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The Comparison Theorem

Calculus 2 · Chapter 2: Indeterminate Forms & Improper IntegralsSection 2.4 · lesson 11 of 5411 min · free

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In this lesson

Settling convergence without ever computing the integral.

  • State both directions of the Comparison Theorem
  • Read off an integrand's large-x behaviour to guess the outcome
  • Choose a benchmark and aim the inequality the right way
  • Handle integrands with no elementary antiderivative at all

Transcript

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MAT-265 - Calculus 2 · Section 2.4

The Comparison Theorem

A physicist studying heat radiation needs to know whether the total energy described by a certain integral is finite. If it is, the model predicts a well-defined total. If it isn't, the model breaks down. The integral is one over x cubed plus one, from one to infinity. So the natural plan is: find an antiderivative, then take a limit. And that plan runs into trouble immediately. The antiderivative of one over x cubed plus one requires partial fractions, a logarithm, and an arctangent — producing a messy expression that's painful to evaluate at infinity. And for many improper integrals, no elementary antiderivative exists at all. But hold on. Step back and ask what question actually needs answering.

The Problem

When the Antiderivative Is Out of Reach

Often the only thing that matters is whether the integral converges or diverges. Does it settle to a finite value, or grow without bound? And for that question, the exact value is completely unnecessary. The Comparison Theorem answers it — by comparing your difficult integrand to a simpler one whose behavior is already understood. And the strategy mirrors ordinary everyday reasoning. To show a stack of coins is worth less than ten dollars, you don't need to count it. It's enough to show it's shorter than a stack you already know is worth less than ten dollars. That's the entire idea. Bound the thing you don't understand by something you do.

The Theorem

Two Directions, Not Interchangeable

The theorem rests on a simple geometric truth. If one nonnegative curve lies entirely below another, the area under the lower curve cannot exceed the area under the upper one. Translate area into improper integral and you get two complementary statements. Suppose f and g are continuous, and zero is at most f, which is at most g, for all x at least a. Part one. If the integral of g converges, then the integral of f also converges. Part two. If the integral of f diverges, then the integral of g also diverges. These express the same idea from opposite ends. Part one is the squeeze from above. A smaller nonnegative area, trapped beneath a finite area, must itself be finite. Part two is the push from below. If even the smaller area is infinite, the larger area sitting on top of it must be infinite too. Now — here's what the theorem does not say, and this is where students go wrong. The logic only runs one way in each part. If the larger function's integral diverges, that tells you nothing about the smaller one. The smaller area could be finite or infinite. Likewise, if the smaller function's integral converges, the larger one is undetermined. A very common error is to bound f above by a divergent g, and then conclude nothing useful. Showing f is at most g, where the integral of g diverges, leaves the question of f wide open. The comparison has to point the right way for the conclusion you're after.

Figure 2.4.1

The Two Directions of Comparison

Look at the two pictures, because they make the asymmetry obvious. On the left, a nonnegative f lies under a convergent g. The shaded area beneath f is trapped under something finite. So the integral of f converges. That's Part one. On the right, a nonnegative f lies above a divergent g. The area beneath f is at least the infinite area beneath g. So the integral of f diverges. That's Part two. In the left picture the useful information flows downward. In the right picture it flows upward. Get the direction backwards and you have a true inequality that proves absolutely nothing.

The Benchmarks

A Comparison Is Only as Good as What You Compare To

A comparison is only as useful as the family of known integrals it draws on. So you need benchmarks. And the most important one is the p-integral, which we established in section two point two. The integral from one to infinity of one over x to the p converges if p is greater than one, and diverges if p is at most one. It's worth seeing once more why the threshold sits exactly at one. For p not equal to one, the antiderivative is a power. And t to the one minus p either shrinks to zero or explodes, depending on the sign of one minus p. At p equals one the antiderivative becomes log t, which also tends to infinity. Hence the clean rule. Convergence is reserved for powers strictly greater than one. There's a second reliable benchmark, and it's often the more powerful one. The exponential integral. The integral from one to infinity of e to the minus x converges, to e to the minus one. And more generally, the integral of e to the minus k x converges for any positive k. Exponential decay is so fast that it dominates — and outpaces — every power of x. Which makes e to the minus k x an excellent ceiling for a very wide range of integrands.

Worked Examples 1 and 2

Aim the Inequality the Right Way

The central skill is selecting g wisely. And a good comparison function has to do two things at once. Bound f in the needed direction. And have known convergence behavior. The guiding instinct is to ask how the integrand behaves for large x — because the tail near infinity is what decides convergence. First example. Does the integral of one over x cubed plus one converge? For large x, that plus one is negligible next to x cubed. So the integrand behaves like one over x cubed. And the integral of x to the minus three converges, since p is three. So the goal is to bound the integrand above by one over x cubed. For every x at least one, x cubed plus one is greater than x cubed. So one over x cubed plus one is less than one over x cubed. Both sides nonnegative. Part one applies. The integral converges. And the physicist's question is answered — the total is finite — without computing the value at all. Second example. The integral of one over root x plus one. For large x this behaves like x to the minus one half, and that diverges since p is one half. So now you want the opposite direction. Bound the integrand below by something divergent. For x at least one, root x plus one is at most two root x. So one over root x plus one is at least one over two root x. That lower bound has a divergent integral — and a nonzero constant multiple of a divergent integral still diverges. Part two applies. The integral diverges. So notice: the directions were not interchangeable. Decide which outcome the large-x behavior suggests first. Then aim the inequality to match.

Worked Example 3

An Integrand With No Antiderivative At All

Now the example that shows why this technique is essential rather than merely convenient. Show that the integral from one to infinity of e to the minus x squared converges. This is the bell-curve integrand from statistics. And as we saw in section one point seven, it has no elementary antiderivative. So direct evaluation is impossible. Not difficult. Impossible. But comparison is easy. For x at least one, x squared is at least x. So minus x squared is at most minus x. And since the exponential function is increasing, e to the minus x squared is at most e to the minus x. Both functions are positive. And the integral of e to the minus x converges, to e to the minus one. By Part one, the integral of e to the minus x squared converges. Done. We now know a definite fact about an integral we cannot evaluate. This is exactly the kind of integral — central to probability and statistics — where comparison isn't merely convenient. It's the only practical route to an answer.

A Practical Strategy

Four Steps, Every Time

The examples share a repeatable procedure. Four steps, every time. Step one. Examine the large-x behavior. Identify the dominant term in the integrand to guess whether the integral converges or diverges. Drop the lower-order terms. One over x cubed plus one behaves like one over x cubed. One over root x plus one behaves like one over root x. Step two. Pick the matching benchmark. Compare against a p-integral, or an exponential, whose convergence you know. And your guess from step one tells you which direction you need. Step three. Establish the inequality. For convergence, show f is between zero and g, with the integral of g convergent. For divergence, show g is between zero and f, with the integral of g divergent. And justify that inequality algebraically, for all x in the domain. Not just for large x. Step four. Invoke the theorem. State which part applies and write the conclusion. One shortcut worth knowing for ratios of polynomials. Look at the degree gap — the denominator's degree minus the numerator's. A gap of two or more makes the integrand behave like one over x to the p with p at least two. That signals convergence. A gap of exactly one behaves like one over x. And that diverges.

Key Takeaways

What to Carry Forward

Five things to carry forward. The Comparison Theorem decides convergence or divergence without evaluating the integral — by bounding the integrand with a simpler function of known behavior. Part one: if f is squeezed between zero and g, and the integral of g converges, then the integral of f converges. Part two: if g is squeezed between zero and f, and the integral of g diverges, then the integral of f diverges. The directions are not symmetric. Bound above by a convergent function to prove convergence. Bound below by a divergent one to prove divergence. A correct inequality aimed the wrong way proves absolutely nothing. The key benchmark is the p-integral, converging exactly when p is greater than one. And exponentials converge and dominate all powers. And the strategy: read off the dominant large-x term, choose the matching benchmark, prove the inequality, then cite the relevant part of the theorem. That closes Chapter Two. Next up is Chapter Three — Hyperbolic Functions, beginning with their definitions and graphs.

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FAQ

Questions students ask us

About The Comparison Theorem

What is the Comparison Theorem for improper integrals?
The Comparison Theorem decides whether an improper integral converges or diverges without evaluating it. If f and g are continuous with 0 \le f(x) \le g(x) for all x \ge a, then convergence of \int_a^\infty g\,dx implies convergence of \int_a^\infty f\,dx. Likewise, if \int_a^\infty f\,dx diverges, then \int_a^\infty g\,dx diverges too. Geometrically, the area under a lower nonnegative curve cannot exceed the area under the upper one.
How do you choose a comparison function for an improper integral?
Choose g by finding the integrand's dominant term for large x, since the tail near infinity decides convergence. Drop lower-order terms to guess the outcome, then pick a known benchmark. The integral \int_1^\infty x^{-p}\,dx converges exactly when p>1, and \int_1^\infty e^{-kx}\,dx converges for any k>0. For example, \frac{1}{\sqrt{x}+1} behaves like x^{-1/2}, so expect divergence and bound it below by \frac{1}{2\sqrt{x}} for x \ge 1.
Does the integral of e^(-x^2) from 1 to infinity converge?
Yes, \int_1^\infty e^{-x^2}\,dx converges, even though e^{-x^2} has no elementary antiderivative. For x \ge 1, x^2 \ge x, so -x^2 \le -x. Because the exponential function is increasing, this gives 0 < e^{-x^2} \le e^{-x}. Since \int_1^\infty e^{-x}\,dx = e^{-1} is finite, Part one of the Comparison Theorem shows that the original integral converges.
What is the most common mistake when using the Comparison Theorem?
The most common mistake is aiming the inequality the wrong way. Showing f \le g where \int g diverges proves nothing about f, and showing f \ge g where \int g converges proves nothing either. To prove convergence, bound f above by a function with a convergent integral. To prove divergence, bound f below by a function with a divergent integral. Also verify the inequality algebraically for every x in the interval, not just for large x.

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