Asymptotic Equivalence and Integral Convergence

Why Asymptotic Equivalence Helps

An improper integral can be difficult because its integrand has a complicated formula near a singular endpoint or far out on an infinite interval. Often, however, the exact formula matters less than its leading behavior in that region. If a positive integrand behaves like a simpler positive function, the simpler function can reveal whether the integral converges.

The precise idea is asymptotic equivalence: two functions are equivalent when their ratio tends to one at the endpoint or along the tail being studied. This is stronger than saying their difference tends to zero. Under the positivity and local integrability conditions stated below, equivalent functions have the same improper-integral convergence behavior.

Definition and Convergence Result

Definition: Let $f$ and $g$ be positive functions near a finite endpoint $a$. We say that $f$ is asymptotically equivalent to $g$ as $x\to a^+$, and write $f(x)\sim g(x)$, if $$ \lim_{x\to a^+}\frac{f(x)}{g(x)}=1. $$ For positive functions on a tail, $f(x)\sim g(x)$ as $x\to\infty$ means $$ \lim_{x\to\infty}\frac{f(x)}{g(x)}=1. $$

Here, positivity is required sufficiently close to the endpoint, or sufficiently far along the tail. We also assume the functions are integrable on every closed subinterval away from the improper endpoint, as is the case for the continuous integrands considered here. The question is then whether the integral over the endpoint region or infinite tail is finite.

Theorem: Suppose $f$ and $g$ are positive and locally integrable near the same improper endpoint, and $f(x)\sim g(x)$ there. Then their improper integrals have the same convergence behavior: either both converge or both diverge. The same statement holds for a tail extending to infinity.

The theorem is a direct consequence of the limit comparison principle: a positive ratio tending to one is eventually bounded above and below by positive constants. The following proof gives the bounds and explains why they settle convergence.

Why the Result Works

Use the ratio limit.
Since $f(x)/g(x)\to1$, there is a region sufficiently close to the endpoint on which the ratio differs from one by less than $1/2$.

Obtain two-sided bounds.
The ratio condition gives $1/2<f(x)/g(x)<3/2$, and positivity of $g$ allows multiplication without changing the inequality directions.

Compare truncated integrals.
Integrating the inequalities shows that either function has a finite improper integral exactly when the other does.

For completeness, consider a finite endpoint $a$. The definition of the limit gives a $\delta>0$ such that, whenever $a<x<a+\delta$,

\left|\frac{f(x)}{g(x)}-1\right|<\frac12.

Thus $1/2<f(x)/g(x)<3/2$. Since $g(x)>0$, this is equivalent to

\frac12g(x)<f(x)<\frac32g(x).

For $a<t<a+\delta$, integrate on the ordinary, nonsingular interval from $t$ to $a+\delta$:

\frac12\int_t^{a+\delta}g(x)\,dx \leq \int_t^{a+\delta}f(x)\,dx \leq \frac32\int_t^{a+\delta}g(x)\,dx.

If the improper integral of $g$ converges near $a$, its truncated integrals have a finite limit as $t\to a^+$. The upper inequality then bounds the truncated integrals of $f$, so the integral of $f$ converges as well. Conversely, if the integral of $f$ converges, the lower inequality gives $\int_t^{a+\delta}g(x)\,dx\leq 2\int_t^{a+\delta}f(x)\,dx$, so the integral of $g$ converges. Therefore either both converge or neither does.

For an infinite tail, the same argument applies on $[A,R]$ for sufficiently large $A$, then we let $R\to\infty$. The finite portions away from the improper endpoint do not affect convergence when the integrands are locally integrable there.

Worked Examples

Worked Example: A Radical Difference on an Infinite Tail

Determine the convergence of $$ \int_1^\infty \bigl(\sqrt{x^2+4}-x\bigr)\,dx. $$ The expression is positive for $x\geq1$, but direct integration is not needed. Compare it with $g(x)=2/x$. Compute the ratio and rationalize the numerator: $$ \frac{\sqrt{x^2+4}-x}{2/x} =\frac{x}{2}\bigl(\sqrt{x^2+4}-x\bigr) =\frac{x}{2}\cdot\frac{4}{\sqrt{x^2+4}+x} =\frac{2x}{\sqrt{x^2+4}+x}. $$ Dividing the denominator by $x>0$ gives $$ \frac{2x}{\sqrt{x^2+4}+x} =\frac{2}{\sqrt{1+4/x^2}+1}\longrightarrow\frac{2}{1+1}=1. $$ Thus the original integrand is asymptotically equivalent to $2/x$. For $R>1$, $$ \int_1^R\frac{2}{x}\,dx=2\ln R-2\ln1=2\ln R. $$ As $R\to\infty$, this tends to $+\infty$. The comparison theorem therefore shows that the given improper integral diverges. The ratio limit checks the model, and the model’s integral checks the convergence conclusion.

Worked Example: A Trigonometric Singularity at Zero

Determine the convergence of $$ \int_0^1\frac{1-\cos x}{x^3}\,dx. $$ The integrand is positive for $0<x\leq1$. Use the model $g(x)=1/(2x)$. The standard limit $\lim_{x\to0}(1-\cos x)/x^2=1/2$ gives $$ \frac{(1-\cos x)/x^3}{1/(2x)} =\frac{2(1-\cos x)}{x^2} \longrightarrow 1. $$ So the two positive integrands are asymptotically equivalent near zero. For $0<t<1$, the model integral is $$ \int_t^1\frac{1}{2x}\,dx =\frac12\bigl(\ln1-\ln t\bigr) =-\frac12\ln t. $$ As $t\to0^+$, $\ln t\to-\infty$, so this integral tends to $+\infty$. The original integral diverges by asymptotic equivalence. In particular, the factor $1-\cos x$ cancels two powers of $x$ in the leading behavior, leaving a nonintegrable $1/(2x)$ model.

Worked Example: An Inverse Tangent Near Zero

Determine the convergence of $$ \int_0^1\frac{\arctan x}{x^{3/2}}\,dx. $$ For $0<x\leq1$, the integrand is positive. Compare it with $g(x)=x^{-1/2}$. The standard limit $\lim_{x\to0^+}\arctan x/x=1$ yields $$ \frac{(\arctan x)/x^{3/2}}{x^{-1/2}} =\frac{\arctan x}{x} \longrightarrow1. $$ The model integral is finite: $$ \int_t^1x^{-1/2}\,dx =\left[2\sqrt{x}\right]_t^1 =2-2\sqrt{t} \longrightarrow2 \qquad(t\to0^+). $$ Therefore the given integral converges. The exact value of the original integral is not needed to determine convergence: the ratio limit identifies its leading behavior, and the model integral is finite. As a final check, the antiderivative used for the model is correct because $\frac{d}{dx}(2\sqrt{x})=x^{-1/2}$ for $x>0$.

How to Choose and Use a Model

The practical task is to find a positive function $g$ whose ratio with the original integrand $f$ has limit one at the relevant endpoint. A useful model should be simple enough that its improper integral can be evaluated or its convergence determined. Common sources of models include standard limits and algebraic simplification, as in the examples.

Keep the endpoint in view. A function may have different useful approximations near a finite singularity and far out on an infinite interval. The ratio must be evaluated in the limit relevant to the integral being tested. A limit at one location does not establish equivalence at another.

Asymptotic equivalence is also a statement about relative size, not exact equality. It licenses a convergence conclusion, not the replacement of one integrand by another as an exact identity or the claim that their integrals have the same value.

A Common Mistake: Confusing a Small Difference with a Ratio Limit

It is not enough that $f(x)-g(x)\to0$. For example, on $x\geq1$, let $f(x)=1/x+1/x^2$ and $g(x)=1/x^2$. Then $$ f(x)-g(x)=\frac1x\longrightarrow0, $$ but $$ \frac{f(x)}{g(x)} =\frac{1/x+1/x^2}{1/x^2} =x+1\longrightarrow\infty, $$ not one. In fact, $$ \int_1^R f(x)\,dx =\ln R+\left(1-\frac1R\right)\longrightarrow\infty, \qquad \int_1^\infty g(x)\,dx =\left[-\frac1x\right]_1^\infty=1. $$ Thus the two integrals have different convergence behavior despite the difference of the functions tending to zero. The correction is to calculate the ratio $f/g$ and verify that it tends to one before invoking asymptotic equivalence.

Why the Method Matters

Asymptotic equivalence separates the part of an integrand that controls convergence from details that do not change its leading behavior. This can turn an unfamiliar expression into a familiar model while preserving the answer to the convergence question. The proof also clarifies the limits of the method: the functions must be positive in the region under study, and the ratio must tend to one at the same improper endpoint.

Check Your Understanding

Use the definition and the convergence result to answer these questions.

What ratio limit defines $f(x)\sim g(x)$ as $x\to\infty$?
Why does a ratio limit of one give upper and lower bounds by constant multiples near the endpoint?
If $f(x)\sim g(x)$ near zero and $\int_0^1 g(x)\,dx$ diverges, what follows about $\int_0^1 f(x)\,dx$?
Why does $f(x)-g(x)\to0$ not by itself establish asymptotic equivalence?
For an improper integral on an infinite interval, which part of the interval must the ratio limit describe?