Remainder Bounds for Exponential Tails

Why a Cutoff Has a Measurable Error

An integral over an infinite interval cannot be evaluated by integrating only to a finite endpoint and then pretending nothing was omitted. The difference between the full integral and its truncated value is the tail beyond the cutoff. For an exponential tail, that remainder can be computed exactly, so a cutoff can be chosen to satisfy a specified error tolerance.

Throughout this tutorial, assume the decay rate \(a\) is positive. This condition matters: \(e^{-ax}\) decays as \(x\) increases only when \(a>0\). We will first consider the basic tail beginning at \(0\), then include a scale factor and a shifted starting point.

Definition: For a nonnegative integrand \(f\) on \([A,\infty)\) whose improper integral converges, the truncation remainder at \(R\ge A\) is the omitted amount \(E(R)=\int_R^\infty f(x)\,dx\). The truncated integral is \(\int_A^R f(x)\,dx\), and its error in approximating the full integral is exactly \(E(R)\).

The Basic Exponential Remainder

For \(a>0\), an antiderivative of \(e^{-ax}\) is \(-e^{-ax}/a\). Evaluating the improper integral by a limit gives its tail from \(R\) onward:

\int_R^\infty e^{-ax}\,dx = \lim_{b\to\infty}\int_R^b e^{-ax}\,dx = \lim_{b\to\infty}\left[-\frac{e^{-ax}}{a}\right]_R^b = \frac{e^{-aR}}{a}.

The limit term at the upper endpoint is zero because \(a>0\) implies \(e^{-ab}\to 0\) as \(b\to\infty\). Thus the amount omitted is not merely bounded by an exponential expression: for this integrand, it equals that expression.

Theorem: If \(a>0\) and \(R\ge 0\), then the remainder after truncating \(\int_0^\infty e^{-ax}\,dx\) at \(R\) is \(E(R)=e^{-aR}/a\). To ensure \(E(R)\le\varepsilon\), where \(\varepsilon>0\), it is enough and necessary to choose \(R\ge \frac{1}{a}\ln\!\left(\frac{1}{a\varepsilon}\right)\) when \(a\varepsilon<1\). If \(a\varepsilon\ge 1\), the cutoff \(R=0\) already meets the tolerance.

To obtain the cutoff condition, start from the exact remainder and solve the inequality. Since \(a>0\), multiplication by \(a\) preserves the inequality. Since the natural logarithm is increasing, taking logarithms also preserves it.

\frac{e^{-aR}}{a}\le\varepsilon \quad\Longleftrightarrow\quad e^{-aR}\le a\varepsilon \quad\Longleftrightarrow\quad -aR\le \ln(a\varepsilon) \quad\Longleftrightarrow\quad R\ge \frac{1}{a}\ln\!\left(\frac{1}{a\varepsilon}\right).

This logarithmic condition is relevant as a positive lower bound when \(a\varepsilon<1\). If \(a\varepsilon\ge1\), then \(E(0)=1/a\le\varepsilon\), so no positive cutoff is required to meet the stated tolerance. In practice, a cutoff also has to lie in the domain of integration.

Derivation of the Error Formula

The exactness of the remainder follows from splitting the convergent improper integral at a finite point. This derivation also explains why the tail is the right error to control.

Split at the cutoff.
For \(b>R\), add the two adjacent finite integrals to recover the integral from \(0\) to \(b\): \(\int_0^R e^{-ax}\,dx+\int_R^b e^{-ax}\,dx=\int_0^b e^{-ax}\,dx\).

Take the upper-endpoint limit.
As \(b\to\infty\), the first term remains fixed and the second converges because its evaluated upper-endpoint contribution \(e^{-ab}/a\) tends to zero.

Subtract the truncated value.
The full integral is the sum of the truncated integral and \(\int_R^\infty e^{-ax}\,dx\). Their difference is therefore exactly \(e^{-aR}/a\).

In detail, the two quantities are

\int_0^\infty e^{-ax}\,dx=\frac{1}{a}, \qquad \int_0^R e^{-ax}\,dx=\frac{1-e^{-aR}}{a}.

Subtracting the truncated value from the full value gives

\frac{1}{a}-\frac{1-e^{-aR}}{a} = \frac{e^{-aR}}{a} = \int_R^\infty e^{-ax}\,dx.

The integrand is nonnegative, so this difference is also the absolute error in using the truncated integral to approximate the full integral. For an integrand that changes sign, a signed difference need not equal the absolute error; the result here uses the positivity of the exponential tail.

Worked Examples

Worked Example: Choosing a Cutoff for \(e^{-2x}\)

Choose a cutoff \(R\ge0\) so that truncating \(\int_0^\infty e^{-2x}\,dx\) leaves an error no greater than \(0.01\).

Here \(a=2\), so the exact omitted tail is

E(R)=\frac{e^{-2R}}{2}.

Set this at most \(0.01\) and solve:

\frac{e^{-2R}}{2}\le 0.01 \quad\Longleftrightarrow\quad e^{-2R}\le 0.02 \quad\Longleftrightarrow\quad R\ge \frac{1}{2}\ln(50).

Since \(\frac12\ln(50)\approx1.956\), choose \(R=2\). Check the actual remainder by substituting this cutoff into the exact formula:

E(2)=\frac{e^{-4}}{2}\approx 0.009158<0.01.

The cutoff meets the tolerance. The inequality is checked using the unrounded exponential value; the displayed decimal is rounded to six decimal places.

Worked Example: A Shifted, Scaled Exponential Tail

Consider \(f(x)=5e^{-0.5(x-1)}\) for \(x\ge1\). Find the cutoff \(R\) that makes the omitted tail equal to \(0.02\), and verify the result.

For \(R\ge1\), integrate from \(R\) to infinity. The substitution \(u=x-1\), or direct evaluation by an antiderivative, gives

E(R)=\int_R^\infty 5e^{-0.5(x-1)}\,dx = 10e^{-0.5(R-1)}.

Set \(E(R)=0.02\) and solve for \(R\):

10e^{-0.5(R-1)}=0.02 \quad\Longleftrightarrow\quad e^{-0.5(R-1)}=0.002 \quad\Longleftrightarrow\quad R=1+2\ln(500).

Numerically, \(R\approx13.429\). Substitution into the exact remainder verifies the tolerance:

E\bigl(1+2\ln(500)\bigr) = 10e^{-\ln(500)} = \frac{10}{500} = 0.02.

This cutoff gives equality. Any larger cutoff gives a smaller remainder, because the exponential factor decreases as \(R\) increases.

Worked Example: Scaled Decay with a Smaller Rate

For \(f(x)=3e^{-4x}\) on \([0,\infty)\), find a cutoff that makes the omitted tail at most \(0.001\).

The remainder is \(E(R)=\frac{3}{4}e^{-4R}\). Solving the tolerance inequality gives

\frac{3}{4}e^{-4R}\le0.001 \quad\Longleftrightarrow\quad e^{-4R}\le\frac{1}{750} \quad\Longleftrightarrow\quad R\ge\frac14\ln(750).

Choose the equality cutoff \(R=\frac14\ln(750)\), approximately \(1.655\). Substitution checks the remainder exactly:

E\left(\frac14\ln(750)\right) = \frac34 e^{-\ln(750)} = \frac{3}{4\cdot750} = 0.001.

Thus the chosen cutoff meets the tolerance exactly. If a decimal cutoff is needed, round it upward rather than downward so that it remains at least \(\frac14\ln(750)\).

Shifted and Scaled Tails

The same calculation applies when the exponential begins at a point \(A\) or has a positive scale factor \(C\). For \(C>0\), \(a>0\), and \(R\ge A\), the antiderivative gives

\int_R^\infty C e^{-a(x-A)}\,dx = \frac{C}{a}e^{-a(R-A)}.

To make this remainder at most \(\varepsilon>0\), solve the inequality for \(R\). If the tolerance is smaller than the full integral \(C/a\), the required condition is

R\ge A+\frac{1}{a}\ln\!\left(\frac{C}{a\varepsilon}\right).

If \(\varepsilon\ge C/a\), taking \(R=A\) already meets the tolerance, since the entire integral is \(C/a\). More generally, the cutoff may be written as \(A\) or the logarithmic threshold, whichever is larger. This form separates the roles of the parameters: \(C\) scales the error, \(a\) controls the rate of decay, and \(A\) shifts the starting location.

A Common Cutoff Mistake

A common error is to omit the factor \(1/a\) when solving for the cutoff, or to reverse the inequality when taking logarithms. For example, with \(e^{-2x}\) and tolerance \(0.01\), the incorrect calculation

\frac{e^{-2R}}{2}\le0.01 \quad\Longrightarrow\quad e^{-2R}\le0.01 \quad\Longrightarrow\quad R\ge\frac12\ln(100)

drops the factor \(2\) on the right-hand side of the first inequality. The corrected step is \(e^{-2R}\le0.02\), which gives \(R\ge\frac12\ln(50)\). A quick substitution exposes the incorrect cutoff: at \(R=\frac12\ln(100)\), the actual remainder is

E(R)=\frac{e^{-\ln(100)}}{2} =\frac{1}{200} =0.005.

This value happens to be below \(0.01\), so this particular dropped factor produces an unnecessarily large cutoff rather than a failed tolerance. In other rearrangements, a sign error can produce a cutoff that is too small. The reliable procedure is to begin with the exact remainder, preserve every factor, solve the inequality one step at a time, and substitute the proposed cutoff back into the remainder formula.

Why Exact Remainders Are Useful

An exact tail formula turns an infinite-interval calculation into a practical error-control problem. Rather than guessing how far to integrate, one specifies an acceptable error and computes a cutoff. Because \(E(R)=\frac{C}{a}e^{-a(R-A)}\) decreases with \(R\), every cutoff larger than a valid threshold also works.

In numerical work, the cutoff may need to be a convenient decimal or grid point. Compute the threshold first, then round the cutoff upward if rounding is necessary. Rounding downward can violate the inequality even when the difference is small. Finally, check the remainder using the exact formula and compare it with the prescribed tolerance.

Check Your Understanding

Use the exact remainder formula, and verify each proposed cutoff by substitution.

For \(\int_0^\infty e^{-3x}\,dx\), write the remainder after cutoff \(R\).
Find the smallest cutoff that makes the remainder of \(\int_0^\infty e^{-x}\,dx\) at most \(0.02\).
For \(4e^{-2(x-3)}\) on \([3,\infty)\), find the exact tail beyond \(R\ge3\).
For the tail in question 3, find a cutoff that makes the omitted amount at most \(0.1\).
Why should a decimal approximation to a required cutoff be rounded upward rather than downward?