Exponential Tails

Introduction

An exponential can become very small as \(x\) increases, but a small integrand does not by itself tell us whether its integral over an unbounded interval is finite. For \(e^{-kx}\), the decisive feature is the sign of \(k\): positive \(k\) produces decay, zero \(k\) gives a constant, and negative \(k\) produces growth.

We will use finite cutoffs to make this distinction precise. The same calculation gives the value of the integral when it converges and shows exactly how the cutoff integrals behave when it does not.

Definition and cutoff setup

An integral over an unbounded interval is defined using a limit of ordinary definite integrals. For a continuous function \(f\) on \([a,\infty)\), where \(a\) is finite, the right-infinite integral is defined by

\int_a^\infty f(x)\,dx = \lim_{T\to\infty}\int_a^T f(x)\,dx,

provided this limit exists as a finite real number. If the cutoff integrals grow without bound, the improper integral diverges. In this tutorial, the function is \(f(x)=e^{-kx}\), where \(k\) is a real constant.

Definition: The exponential tail with rate \(k\), beginning at \(a\), is the improper integral \(\int_a^\infty e^{-kx}\,dx\). It converges only if the cutoff limit \(\lim_{T\to\infty}\int_a^T e^{-kx}\,dx\) exists and is finite.

The most common case begins at \(a=0\). We first compute the cutoff integral for a general \(k\), then consider separately the positive, zero, and negative cases. Separating the cases matters: the antiderivative formula involving division by \(k\) cannot be used when \(k=0\).

The cutoff calculation

When \(k\ne 0\), an antiderivative of \(e^{-kx}\) is \(-\frac{1}{k}e^{-kx}\). Indeed, differentiating gives \(-\frac{1}{k}(-k)e^{-kx}=e^{-kx}\). Therefore, for any finite \(T\geq a\),

\int_a^T e^{-kx}\,dx = \left[-\frac{1}{k}e^{-kx}\right]_a^T = \frac{e^{-ka}-e^{-kT}}{k}.

This is a finite-interval calculation, valid for either sign of \(k\), provided \(k\ne 0\). Whether the improper integral exists depends on the limit of this expression as \(T\to\infty\).

Positive rate, \(k>0\).
Then \(e^{-kT}\to 0\), so the cutoff integral approaches a finite value.

Zero rate, \(k=0\).
The integrand is constantly \(1\), so the cutoff integral increases without bound.

Negative rate, \(k<0\).
Writing \(k=-|k|\) shows that the integrand grows as \(e^{|k|x}\), and the cutoff integral increases without bound.

Convergence when \(k>0\)

If \(k>0\), then \(e^{-kT}\to 0\) as \(T\to\infty\). Substituting this limit into the cutoff formula gives the general result for a finite starting point \(a\):

\int_a^\infty e^{-kx}\,dx = \lim_{T\to\infty}\frac{e^{-ka}-e^{-kT}}{k} = \frac{e^{-ka}}{k}, \qquad k>0.

For the half-line beginning at zero, \(e^{-k\cdot 0}=1\), so this becomes

\int_0^\infty e^{-kx}\,dx=\frac{1}{k}, \qquad k>0.

This derivation explains why the answer is finite: the cutoff contribution from the upper endpoint tends to zero, while the lower-endpoint contribution remains fixed. The formula also shows that moving the starting point to the right multiplies the tail by \(e^{-ka}\), a positive factor that decreases as \(a\) increases.

Theorem: For real \(k\), \(\int_0^\infty e^{-kx}\,dx\) converges if and only if \(k>0\). When it converges, its value is \(1/k\). More generally, for \(a\geq 0\) and \(k>0\), \(\int_a^\infty e^{-kx}\,dx=e^{-ka}/k\).

Worked examples

Worked Example: A positive rate on the half-line

Evaluate \(\int_0^\infty e^{-4x}\,dx\) using a cutoff.

For a finite \(T\geq 0\), \(e^{-4x}\) has antiderivative \(-\frac14 e^{-4x}\). Thus

\int_0^T e^{-4x}\,dx = \left[-\frac14 e^{-4x}\right]_0^T = -\frac14e^{-4T}+\frac14e^0 = \frac{1-e^{-4T}}{4}.

Since \(4>0\), \(e^{-4T}\to0\). Therefore

\int_0^\infty e^{-4x}\,dx = \lim_{T\to\infty}\frac{1-e^{-4T}}{4} = \frac14.

Check: the antiderivative differentiates to \(-\frac14(-4)e^{-4x}=e^{-4x}\), and the cutoff expression tends to \(1/4\), as required.

Worked Example: A tail beginning after zero

Evaluate \(\int_2^\infty e^{-x/2}\,dx\).

Here the rate is \(k=\frac12>0\). For \(T\geq2\), an antiderivative is \(-2e^{-x/2}\), because its derivative is \(-2(-\frac12)e^{-x/2}=e^{-x/2}\). Hence

\int_2^T e^{-x/2}\,dx = \left[-2e^{-x/2}\right]_2^T = -2e^{-T/2}+2e^{-1}.

As \(T\to\infty\), \(e^{-T/2}\to0\), giving

\int_2^\infty e^{-x/2}\,dx = 2e^{-1} = \frac{2}{e}.

Check: the general formula gives \(e^{-ka}/k=e^{-(1/2)(2)}/(1/2)=e^{-1}\cdot2=2/e\), matching the cutoff calculation.

What happens when \(k=0\) or \(k<0\)?

When \(k=0\), the integrand is \(e^0=1\). The finite cutoff integral from zero is \(\int_0^T 1\,dx=T\), which tends to \(+\infty\). Thus the improper integral does not converge to a finite value.

When \(k<0\), set \(c=-k\), so \(c>0\). Then \(e^{-kx}=e^{cx}\), which grows rather than decays. For \(T\geq0\),

\int_0^T e^{-kx}\,dx = \frac{1-e^{-kT}}{k} = \frac{e^{cT}-1}{c}.

Because \(e^{cT}\to+\infty\) and \(c>0\), this cutoff integral tends to \(+\infty\). The half-line integral therefore diverges. The same reasoning applies when the lower endpoint is any fixed finite \(a\): the integral over \([a,T]\) also grows without bound as \(T\to\infty\).

Worked Example: The zero-rate case

Determine whether \(\int_1^\infty e^{0x}\,dx\) converges.

Since \(e^{0x}=1\), use a finite upper cutoff \(T\geq1\):

\int_1^T e^{0x}\,dx = \int_1^T 1\,dx = T-1.

As \(T\to\infty\), \(T-1\to+\infty\). Therefore \(\int_1^\infty e^{0x}\,dx\) diverges. Check: the integrand is constant, so each additional unit of interval adds exactly \(1\) to the cutoff integral; the values cannot approach a finite limit.

Worked Example: A negative rate

Determine whether \(\int_0^\infty e^{-(-2)x}\,dx\) converges.

The exponent simplifies to \(2x\), so the integrand is \(e^{2x}\). For a finite cutoff \(T\),

\int_0^T e^{2x}\,dx = \left[\frac12e^{2x}\right]_0^T = \frac{e^{2T}-1}{2}.

As \(T\to\infty\), \(e^{2T}\to+\infty\), and therefore \((e^{2T}-1)/2\to+\infty\). The improper integral diverges. Check: the derivative of \(\frac12e^{2x}\) is \(\frac12(2)e^{2x}=e^{2x}\), and the cutoff values grow without bound.

A common sign mistake

A frequent error is to assume that \(e^{-kT}\to0\) for every value of \(k\). That limit is true only when \(k>0\). For example, with \(k=-1\), the factor is \(e^{-kT}=e^T\), which increases without bound, not toward zero.

The incorrect calculation might say, “\(\lim_{T\to\infty}e^{-kT}=0\), so \(\int_0^\infty e^{-kx}\,dx=1/k\),” and then apply the formula to \(k=-1\) to claim a value of \(-1\). This cannot be right: the integrand \(e^x\) is positive, so every finite cutoff integral is positive.

\int_0^T e^x\,dx=e^T-1\longrightarrow+\infty, \qquad\text{not }-1.

The correction is to check the sign of \(k\) before taking the limit. If \(k>0\), the upper-endpoint exponential tends to zero and the integral has value \(1/k\). If \(k=0\) or \(k<0\), evaluate the cutoff behavior separately; in both cases the cutoff integral diverges to \(+\infty\).

Why the sign test is useful

The calculation gives a simple diagnostic for this specific family of improper integrals. A positive rate means exponential decay and a finite tail area; a zero rate leaves a constant amount of height over an interval that keeps growing; a negative rate makes the height itself grow. In the last two cases the cutoff areas increase without bound.

Keep the conclusion tied to its conditions. The formula \(1/k\) is not a general rule for any exponential tail: it applies to \(\int_0^\infty e^{-kx}\,dx\) only when \(k>0\). At \(k=0\), division by \(k\) is undefined and the integrand must be integrated directly. At \(k<0\), the finite-cutoff formula remains valid, but its limit diverges rather than giving a convergent improper integral.

Check Your Understanding

Use finite cutoffs and limits to answer the following questions.

Evaluate \(\int_0^\infty e^{-3x}\,dx\), and state why the cutoff limit exists.
Find \(\int_1^\infty e^{-2x}\,dx\) using the formula for a finite starting point.
Does \(\int_0^\infty e^{0x}\,dx\) converge? Show the cutoff integral.
For \(k=-\frac12\), describe the behavior of \(\int_0^T e^{-kx}\,dx\) as \(T\to\infty\).
Why is it invalid to use \(1/k\) as the value of the integral when \(k\leq0\)?