Remainder Bounds for Power Tails

Why Bound the Part Beyond a Cutoff?

An improper integral extends over an unbounded interval, so a numerical calculation cannot usually include the entire interval at once. Instead, one computes up to a finite point $R$ and asks how much may be left beyond it. That remaining part is the remainder, or tail.

A useful strategy is to compare the integrand in the tail with a power $x^{-p}$. When $p>1$, the integral of that power from $R$ to infinity has an exact value. If the integrand is bounded by a constant multiple of $x^{-p}$, this exact value gives a rigorous bound on the truncation error.

The Exact Power-Tail Integral

Let $R>0$ and $p>1$. The improper integral $\int_R^\infty x^{-p}\,dx$ is defined as the limit of the finite integrals $\int_R^B x^{-p}\,dx$ as $B$ tends to infinity. Since $p\ne 1$, an antiderivative is $x^{1-p}/(1-p)$. Therefore,

\int_R^B x^{-p}\,dx = \left[\frac{x^{1-p}}{1-p}\right]_R^B = \frac{R^{1-p}-B^{1-p}}{p-1}.

The condition $p>1$ means $1-p<0$, so $B^{1-p}$ tends to zero as $B$ tends to infinity. Taking the limit gives the exact tail:

Power-tail formula: For $R>0$ and $p>1$, $$ \int_R^\infty x^{-p}\,dx = \frac{R^{1-p}}{p-1} = \frac{1}{(p-1)R^{p-1}}. $$

The denominator is $p-1$, and the cutoff appears with exponent $1-p$. Both features matter: increasing $R$ decreases the tail, and increasing $p$ makes the power decay more rapidly. The restriction $p>1$ is essential. At $p=1$, the antiderivative is $\ln x$, which does not have a finite limit at infinity; for $p<1$, the power-antiderivative limit also fails to be finite.

Worked Example: Evaluate an Exact Power Tail

Evaluate $\int_4^\infty x^{-3/2}\,dx$. Here $R=4$ and $p=3/2>1$, so the power-tail formula applies:

\int_4^\infty x^{-3/2}\,dx = \frac{4^{1-3/2}}{(3/2)-1} = \frac{4^{-1/2}}{1/2} = \frac{1/2}{1/2} = 1.

To check directly, an antiderivative is $-2x^{-1/2}$, because \[ \frac{d}{dx}\left(-2x^{-1/2}\right) = (-2)(-1/2)x^{-3/2} = x^{-3/2}. \] Thus

\int_4^\infty x^{-3/2}\,dx = \lim_{B\to\infty} \left[-2x^{-1/2}\right]_4^B = 0-\left(-2\cdot\frac12\right) = 1.

The direct limit agrees with the formula.

Using Comparison to Bound a Remainder

Suppose $f$ is continuous on $[R,\infty)$, and for every $x\ge R$ it satisfies \[ 0\le f(x)\le Cx^{-p}, \] where $C>0$ and $p>1$. Integrating this inequality over a finite interval $[R,B]$ preserves its order. Letting $B$ tend to infinity then bounds the tail of $f$ by the exact power tail.

Remainder bound for a nonnegative integrand: If $0\le f(x)\le Cx^{-p}$ for all $x\ge R>0$, with $C>0$ and $p>1$, then $$ 0\le \int_R^\infty f(x)\,dx \le \frac{C R^{1-p}}{p-1}. $$ In particular, the tail converges.

For an integral beginning at a finite point $a\le R$, write its value as $I=\int_a^\infty f(x)\,dx$, and its truncation at $R$ as $I_R=\int_a^R f(x)\,dx$. When the tail is convergent, \[ I-I_R=\int_R^\infty f(x)\,dx. \] The remainder bound therefore estimates the difference between the full integral and its truncated value.

If $f$ can be positive or negative, a one-sided comparison $f(x)\le Cx^{-p}$ is not enough: negative values could make the integral small through cancellation without controlling its absolute size. Instead, if $|f(x)|\le Cx^{-p}$ for $x\ge R$, then \[ \left|\int_R^\infty f(x)\,dx\right| \le \int_R^\infty |f(x)|\,dx \le \frac{C R^{1-p}}{p-1}. \] This also proves absolute convergence of the tail.

Why the Comparison Bound Works

The key point is that the comparison can first be made on finite intervals, where ordinary integral inequalities apply. The infinite-interval conclusion follows by taking limits.

Compare on a finite interval.
From $0\le f(x)\le Cx^{-p}$, integration from $R$ to $B$, for $B>R$, gives $$ 0\le \int_R^B f(x)\,dx\le C\int_R^B x^{-p}\,dx. $$

Evaluate the upper bound.
The power-tail calculation on a finite interval gives $$ C\int_R^B x^{-p}\,dx = \frac{C}{p-1}\left(R^{1-p}-B^{1-p}\right) \le \frac{C R^{1-p}}{p-1}. $$

Pass to the improper limit.
Because $f\ge0$, the finite integrals $\int_R^B f(x)\,dx$ do not decrease as $B$ increases. They are bounded above by the finite constant $C R^{1-p}/(p-1)$, so they have a finite limit. Taking that limit proves the claimed tail bound.

Worked Remainder Estimates

Worked Example: Bound a Rational Tail

Let $f(x)=3/(x^3+1)$, and estimate the remainder $\int_4^\infty f(x)\,dx$. For $x\ge4$, we have $x^3+1\ge x^3>0$. Taking reciprocals and multiplying by $3$ gives \[ 0\le \frac{3}{x^3+1}\le 3x^{-3}. \] Here $C=3$, $p=3$, and $R=4$. Therefore,

0\le \int_4^\infty \frac{3}{x^3+1}\,dx \le 3\int_4^\infty x^{-3}\,dx = 3\frac{4^{1-3}}{3-1} = 3\frac{4^{-2}}{2} = 3\cdot\frac{1}{16}\cdot\frac12 = \frac{3}{32}.

For a direct check of the comparison integral, an antiderivative of $3x^{-3}$ is $-3/(2x^2)$, since \[ \frac{d}{dx}\left(-\frac{3}{2x^2}\right)=3x^{-3}. \] Its tail is \[ \lim_{B\to\infty}\left[-\frac{3}{2x^2}\right]_4^B = 0-\left(-\frac{3}{32}\right) = \frac{3}{32}, \] as in the bound. Thus truncating the integral of $f$ at $4$ leaves at most $3/32$ beyond the cutoff.

Worked Example: Choose a Cutoff for a Tolerance

Let $f(x)=5/(x^4+1)$ for $x\ge1$, and suppose we want the remainder after truncation to be no greater than $1/100$. Since $x^4+1\ge x^4$, \[ 0\le f(x)\le 5x^{-4}. \] The power-tail bound at a cutoff $R\ge1$ is

\int_R^\infty f(x)\,dx \le 5\int_R^\infty x^{-4}\,dx = \frac{5R^{1-4}}{4-1} = \frac{5}{3R^3}.

To make this upper bound at most $1/100$, solve \[ \frac{5}{3R^3}\le\frac{1}{100}. \] Since $3R^3>0$, multiplying both sides by $300R^3$ gives $500\le3R^3$, or $R^3\ge500/3$. The integer cutoff $R=6$ satisfies this because $6^3=216$ and $216\ge500/3$. At that cutoff,

\int_6^\infty f(x)\,dx \le \frac{5}{3\cdot6^3} = \frac{5}{648} \le \frac{1}{100},

where the last inequality follows from $500\le648$. The antiderivative used for the comparison tail is $-5/(3x^3)$, whose derivative is $5x^{-4}$. Evaluating it at $6$ and infinity gives $0-(-5/648)=5/648$, confirming the bound. Thus truncation at $6$ guarantees an error no greater than $1/100$.

A Common Mistake: Using the Wrong Denominator

In the power-tail formula, the denominator is $p-1$, not $p$. For example, when integrating $3x^{-3}$ from $4$ to infinity, the exponent parameter is $p=3$, so the denominator is $3-1=2$. Replacing it by $3$ gives an incorrect equality:

\text{Incorrect:}\qquad 3\int_4^\infty x^{-3}\,dx \ne 3\frac{4^{-2}}{3}.

The correct calculation is

3\int_4^\infty x^{-3}\,dx = 3\frac{4^{1-3}}{3-1} = 3\frac{4^{-2}}{2} = \frac{3}{32}.

A quick way to avoid this error is to write the full formula $R^{1-p}/(p-1)$ before substituting values. Then check that the denominator comes from the exponent $1-p$ in the antiderivative.

Interpreting the Bound

A remainder bound is a guarantee, not generally an exact value. In the rational-tail example, $3/32$ is an upper bound for the remaining integral; the comparison does not assert that the remainder equals $3/32$. A sharper comparison, if available, may produce a smaller bound.

The formula also helps plan a computation. If the integrand is bounded by $Cx^{-p}$, then the remainder bound is \[ \frac{C}{(p-1)R^{p-1}}. \] To guarantee an error at most $\varepsilon>0$, it is sufficient to choose $R$ so that \[ \frac{C}{(p-1)R^{p-1}}\le\varepsilon. \] This condition controls how far out the integral must be computed; it does not require knowing the exact value of the full improper integral.

Key takeaway: A pointwise power bound on the tail becomes an integral-error bound because the comparison power has a known finite tail. Check that $p>1$, that the comparison holds for every $x\ge R$, and that the denominator is $p-1$.

Check Your Understanding

Use the power-tail formula and the comparison conditions to answer these questions.

Evaluate $\int_9^\infty x^{-2}\,dx$ using the formula.
If $0\le f(x)\le 4x^{-5}$ for $x\ge2$, what upper bound does comparison give for $\int_2^\infty f(x)\,dx$?
Why does a bound $f(x)\le Cx^{-p}$ not by itself control the absolute remainder when $f$ can be negative?
For $0\le f(x)\le 2x^{-3}$, find a cutoff $R$ that makes the remainder bound at most $1/50$.
In the power-tail formula, what condition on $p$ ensures that the tail is finite, and what is the denominator in the formula?