Exponential Decay Compared With Powers

Why Compare Exponentials With Powers?

Improper integrals on an infinite interval often contain both a polynomial factor and an exponential factor. For instance, the factor $x^6$ grows without bound, while $e^{-2x}$ tends to zero. Looking at either factor alone does not determine what their product does.

The decisive fact is that, at infinity, exponential decay outpaces every fixed power. This lets us compare a product such as $x^6e^{-2x}$ with a familiar power tail such as $x^{-2}$. Since the latter has a convergent integral, direct comparison can establish convergence of the original integral.

Throughout, we study tails beginning at $1$, so the integrands considered are continuous there unless stated otherwise. This keeps the question focused on behavior at infinity rather than on possible finite-endpoint singularities.

The Exponential-Dominance Result

Theorem: If $a>0$ and $r\in\mathbb{R}$, then $$ \lim_{x\to\infty}x^r e^{-ax}=0. $$ Consequently, for any real $p$ and any $q>1$, there is a number $X\ge1$ such that $$ 0\le x^p e^{-ax}\le x^{-q}\qquad (x\ge X). $$

The first statement says that the exponential factor wins even if the power grows, as it does when $r>0$. The second statement is especially useful for integration: it says that an exponentially decaying power product is eventually bounded above by a convergent power tail. The threshold $X$ can depend on $a$, $p$, and $q$.

Why Exponential Decay Wins

Take logarithms of the positive expression $x^r e^{-ax}$, for $x\ge1$:

\ln\!\left(x^r e^{-ax}\right)=r\ln x-ax.

To see which term controls this expression, first verify that $\ln x/x\to0$. For $x\ge1$, the integral formula for the natural logarithm gives

\ln x=\int_1^x\frac{1}{t}\,dt \le \int_1^x\frac{1}{\sqrt{t}}\,dt =2(\sqrt{x}-1).

Here $1/t\le1/\sqrt{t}$ for $t\ge1$. Dividing by $x$ shows that $0\le \ln x/x\le2/\sqrt{x}$, so $\ln x/x\to0$.

Control the logarithmic term.
Since $\ln x/x\to0$, for the fixed $r$ and $a>0$, there is a threshold beyond which $|r|\ln x\le ax/2$.

Bound the logarithm of the product.
For all sufficiently large $x$, $r\ln x-ax\le |r|\ln x-ax\le-ax/2$.

Exponentiate the bound.
It follows that $0<x^r e^{-ax}\le e^{-ax/2}$, and $e^{-ax/2}\to0$. Thus $x^r e^{-ax}\to0$.

To obtain the comparison with $x^{-q}$, divide the proposed bound by that power:

\frac{x^p e^{-ax}}{x^{-q}}=x^{p+q}e^{-ax}\longrightarrow0.

The limit is zero by the result just proved, with $r=p+q$. Therefore the ratio is at most $1$ for all sufficiently large $x$, which gives $x^p e^{-ax}\le x^{-q}$ there.

From the Bound to Integral Convergence

The power-tail criterion states that $\int_1^\infty x^{-q}\,dx$ converges when $q>1$. In fact, for $B>1$,

\int_1^B x^{-q}\,dx =\frac{1-B^{1-q}}{q-1} \longrightarrow \frac{1}{q-1}\qquad (B\to\infty).

If a nonnegative integrand $f$ satisfies $f(x)\le x^{-q}$ for all sufficiently large $x$, the direct comparison theorem implies that $\int_1^\infty f(x)\,dx$ converges, provided $f$ is integrable on the remaining finite interval. For the continuous functions used below, that finite-interval condition holds.

Consequence: For every $a>0$ and every real $p$, $$ \int_1^\infty x^p e^{-ax}\,dx $$ converges. To apply direct comparison, choose any $q>1$; eventually $x^p e^{-ax}\le x^{-q}$.

This conclusion includes negative powers as well as positive ones. The exponential decay controls any fixed power in either direction on the interval $[1,\infty)$.

Worked Example: Comparing a Growing Power Product With a Power Tail

Classify $$ \int_1^\infty x^6 e^{-2x}\,dx. $$

The integrand is nonnegative and continuous on $[1,\infty)$. Choose the convergent comparison function $g(x)=x^{-2}$. The ratio is

\frac{x^6e^{-2x}}{x^{-2}}=x^8e^{-2x}.

The exponential-dominance result, with $r=8$ and $a=2$, gives $x^8e^{-2x}\to0$. Therefore this ratio is at most $1$ for all sufficiently large $x$, so $0\le x^6e^{-2x}\le x^{-2}$ on that tail.

The comparison integral is

\int_1^\infty x^{-2}\,dx =\lim_{B\to\infty}\left[-\frac1x\right]_1^B =\lim_{B\to\infty}\left(1-\frac1B\right)=1.

It converges. Direct comparison now shows that $\int_1^\infty x^6e^{-2x}\,dx$ converges as well. The classification is confirmed: the growing factor $x^6$ does not overcome the exponential decay.

Applying Limit Comparison

Direct comparison is useful when an integrand can be bounded above by a known convergent function. Limit comparison is useful when the ratio of two positive functions approaches a finite positive number. In that case, the functions have the same convergence behavior on the tail.

Limit Comparison Theorem: Suppose $f$ and $g$ are positive and continuous for all sufficiently large $x$, and $$ \lim_{x\to\infty}\frac{f(x)}{g(x)}=L \qquad\text{where }0<L<\infty. $$ Then $\int^\infty f(x)\,dx$ and $\int^\infty g(x)\,dx$ either both converge or both diverge.

The reason is that, once the ratio is sufficiently close to $L$, it lies between two positive constants. Thus $f$ is bounded above and below by positive constant multiples of $g$ on a tail, and the two integrals have the same convergence behavior.

Worked Example: Limit Comparison With an Exponential-Power Product

Classify $$ \int_1^\infty (x^2+1)\left(1+\frac{2}{x}\right)e^{-x}\,dx. $$

Set $$ f(x)=(x^2+1)\left(1+\frac{2}{x}\right)e^{-x}, \qquad g(x)=x^2e^{-x}. $$ Both functions are positive and continuous for $x\ge1$. Their ratio simplifies as follows:

\frac{f(x)}{g(x)} =\frac{(x^2+1)(1+2/x)e^{-x}}{x^2e^{-x}} =\left(1+\frac{1}{x^2}\right)\left(1+\frac{2}{x}\right) \longrightarrow 1\cdot1=1.

Since the limit is finite and positive, limit comparison reduces the question to $\int_1^\infty g(x)\,dx$. Compare $g(x)=x^2e^{-x}$ with $x^{-2}$:

\frac{x^2e^{-x}}{x^{-2}}=x^4e^{-x}\longrightarrow0.

Hence $0\le g(x)\le x^{-2}$ for all sufficiently large $x$. Since $\int_1^\infty x^{-2}\,dx=1$, direct comparison proves that $\int_1^\infty g(x)\,dx$ converges. Limit comparison with ratio $1$ then proves that the original integral converges.

This example uses the tests in sequence: first limit comparison identifies a simpler integrand with the same behavior, and then direct comparison establishes convergence of that simpler integral.

Polynomial Numerators and Exponential Denominators

A common form is a polynomial divided by an increasing exponential. If $P$ is a polynomial of degree $d$, then its terms grow no faster than a constant multiple of $x^d$ for $x\ge1$. Consequently, an integrand such as $P(x)e^{-ax}$, with $a>0$, can be compared with a power tail: the ratio of $x^d e^{-ax}$ to $x^{-q}$ is $x^{d+q}e^{-ax}$, which tends to zero for every fixed $q>1$.

The same reasoning applies to $P(x)/e^{ax}$, because $1/e^{ax}=e^{-ax}$. Positivity matters when applying direct comparison; if a polynomial changes sign, convergence of its absolute value is a separate question and is not needed for the nonnegative examples here.

Worked Example: A Polynomial Divided by an Exponential

Classify $$ \int_1^\infty \frac{3x^2+2x+1}{e^x}\,dx. $$

For $x\ge1$, $3x^2+2x+1\le6x^2$, since $2x\le2x^2$ and $1\le x^2$. Thus the integrand is nonnegative and satisfies

0\le \frac{3x^2+2x+1}{e^x}\le 6x^2e^{-x}.

Compare the upper bound with $x^{-2}$. The ratio is

\frac{6x^2e^{-x}}{x^{-2}}=6x^4e^{-x}\longrightarrow0.

Therefore $6x^2e^{-x}\le x^{-2}$ for all sufficiently large $x$. As $\int_1^\infty x^{-2}\,dx=1$, direct comparison proves convergence of the upper-bound integral, and a second direct comparison proves convergence of the given integral.

The inequalities used in the argument have the correct direction: the integrand is bounded above by a function whose tail is integrable.

A Common Mistake

A frequent error is to inspect the power and ignore the exponential. For example, one might claim that $x^6e^{-x}$ grows because $x^6$ grows. That conclusion does not follow: the product includes $e^{-x}$, and its effect must be accounted for.

The comparison calculation makes the error visible. To compare $x^6e^{-x}$ with $x^{-2}$, form the ratio:

\frac{x^6e^{-x}}{x^{-2}}=x^8e^{-x}.

A mistaken calculation might say that this ratio tends to infinity because $x^8\to\infty$. But the ratio contains $e^{-x}$ as well. Taking its logarithm gives

\ln(x^8e^{-x})=8\ln x-x\longrightarrow-\infty,

so the ratio actually tends to zero. The corrected comparison is $x^6e^{-x}\le x^{-2}$ eventually, which proves convergence by direct comparison. The general rule is to evaluate the whole product or ratio, not just one factor.

Summary

For every positive $a$ and fixed real $r$, the product $x^r e^{-ax}$ tends to zero as $x\to\infty$. Choosing a power tail $x^{-q}$ with $q>1$ gives an eventual integrable upper bound. This proves convergence of integrals containing a fixed power multiplied by exponential decay.

Use direct comparison when an eventual upper bound is available. Use limit comparison when a positive integrand has a ratio tending to a finite, positive number with a simpler comparison function. In both cases, check positivity and apply the comparison on a tail where the required conditions hold.

Check Your Understanding

Use the exponential-dominance result and the comparison tests to answer these questions.

For $a>0$ and $r\in\mathbb{R}$, what is the limit of $x^r e^{-ax}$ as $x\to\infty$, and what feature of the logarithm proves it?
Which power tail $x^{-q}$ can be used to show convergence of $\int_1^\infty x^3e^{-2x}\,dx$, and why must $q>1$?
Explain why $x^5e^{-x}$ is eventually smaller than $x^{-2}$.
If $f(x)>0$, $g(x)>0$, and $f(x)/g(x)\to4$, what does limit comparison say about their improper integrals?
What is the error in deciding the behavior of $x^6e^{-x}$ by considering only the growth of $x^6$?