Exercises — Convergence tests for improper integrals — comparison

Here $f(x)=\frac{1}{x^3+7}$. As $x\to \infty$ the $+7$ is dwarfed by $x^3$, so $f$ behaves like $==1/x^3==$. Take benchmark $g(x)=1/x^3$. Since $x^3+7>x^3$, we get $\frac{1}{x^3+7}<\frac{1}{x^3}$ — that is, $f$ sits below $g$. ${\int }_1^{\infty }{x}^{-3}dx$ converges ($p=3>1$), and we bound above by a convergent benchmark. Verdict: converges (left panel of the top figure).

Integrand $f(x)=\frac{x}{x^2+1}$. For large $x$, numerator $\approx x$, denominator $\approx x^2$, so $f\approx x/x^2=1/x$. Benchmark $g(x)=1/x$. ${\int }_1^{\infty }{x}^{-1}dx$ diverges ($p=1$). We expect our integral to diverge — to prove it we'll need to bound $f$ below by a multiple of $1/x$ (right panel of the top figure). (Full proof in L2.)

Integrand $f(x)=\frac{1}{x^{3/2}+\sqrt{x}}$. Compare the two powers in the denominator: $x^{3/2}$ vs $x^{1/2}$. The larger power dominates as $x\to \infty$, so denominator $\approx x^{3/2}$ and $f\approx x^{-3/2}$. Benchmark $g(x)=x^{-3/2}$, which converges ($p=3/2>1$). Expect convergence.

Integrand $f(x)=\frac{1}{x^3+7}$, benchmark $g(x)=\frac{1}{x^3}$. On $[1,\infty )$, $f\ge 0$ ✅ (DCT requires nonnegativity). $x^3+7>x^3⟹0\le \frac{1}{x^3+7}<\frac{1}{x^3}.$ ${\int }_1^{\infty }\frac{dx}{x^3}$ converges ($p=3>1$). So the area-so-far $F(t)={\int }_1^{t}f$ is increasing (because $f\ge 0$) and bounded above by the finite area of $g$. By the Monotone Convergence Theorem (stated at the top), $F(t)$ settles to a finite limit. Converges. ✅ (DCT is the strategy; Monotone Convergence is why it works.)

Integrand $f(x)=\frac{x}{x^2+1}$, benchmark $g(x)=1/x$ (both $f,g>0$). $L={\lim }_{x\to \infty }\frac{f}{g}={\lim }_{x\to \infty }\frac{x/(x^2+1)}{1/x}={\lim }_{x\to \infty }\frac{x^2}{x^2+1}=1.$ Since $0<L=1<\infty$, $f$ and $g$ share the same fate. ${\int }_1^{\infty }1/xdx$ diverges, so our integral diverges. ❌ (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 $g(x)=x^{-3/2}$. $L={\lim }_{x\to \infty }\frac{1/(x^{3/2}+\sqrt{x})}{x^{-3/2}}={\lim }_{x\to \infty }\frac{x^{3/2}}{x^{3/2}+x^{1/2}}={\lim }_{x\to \infty }\frac{1}{1+x^{-1}}=1.$ (Why divide by $x^{3/2}$? To expose the leading term.) $0<L=1<\infty$, same fate as $\int x^{-3/2}$, which converges ($p=3/2>1$). Converges. ✅

Integrand $f(x)=\frac{2+\cos x}{x^3}$. $\cos x$ wiggles but is caged: $-1\le \cos x\le 1$, so $1\le 2+\cos x\le 3$. Thus $0\le \frac{2+\cos x}{x^3}\le \frac{3}{x^3}.$ Bound above (proving convergence) by benchmark $g(x)=3/x^3$; ${\int }_1^{\infty }3x^{-3}dx$ converges ($p=3>1$). By DCT, converges. ✅

Powers of $x$ won't settle this — $\ln x$ grows slower than every positive power, so DCT against $x^{-p}$ is inconclusive. Integrate directly via substitution $u=\ln x$, $du=dx/x$: $\int \frac{dx}{x\ln x}=\int \frac{du}{u}=\ln |u|=\ln (\ln x).$ ${\int }_2^t\frac{dx}{x\ln x}=\ln (\ln t)-\ln (\ln 2)\underset{t\to \infty }{\to }\infty .$ Diverges. ❌ (This is the integral-test cousin of the log series in Integral Test for series.)

Integrand $f(x)=\frac{\ln x}{x^2}$. Careful: $\ln x\to \infty$, so is $f$ really small? Yes — $\ln x$ grows slower than any power, so we can sacrifice a sliver of the $x^2$ to swallow the log. The key inequality $\ln x\le \sqrt{x}$ for ALL $x\ge 1$ (not just asymptotically): define $h(x)=\sqrt{x}-\ln x$. Then $h(1)=1-0=1>0$, and $h^{\prime }(x)=\frac{1}{2\sqrt{x}}-\frac{1}{x}=\frac{\sqrt{x}-2}{2x}$. So $h^{\prime }(x)<0$ on $[1,4)$ and $h^{\prime }(x)>0$ on $(4,\infty )$: $h$ dips to a minimum at $x=4$ where $h(4)=2-\ln 4\approx 2-1.386=0.614>0$. Since the global minimum on $[1,\infty )$ is positive, $h(x)>0$ everywhere, i.e. $==\ln x<\sqrt{x}\text{for all }x\ge 1==$. No hand-waving needed. Therefore $0\le \frac{\ln x}{x^2}\le \frac{x^{1/2}}{x^2}=\frac{1}{x^{3/2}}.$ Bound above by benchmark $g(x)=x^{-3/2}$; ${\int }_1^{\infty }{x}^{-3/2}dx$ converges ($p=3/2>1$). By DCT, converges. ✅ Why exactly $1$? (the direct check). Convergence is now proved, but we can pin the value by integration by parts with $u=\ln x,dv=x^{-2}dx$, so $du=\frac{dx}{x},v=-x^{-1}$: $\int \frac{\ln x}{x^2}dx=-\frac{\ln x}{x}+\int \frac{dx}{x^2}=-\frac{\ln x}{x}-\frac{1}{x}.$ Evaluate from $1$ to $t$: at the lower limit ($x=1$) this is $-0-1=-1$; at the upper limit, $-\frac{\ln t}{t}-\frac{1}{t}\to 0$ as $t\to \infty$ (both terms $\to 0$ since $x$ beats $\ln x$). So the integral $=0-(-1)=\boxed{1}$. ✅ (checked in VERIFY.)

Two dangers, so split at $x=1$: near $x=0^{+}$ the term $\sqrt{x}\to 0$ makes the integrand blow up (a type-2 improper point, see Improper integrals — infinite discontinuities (type 2)); near $x=\infty$ it's type-1. ${\int }_0^{\infty }=\underset{\text{type 2 at }0}{\underbrace{{\int }_0^1}}+\underset{\text{type 1}}{\underbrace{{\int }_1^{\infty }}}.$ Piece near 0. First note that for $0<x\le 1$ we have $x^2\le \sqrt{x}$, hence $x^2+\sqrt{x}\le 2\sqrt{x}$; this is a lower bound on the integrand ($\frac{1}{x^2+\sqrt{x}}\ge \frac{1}{2\sqrt{x}}$). Why is a lower bound useless here? To prove convergence we need to cap the area from above (squeeze $f$ under something finite); a lower bound only says the area is at least something, which can never guarantee finiteness. So we discard it and reach for the upper bound instead: $x^2+\sqrt{x}\ge \sqrt{x}$ gives $\frac{1}{x^2+\sqrt{x}}\le \frac{1}{\sqrt{x}}=x^{-1/2}$. Now ${\int }_0^1{x}^{-1/2}dx=2$ converges (${\int }_0^1{x}^{-p}$ converges when $p<1$; here $p=\frac{1}{2}$ — the type-2 $p$-test flips!). So the near-0 piece converges. Piece near ∞: for $x\ge 1$, $x^2+\sqrt{x}\ge x^2$, so integrand $\le x^{-2}$; ${\int }_1^{\infty }{x}^{-2}dx$ converges. Both pieces finite $\Rightarrow$ whole integral converges. ✅

Substitute $u=\ln x$, $du=dx/x$; as $x:2\to \infty$, $u:\ln 2\to \infty$: ${\int }_2^{\infty }\frac{dx}{x(\ln x{)}^p}={\int }_{\ln 2}^{\infty }\frac{du}{u^p}.$ This is exactly the p-integral in $u$! It converges $⟺p>1$. $\boxed{\text{Converges}⟺p>1.}$ (For $p=1$ we recover L3.2, which diverges — consistent.)

Integrand $f(x)=\frac{1}{x^2\ln (x+1)}$, benchmark $g(x)=1/x^2$. $L={\lim }_{x\to \infty }\frac{1/(x^2\ln (x+1))}{1/x^2}={\lim }_{x\to \infty }\frac{1}{\ln (x+1)}=0.$ $L=0$ is the degenerate LCT case: it gives only the one-way implication "if $\int g$ converges then $\int f$ converges." Here ${\int }_1^{\infty }{x}^{-2}dx$ converges, and we still need $f\le g$ for the DCT step to close. When is $\ln (x+1)\ge 1$? Exactly when $x+1\ge e$, i.e. $x\ge e-1\approx 1.718$. On $[1,\infty )$ this holds for all $x\ge 2$ (and $\ln (x+1)$ is increasing, so it never dips back below $1$). Thus for $x\ge 2$, $\frac{1}{x^2\ln (x+1)}\le \frac{1}{x^2}$, and ${\int }_1^{2}f$ is a proper (finite) integral of a continuous function. So $f$ converges too. ✅ (Why one-way? $L=0$ means $f$ is eventually much smaller than $g$. Small-under-convergent ⇒ convergent, but "small" tells us nothing if $g$ diverges.)

Integrand $f(x)=\frac{\ln (x+1)}{x^2}$, benchmark $g(x)=1/x^2$. $L={\lim }_{x\to \infty }\frac{f}{g}={\lim }_{x\to \infty }\frac{\ln (x+1)/x^2}{1/x^2}={\lim }_{x\to \infty }\ln (x+1)=\infty .$ $L=\infty$ is the other degenerate case: it gives only the one-way implication "if $\int g$ diverges then $\int f$ diverges" (because $f$ is eventually much bigger than $g$). But here $\int g=\int x^{-2}$ converges, so this rule is silent — it tells us nothing! Fix: switch to a bigger benchmark that still converges. Take $g_2(x)=x^{-3/2}$. The bound. For $x\ge 1$, $x+1\le x^2$... no — instead bound the log directly. Since $\ln$ is increasing and $x+1\le 2x$ for $x\ge 1$, $\ln (x+1)\le \ln (2x)=\ln 2+\ln x$. From L3.3, $\ln x\le \sqrt{x}$ for $x\ge 1$, and also $\ln 2\le \sqrt{x}$ once $x\ge (\ln 2{)}^2\approx 0.48$ (true for all $x\ge 1$). Hence for $x\ge 1$, $\ln (x+1)\le \ln 2+\ln x\le \sqrt{x}+\sqrt{x}=2\sqrt{x}.$ Therefore $0\le \frac{\ln (x+1)}{x^2}\le \frac{2\sqrt{x}}{x^2}=\frac{2}{x^{3/2}}.$ Bound above by $2x^{-3/2}$ (left panel of the top figure); ${\int }_1^{\infty }2x^{-3/2}dx$ converges ($p=3/2>1$). By DCT, ${\int }_1^{\infty }\frac{\ln (x+1)}{x^2}dx$ converges. ✅ (Moral: $L=\infty$ against a convergent $g$ is useless; you must find a bigger benchmark that still converges — and $x^{-3/2}$, with $p$ still above $1$, has exactly the room to absorb the slow log.)

DCT needs $f\ge 0$; here $\sin x$ oscillates in sign, so DCT does not apply directly. Use absolute comparison (see Absolute vs conditional convergence): $\left|\frac{\sin x}{x^2}\right|=\frac{|\sin x|}{x^2}\le \frac{1}{x^2}.$ ${\int }_1^{\infty }{x}^{-2}dx$ converges $\Rightarrow {\int }_1^{\infty }\left|\frac{\sin x}{x^2}\right|dx$ converges (DCT on the nonnegative $|f|$). Absolute convergence $\Rightarrow$ convergence, so ${\int }_1^{\infty }\frac{\sin x}{x^2}dx$ converges. ✅

The denominator has two powers: $x^a$ and $x^1$. The larger exponent dominates the tail — the red curve in the figure below tracks whichever benchmark wins.

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

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

Case $0<a<1$: now the $x^1$ term wins. LCT with $g=x^{-1}$: $L={\lim }_{x\to \infty }\frac{x}{x^a+x}={\lim }_{x\to \infty }\frac{1}{x^{a-1}+1}=1(\text{since }a-1<0).$ Same fate as $\int x^{-1}$ $\Rightarrow$ diverges.

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

The figure below plots $f(x)=1/(x^2+x)$ (red) hugging its winning benchmark $1/x^2$ for the convergent case $a=2$ — the geometric picture behind the LCT limit $L=1$.

Where is the trouble? ::: At $\infty$ (type 1) or at a finite blow-up point (type 2) — split the integral if both. Tail $p$-rule vs finite-point $p$-rule? ::: ${\int }_1^{\infty }{x}^{-p}$ needs $p>1$; ${\int }_0^1{x}^{-p}$ needs $p<1$ (flipped). Which benchmark power? ::: The fastest-growing term in the denominator controls the tail. $f$ changes sign? ::: Use $|f|\le g$ (absolute comparison), not raw DCT. LCT gives $L=0$? ::: Only "$\int g$ converges $\Rightarrow \int f$ converges" survives (L5.1). LCT gives $L=\infty$? ::: Only "$\int g$ diverges $\Rightarrow \int f$ diverges" survives; against a convergent $g$ it is silent — change benchmark (L5.2). Direction of DCT inequality? ::: Squeeze to please (above a convergent), push to die (below a divergent) — the picture at the top.