Comparison Test for Improper Integrals

For nonnegative integrands, domination decides convergence: if

0\le f\le g

then

\int g<\infty

forces

\int f<\infty

, and

\int f=\infty

forces

\int g=\infty

کارت قضیه کنکور ارشد علوم کامپیوتر

اهمیت

۰/۵

Definition / intuition

An improper integral $\int_a^{\infty} f(x)\,dx$ is defined as $\lim_{b\to\infty}\int_a^{b} f(x)\,dx$ . When $f\ge 0$ the partial integrals $F(b)=\int_a^{b} f$ are monotone increasing in $b$ , so the limit either is a finite number (the integral converges) or is $+\infty$ (it diverges) — it can never oscillate. That monotonicity is exactly what makes a comparison test work: you only need to control the size of the tail, never its sign.

The theorem

Let $f,g$ be continuous on $[a,\infty)$ with $0\le f(x)\le g(x)$ for all large $x$ . Then

$\int_a^{\infty} g(x)\,dx\ \text{converges}\ \Longrightarrow\ \int_a^{\infty} f(x)\,dx\ \text{converges},$ $\int_a^{\infty} f(x)\,dx\ \text{diverges}\ \Longrightarrow\ \int_a^{\infty} g(x)\,dx\ \text{diverges}.$

The contrapositive is the second line: the smaller function dragging the bigger one up. The benchmark family to compare against is the $p$ -integral

$\int_1^{\infty}\frac{dx}{x^{p}}\ \text{converges}\iff p>1.$

How to apply it

Check the integrand is (eventually) nonnegative — the test fails for sign-changing $f$ (there you'd test $|f|$ for absolute convergence instead).
Find a clean dominating $g$ (to prove convergence) or a clean dominated $f$ (to prove divergence), usually a power $x^{-p}$ or an exponential.
Confirm the inequality holds for all large $x$ — finitely many early exceptions never affect convergence.
Read off the verdict from the benchmark.

A useful limit form: if $f,g>0$ and $\lim_{x\to\infty} f(x)/g(x)=L$ with $0<L<\infty$ , then $\int f$ and $\int g$ share the same fate.

Mini-example

Does $\displaystyle\int_1^{\infty}\frac{dx}{x^{2}+\sqrt{x}}$ converge? For $x\ge 1$ the denominator exceeds $x^{2}$ , so

$0<\frac{1}{x^{2}+\sqrt{x}}\le\frac{1}{x^{2}},\qquad \int_1^{\infty}\frac{dx}{x^{2}}=1<\infty.$

Domination by a convergent $p$ -integral ( $p=2>1$ ) gives convergence.

Domination as a one-line trick

The konkur favorite: on $[1,\infty)$ you have $x\ge 1$ , hence $0<f(x)\le x\,f(x)$ for any positive $f$ . So

$\int_1^{\infty} x\,f(x)\,dx\ \text{converges}\ \Longrightarrow\ \int_1^{\infty} f(x)\,dx\le\int_1^{\infty} x\,f(x)\,dx<\infty.$

No such implication holds for $f^{2}$ versus $f$ in either direction:

Claim	Verdict	Witness on $[1,\infty)$
$\int f<\infty\Rightarrow\int f^{2}<\infty$	false	tall thin triangular spikes: area $\sum 1/n^{2}$ , squared-area $\sum 2/3n$
$\int f^{2}<\infty\Rightarrow\int f<\infty$	false	$f(x)=1/x$ : $\int x^{-2}=1$ but $\int x^{-1}=\infty$
$\int xf<\infty\Rightarrow\int f<\infty$	true	$f\le xf$ on $[1,\infty)$

Pitfall: Comparison needs a nonnegative integrand and a domination in the right direction — a bigger function that converges, or a smaller one that diverges. The reverse pairings (smaller convergent, bigger divergent) say nothing. And squaring or reciprocating an integrand changes its tail unpredictably: there is no general rule linking $\int f$ to $\int f^{2}$ or to $\int 1/f$ , so reach only for monotone domination like $f\le xf$ .

فرمول‌های کلیدی

Direct comparison (convergence)

0\le f\le g \ \text{and}\ \int_a^{\infty} g<\infty \ \Longrightarrow\ \int_a^{\infty} f<\infty

A nonnegative integrand dominated by a convergent one also converges; contrapositive gives the divergence half.

p-integral benchmark

\int_1^{\infty}\frac{dx}{x^{p}}<\infty \iff p>1

The standard yardstick for tail behavior at infinity used inside comparisons.

Limit comparison

\lim_{x\to\infty}\frac{f(x)}{g(x)}=L,\ 0<L<\infty \ \Longrightarrow\ \int f,\ \int g \ \text{converge together}

When two positive integrands are asymptotically proportional, they share the same convergence verdict.