piece. Pehle note karo ki $0<x\le 1$ ke liye $x^2\le \sqrt{x}$ hai, isliye $x^2+\sqrt{x}\le 2\sqrt{x}$ hai; yeh integrand par ek lower bound hai ($\frac{1}{x^2+\sqrt{x}}\ge \frac{1}{2\sqrt{x}}$ hai). Lower bound yahaan useless kyun hai? Convergence prove karne ke liye hume area ko upar se cap karna hota hai ($f$ ko kisi finite cheez ke neeche squeeze karo); lower bound sirf yeh kehta hai ki area at least kuch hai, jo kabhi finiteness guarantee nahi kar sakta. Isliye hum ise discard karte hain aur upper bound ki taraf jaate hain: $x^2+\sqrt{x}\ge \sqrt{x}$ deta hai $\frac{1}{x^2+\sqrt{x}}\le \frac{1}{\sqrt{x}}=x^{-1/2}$. Ab ${\int }_0^1{x}^{-1/2}dx=2$ converges (${\int }_0^1{x}^{-p}$ converge karta hai jab $p<1$ ho; yahaan $p=\frac{1}{2}$ hai — type-2 $p$-test flip ho jaata hai!). Isliye 0 ke paas wala piece converge karta hai. $\infty$ ke paas wala piece: $x\ge 1$ ke liye, $x^2+\sqrt{x}\ge x^2$ hai, isliye integrand $\le x^{-2}$ hai; ${\int }_1^{\infty }{x}^{-2}dx$ converges. Dono pieces finite $\Rightarrow$ poora integral converges. ✅

Substitute karo $u=\ln x$, $du=dx/x$; jab $x:2\to \infty$, $u:\ln 2\to \infty$ hota hai: ${\int }_2^{\infty }\frac{dx}{x(\ln x{)}^p}={\int }_{\ln 2}^{\infty }\frac{du}{u^p}.$ Yeh $u$ mein exactly p-integral hai! Yeh converge karta hai $⟺p>1$. $\boxed{\text{Converges}⟺p>1.}$ ($p=1$ ke liye hum L3.2 wapas paate hain, jo diverge karta hai — consistent.)

Integrand $f(x)=\frac{1}{x^2\ln (x+1)}$, benchmark $g(x)=1/x^2$ hai. $L={\lim }_{x\to \infty }\frac{1/(x^2\ln (x+1))}{1/x^2}={\lim }_{x\to \infty }\frac{1}{\ln (x+1)}=0.$ $L=0$ LCT ka degenerate case hai: yeh sirf one-way implication deta hai "agar $\int g$ converge kare toh $\int f$ converge karta hai." Yahaan ${\int }_1^{\infty }{x}^{-2}dx$ converges, aur DCT step close karne ke liye hume abhi bhi $f\le g$ chahiye. $\ln (x+1)\ge 1$ kab hota hai? Exactly tab jab $x+1\ge e$ ho, yaani $x\ge e-1\approx 1.718$ ho. $[1,\infty )$ par yeh sab $x\ge 2$ ke liye hold karta hai (aur $\ln (x+1)$ increasing hai, isliye yeh kabhi $1$ se neeche nahi jaata). Isliye $x\ge 2$ ke liye, $\frac{1}{x^2\ln (x+1)}\le \frac{1}{x^2}$ hai, aur ${\int }_1^{2}f$ ek continuous function ka proper (finite) integral hai. Isliye $f$ bhi converge karta hai. ✅ (One-way kyun? $L=0$ ka matlab hai $f$ eventually $g$ se bahut chhota hai. Chhota-under-convergent ⇒ convergent, lekin "chhota" kuch nahi kehta agar $g$ diverge kare.)

Integrand $f(x)=\frac{\ln (x+1)}{x^2}$, benchmark $g(x)=1/x^2$ hai. $L={\lim }_{x\to \infty }\frac{f}{g}={\lim }_{x\to \infty }\frac{\ln (x+1)/x^2}{1/x^2}={\lim }_{x\to \infty }\ln (x+1)=\infty .$ $L=\infty$ doosra degenerate case hai: yeh sirf one-way implication deta hai "agar $\int g$ diverge kare toh $\int f$ diverge karta hai" (kyunki $f$ eventually $g$ se bahut bada hai). Lekin yahaan $\int g=\int x^{-2}$ converges, isliye yeh rule silent hai — yeh kuch nahi kehta! Fix: ek bade benchmark ki taraf switch karo jo abhi bhi converge kare. Lo $g_2(x)=x^{-3/2}$. Bound. $x\ge 1$ ke liye, kyunki $\ln$ increasing hai aur $x+1\le 2x$ hai, $\ln (x+1)\le \ln (2x)=\ln 2+\ln x$ hai. L3.3 se, $\ln x\le \sqrt{x}$ $x\ge 1$ ke liye, aur $\ln 2\le \sqrt{x}$ bhi tab jab $x\ge (\ln 2{)}^2\approx 0.48$ ho (sab $x\ge 1$ ke liye true hai). Isliye $x\ge 1$ ke liye, $\ln (x+1)\le \ln 2+\ln x\le \sqrt{x}+\sqrt{x}=2\sqrt{x}.$ Isliye $0\le \frac{\ln (x+1)}{x^2}\le \frac{2\sqrt{x}}{x^2}=\frac{2}{x^{3/2}}.$ $2x^{-3/2}$ se upar se bound karo (top figure ka left panel); ${\int }_1^{\infty }2x^{-3/2}dx$ converges ($p=3/2>1$). DCT se, ${\int }_1^{\infty }\frac{\ln (x+1)}{x^2}dx$ converges. ✅ (Moral: ek convergent $g$ ke against $L=\infty$ useless hai; tumhe ek bada benchmark dhundhna hai jo abhi bhi converge kare — aur $x^{-3/2}$, jiska $p$ abhi bhi $1$ se upar hai, exactly slow log ko absorb karne ki room rakhta hai.)

DCT ko $f\ge 0$ chahiye; yahaan $\sin x$ sign mein oscillate karta hai, isliye DCT directly apply nahi hota. Absolute comparison use karo (dekho Absolute vs conditional convergence): $\left|\frac{\sin x}{x^2}\right|=\frac{|\sin x|}{x^2}\le \frac{1}{x^2}.$ ${\int }_1^{\infty }{x}^{-2}dx$ converge karta hai $\Rightarrow {\int }_1^{\infty }\left|\frac{\sin x}{x^2}\right|dx$ converge karta hai (nonnegative $|f|$ par DCT). Absolute convergence $\Rightarrow$ convergence, isliye ${\int }_1^{\infty }\frac{\sin x}{x^2}dx$ converges. ✅

Denominator mein do powers hain: $x^a$ aur $x^1$. Bada exponent tail dominate karta hai — neeche di gayi figure mein red curve track karta hai ki kaunsa benchmark jeetta hai.

Case $a>1$: $x^a$ term jeetta hai. $g=x^{-a}$ ke saath LCT: $L={\lim }_{x\to \infty }\frac{x^a}{x^a+x}={\lim }_{x\to \infty }\frac{1}{1+x^{1-a}}=1(\text{since }1-a<0).$ Same fate as $\int x^{-a}$, $a>1$ $\Rightarrow$ converges.

Case $a=1$: denominator $=x+x=2x$ hai, integrand $=\frac{1}{2x}$ hai; ${\int }_1^{\infty }\frac{dx}{2x}=\frac{1}{2}\ln x{|}_1^{\infty }=\infty$ $\Rightarrow$ diverges.

Case $0<a<1$: ab $x^1$ term jeetta hai. $g=x^{-1}$ ke saath LCT: $L={\lim }_{x\to \infty }\frac{x}{x^a+x}={\lim }_{x\to \infty }\frac{1}{x^{a-1}+1}=1(\text{since }a-1<0).$ Same fate as $\int x^{-1}$ $\Rightarrow$ diverges.

Summary: $\boxed{{\int }_1^{\infty }\frac{dx}{x^a+x}\text{converges}⟺a>1.}$

Neeche di gayi figure $f(x)=1/(x^2+x)$ (red) ko convergent case $a=2$ ke liye apne winning benchmark $1/x^2$ ko hug karte dikhati hai — LCT limit $L=1$ ke peeche geometric picture.

Trouble kahan hai? ::: $\infty$ par (type 1) ya ek finite blow-up point par (type 2) — agar dono hain toh integral split karo. Tail $p$-rule vs finite-point $p$-rule? ::: ${\int }_1^{\infty }{x}^{-p}$ ko $p>1$ chahiye; ${\int }_0^1{x}^{-p}$ ko $p<1$ chahiye (flipped). Kaunsi benchmark power? ::: Denominator mein fastest-growing term tail ko control karta hai. $f$ sign change karta hai? ::: Raw DCT nahi, $|f|\le g$ (absolute comparison) use karo. LCT $L=0$ deta hai? ::: Sirf "$\int g$ converges $\Rightarrow \int f$ converges" survive karta hai (L5.1). LCT $L=\infty$ deta hai? ::: Sirf "$\int g$ diverges $\Rightarrow \int f$ diverges" survive karta hai; convergent $g$ ke against yeh silent hai — benchmark badlo (L5.2). Direction of DCT inequality? ::: Squeeze to please (convergent ke neeche), push to die (divergent ke upar) — page ke upar wali picture.