Power-Law Tails on Infinite Intervals

Why Power-Law Tails Matter

A function can become small as $x$ grows and still have an infinite total area under its graph. The rate at which it decreases matters: a tail that shrinks too slowly may contribute an unbounded amount, while a faster-decaying tail can have finite total area.

The family $x^{-p}$, where $p$ is a real number, gives a precise way to study this distinction. Its integral over $[1,\infty)$ provides a benchmark for classifying a basic class of tails, including constant multiples and shifted powers.

Definition: For a function $f$ continuous on $[A,\infty)$, its improper integral over that interval is defined by $$ \int_A^\infty f(x)\,dx = \lim_{R\to\infty}\int_A^R f(x)\,dx, $$ provided the limit exists as a finite real number. In that case, the improper integral is said to converge. If the limit does not exist as a finite real number, the integral diverges.

The integral on $[A,\infty)$ is therefore not evaluated by substituting infinity into an antiderivative. Instead, one first integrates over a finite interval ending at $R$, then examines the limit as $R$ increases without bound.

The Power-Law Benchmark

Consider $\int_1^\infty x^{-p}\,dx$. The exponent $p$ may be any real number. The antiderivative depends on whether $p$ equals $1$: for $p\ne 1$, the power rule applies, while for $p=1$, the integrand is $1/x$ and its antiderivative is logarithmic.

\int_1^R x^{-p}\,dx = \begin{cases} \dfrac{R^{1-p}-1}{1-p}, & p\ne 1,\\[6pt] \ln R, & p=1. \end{cases}

Taking the limit of these expressions determines convergence. When $p>1$, $1-p<0$, so $R^{1-p}$ tends to zero. When $p<1$, $1-p>0$, so $R^{1-p}$ grows without bound. At $p=1$, $\ln R$ also grows without bound.

Theorem: For real $p$, $$ \int_1^\infty x^{-p}\,dx \begin{cases} \text{converges to } \dfrac{1}{p-1}, & p>1,\\[6pt] \text{diverges}, & p\le 1. \end{cases} $$

Why the Threshold Is $p=1$

The threshold follows directly from the finite-interval antiderivative and its limit. The key point is the sign of the exponent $1-p$ in $R^{1-p}$. A negative exponent makes this term tend to zero; a positive exponent makes it grow. The separate logarithmic case at $p=1$ also grows, rather than approaching a finite value.

Integrate to a finite endpoint.
For $p\ne 1$, use the power rule to obtain $\dfrac{R^{1-p}-1}{1-p}$.

Determine the exponent's sign.
If $p>1$, then $1-p<0$ and $R^{1-p}\to 0$. If $p<1$, then $1-p>0$ and $R^{1-p}\to\infty$.

Handle the critical exponent.
When $p=1$, the finite integral is $\ln R$, which tends to infinity.

Evaluate the finite case.
For $p>1$, the limit is $\dfrac{0-1}{1-p}=\dfrac{1}{p-1}$.

This calculation proves the threshold, rather than merely suggesting it from a few examples. It also identifies the value of every convergent benchmark integral.

Worked Example: A Convergent Power Tail

Evaluate $\int_1^\infty x^{-3/2}\,dx$. Here $p=3/2>1$, so the benchmark theorem predicts convergence. For finite $R>1$, an antiderivative is $-2x^{-1/2}$, giving

\int_1^R x^{-3/2}\,dx = \left[-2x^{-1/2}\right]_1^R = -\frac{2}{\sqrt R}+2 = 2-\frac{2}{\sqrt R}.

As $R\to\infty$, $2/\sqrt R\to0$, so the improper integral equals $2$. Check: differentiating the antiderivative gives $\dfrac{d}{dx}(-2x^{-1/2})=x^{-3/2}$, and the limiting evaluation gives the stated value.

Worked Example: The Critical Exponent

Determine whether $\int_1^\infty \dfrac{1}{x}\,dx$ converges. For finite $R>1$,

\int_1^R \frac{1}{x}\,dx = [\ln x]_1^R = \ln R-\ln 1 = \ln R.

Since $\ln R\to\infty$, the improper integral diverges. Check: the derivative of $\ln x$ is $1/x$, and the finite-interval expression increases without bound as its upper endpoint increases. Thus the critical case $p=1$ is divergent, not convergent.

Constant Multiples

Multiplying a power tail by a fixed constant changes its scale, but not the exponent that governs its convergence. For $C>0$, linearity on every finite interval gives $\int_1^R Cx^{-p}\,dx=C\int_1^R x^{-p}\,dx$. If the benchmark has a finite limit, multiplying by $C$ keeps it finite. If the benchmark grows without bound, multiplying by a positive constant keeps it unbounded.

More explicitly, for $C>0$, the integral $\int_1^\infty Cx^{-p}\,dx$ converges exactly when $p>1$, and then its value is $C/(p-1)$. If $C=0$, the integral is zero for every $p$. If $C<0$, the same threshold applies for finite convergence: for $p>1$ the value is $C/(p-1)$, while for $p\le1$ the finite-interval integrals tend to negative infinity.

Worked Example: A Constant Multiple of a Convergent Tail

Evaluate $\int_1^\infty 5x^{-2}\,dx$. Since $p=2>1$, the integral converges. For finite $R>1$,

\int_1^R 5x^{-2}\,dx = \left[-\frac{5}{x}\right]_1^R = -\frac{5}{R}-(-5) = 5-\frac{5}{R}.

As $R\to\infty$, this tends to $5$. Check: differentiating $-5/x$ gives $5/x^2$, and the endpoint expression tends to $5$. This also agrees with the benchmark value $C/(p-1)=5/(2-1)=5$.

Shifted Powers

A horizontal shift changes where a power tail starts, but it does not change the convergence threshold, provided the shifted base remains positive throughout the interval. Let $A$ and $a$ be real numbers with $A+a>0$. Then $x+a>0$ for every $x\ge A$. For $C>0$, consider $\int_A^\infty C(x+a)^{-p}\,dx$.

Use the substitution $u=x+a$, so $du=dx$. The lower endpoint becomes $A+a$, and the upper endpoint $R+a$ tends to infinity as $R\to\infty$. Thus the question reduces to evaluating a power tail starting at the positive number $A+a$. For $p\ne1$, its finite-interval integral is

\int_A^R C(x+a)^{-p}\,dx = \frac{C}{1-p} \left((R+a)^{1-p}-(A+a)^{1-p}\right).

For $p>1$, the first power in parentheses tends to zero, leaving the finite value $C(A+a)^{1-p}/(p-1)$. For $p<1$, that term grows without bound, so the integral diverges. When $p=1$, the finite-interval integral is $C\bigl(\ln(R+a)-\ln(A+a)\bigr)$, which also grows without bound. Hence a positive constant and a valid fixed shift change the value in the convergent case, but not the threshold $p>1$.

Worked Example: A Shifted Convergent Tail

Evaluate $\int_1^\infty \dfrac{6}{(x+2)^2}\,dx$. The base $x+2$ is positive for $x\ge1$, and the exponent is $p=2>1$. For finite $R>1$, an antiderivative is $-6/(x+2)$, so

\int_1^R \frac{6}{(x+2)^2}\,dx = \left[-\frac{6}{x+2}\right]_1^R = -\frac{6}{R+2}+\frac{6}{3} = 2-\frac{6}{R+2}.

As $R\to\infty$, the last term tends to zero, and the integral equals $2$. Check: differentiating $-6/(x+2)$ gives $6/(x+2)^2$. The general shifted formula gives $C(A+a)^{1-p}/(p-1)=6(3)^{-1}/1=2$, in agreement.

A Common Mistake: Treating $p=1$ as an Ordinary Power

The power antiderivative $\dfrac{x^{1-p}}{1-p}$ is valid only for $p\ne1$. Substituting $p=1$ into this expression produces division by zero; it does not provide an antiderivative for $1/x$. The correct antiderivative at $p=1$ is $\ln x$.

Incorrect calculation: Setting $p=1$ in $\dfrac{x^{1-p}}{1-p}$ and concluding that $\int_1^\infty \dfrac{1}{x}\,dx$ is finite because $x^0=1$ is not valid. The formula has denominator $1-p=0$, so the substitution is undefined.

The correction is to integrate the critical case separately: $\int_1^R \dfrac{1}{x}\,dx=\ln R$, and $\ln R\to\infty$. Therefore the integral diverges. More generally, do not infer convergence merely because a tail tends to zero: the integral depends on the accumulated area, and the exponent determines whether that accumulation is finite.

Using the Benchmark

The power-law test is useful because it gives a sharp answer for an entire family with very little computation: once the tail has the form $C(x+a)^{-p}$, with $C>0$ and $x+a$ positive on the interval, the exponent decides convergence. The constants and shift determine the finite value when the integral converges, but they do not move the threshold.

For $p>1$, the benchmark also gives the exact remaining tail beyond any $R\ge1$:

\int_R^\infty x^{-p}\,dx = \frac{R^{1-p}}{p-1}.

This follows by applying the same antiderivative from $R$ to a finite upper endpoint and taking its limit. It describes how much area remains after stopping at $R$: increasing $R$ reduces the remainder, while exponents closer to $1$ give slower decay of that remainder. At the critical exponent or below, there is no finite total tail to approximate in this way.

The benchmark supplies a precise boundary: powers with exponent greater than $1$ have finite tails, and powers with exponent at most $1$ do not. That is the key fact to carry into the study of more general integrands.

Check Your Understanding

Use finite-interval integrals and their limits to answer each question.

Does $\int_1^\infty x^{-4/3}\,dx$ converge, and if so, what is its value?
Does $\int_1^\infty 7x^{-1}\,dx$ converge? Show the finite-$R$ expression that determines your answer.
Evaluate $\int_2^\infty 3(x+1)^{-2}\,dx$.
For what real values of $p$ does $\int_1^\infty (x+4)^{-p}\,dx$ converge?
Why must the logarithmic case $p=1$ be handled separately from the power-antiderivative formula?

Power-Law Tails on Infinite Intervals

What You'll Learn

Why Power-Law Tails Matter

The Power-Law Benchmark