Choosing the Direction of a Comparison

Why the Direction Matters

An inequality between two nonnegative functions can control the size of an improper integral. But the conclusion depends on which function is larger. To prove that an integral is finite, it is useful to bound its integrand above by one whose integral is finite. To prove that it is infinite, it is useful to bound its integrand below by one whose integral is infinite.

The distinction is easy to reverse when comparing expressions quickly. A bound by a function with a divergent integral does not, by itself, show that the smaller function also has a divergent integral. The smaller function might still have a finite integral.

Key idea: To prove convergence, compare the integrand from above with a convergent integrand. To prove divergence, compare it from below with a divergent integrand.

The Comparison Rules

Consider two nonnegative functions \(f\) and \(g\) on the same interval with an improper endpoint. Assume they are integrable on every truncated interval away from that endpoint, and that the stated inequality holds throughout the interval being compared. The integral is improper because it is defined by a limit of integrals over truncated intervals.

Theorem: Direct comparison for nonnegative improper integrals. If \(0\le f(x)\le g(x)\) throughout the interval and \(\int g(x)\,dx\) converges, then \(\int f(x)\,dx\) converges. If \(0\le g(x)\le f(x)\) throughout the interval and \(\int g(x)\,dx\) diverges, then \(\int f(x)\,dx\) diverges.

These two statements are the useful directions of direct comparison. In the first, the target integrand \(f\) is no larger than a function with finite total area. In the second, \(f\) is at least as large as a function whose accumulated area is unbounded.

The theorem requires nonnegative integrands. It also requires the inequality on the interval relevant to the improper limit, not merely at a few points or near some unrelated part of the interval. If an integral has more than one improper point, each relevant piece must be treated separately.

Why the Rules Work

The proof comes from first comparing ordinary integrals on truncated intervals. For clarity, take an endpoint \(b\) approached from the left. For \(c<b\), integrate over the proper interval from \(a\) to \(c\). Nonnegativity and the pointwise inequality give an inequality between these truncated integrals.

Compare truncated integrals.
If \(0\le f\le g\), then for every \(c<b\), \(\int_a^c f(x)\,dx\le\int_a^c g(x)\,dx\), because integrating a nonnegative difference preserves the inequality.

Pass to the endpoint limit for convergence.
If \(\int_a^b g(x)\,dx\) is finite, the truncated integrals of \(g\) are bounded above by that finite value. The truncated integrals of \(f\) are nonnegative, increasing as \(c\) increases, and bounded above by the corresponding integrals of \(g\). They therefore have a finite limit, which is \(\int_a^b f(x)\,dx\).

Pass to the endpoint limit for divergence.
If \(0\le g\le f\), then \(\int_a^c g(x)\,dx\le\int_a^c f(x)\,dx\) for every \(c<b\). If the truncated integrals of \(g\) grow without bound as \(c\) approaches \(b\), those of \(f\) must also grow without bound.

This argument explains the direction directly: a finite upper bound controls the target from above, while an unbounded lower bound forces the target to be unbounded as well.

Worked Examples: An Upper Bound Proves Convergence

Worked Example: A singularity at zero

Determine whether \(\int_0^1 \frac{1}{\sqrt{x}(1+\sqrt{x})}\,dx\) converges.

For \(0<x\le1\), we have \(1+\sqrt{x}\ge1\). Since the denominator is positive, taking reciprocals gives

0\le \frac{1}{\sqrt{x}(1+\sqrt{x})} \le \frac{1}{\sqrt{x}}.

The comparison integral is finite:

\int_0^1 \frac{1}{\sqrt{x}}\,dx =\lim_{\delta\to0^+}\int_\delta^1 x^{-1/2}\,dx =\lim_{\delta\to0^+}\left[2\sqrt{x}\right]_\delta^1 =\lim_{\delta\to0^+}(2-2\sqrt{\delta}) =2.

Thus the target integral converges by comparison. As a check, it can also be evaluated: set \(t=\sqrt{x}\), so \(x=t^2\) and \(dx=2t\,dt\). The truncated integral becomes

\int_{\delta}^1 \frac{1}{\sqrt{x}(1+\sqrt{x})}\,dx =\int_{\sqrt{\delta}}^1 \frac{2}{1+t}\,dt =2\ln 2-2\ln(1+\sqrt{\delta}).

As \(\delta\to0^+\), this tends to \(2\ln 2\), a finite value. The comparison and the direct evaluation agree.

Worked Examples: A Lower Bound Proves Divergence

Worked Example: A lower bound by a divergent integrand

Determine whether \(\int_0^1 \frac{1}{x(1+x)}\,dx\) converges.

For \(0<x\le1\), \(1+x\le2\). Multiplying by the positive quantity \(x\) gives \(x(1+x)\le2x\); taking reciprocals gives

\frac{1}{x(1+x)}\ge\frac{1}{2x}>0.

The lower comparison has a divergent integral:

\int_\delta^1\frac{1}{2x}\,dx =\frac12\left[\ln x\right]_\delta^1 =-\frac12\ln\delta \longrightarrow+\infty \qquad(\delta\to0^+).

Therefore the target integral diverges. To check the conclusion directly, decompose the integrand and evaluate a truncated integral:

\frac{1}{x(1+x)}=\frac1x-\frac{1}{1+x}, \qquad \int_\delta^1\frac{1}{x(1+x)}\,dx =-\ln\delta-\ln2+\ln(1+\delta).

As \(\delta\to0^+\), the term \(-\ln\delta\) tends to \(+\infty\), while \(\ln(1+\delta)\) tends to \(0\). The truncated target integrals are unbounded, as the comparison predicted.

Worked Examples: Check the Bound on the Whole Interval

Worked Example: Convergence on an infinite interval

Determine whether \(\int_1^\infty \frac{1}{1+x^2}\,dx\) converges.

For every \(x\ge1\), \(1+x^2\ge x^2>0\). Hence

0\le\frac{1}{1+x^2}\le\frac{1}{x^2}.

The upper comparison converges:

\int_1^\infty\frac{1}{x^2}\,dx =\lim_{R\to\infty}\left[-\frac1x\right]_1^R =\lim_{R\to\infty}\left(1-\frac1R\right) =1.

Thus the target integral converges. Its direct evaluation confirms the result:

\int_1^R\frac{1}{1+x^2}\,dx =\left[\arctan x\right]_1^R =\arctan R-\frac{\pi}{4} \longrightarrow\frac{\pi}{2}-\frac{\pi}{4} =\frac{\pi}{4}.

The derivative \(\frac{d}{dx}\arctan x=\frac{1}{1+x^2}\) verifies the antiderivative used. The comparison bound holds for every \(x\ge1\), as required.

Common Mistakes: An Inequality Is Not Enough

A frequent error is to use the right comparison function but draw the conclusion permitted by the opposite inequality. For example, on \(0<x\le1\), \(1\le1/x\), and \(\int_0^1 1/x\,dx\) diverges. It would be incorrect to conclude that \(\int_0^1 1\,dx\) diverges: in fact,

\int_0^1 1\,dx=1.

The inequality says the target \(1\) is below the divergent function \(1/x\). A divergent upper comparison gives no conclusion about whether the smaller target integral converges. To prove convergence here, use the upper bound \(1\le1\), whose comparison integral is finite.

The reverse mismatch also fails. On \(0<x\le1\), \(1/x\ge1\), and \(\int_0^1 1\,dx=1\) is finite. This does not show that \(\int_0^1 1/x\,dx\) is finite. A finite lower comparison cannot control how large the target becomes. In that case, the target itself is a divergent lower comparison for its own integral.

Check before concluding: Write the inequality with the target integrand clearly identified. For convergence, the target must be on the smaller side of a convergent comparison. For divergence, it must be on the larger side of a divergent comparison.

A Practical Choice of Direction

When applying direct comparison, begin with the conclusion you want rather than with an inequality that happens to look familiar. Then select a comparison function whose integral is already known to converge or diverge, and prove the inequality in the direction that supports that conclusion.

Goal	Required bound	Comparison integral
Prove \(\int f\) converges	\(0\le f\le g\)	\(\int g\) converges
Prove \(\int f\) diverges	\(0\le g\le f\)	\(\int g\) diverges

Finally, check the domain. If an inequality only holds for \(x\ge A\), it can support a comparison on that tail, but it does not automatically establish a bound on an entire interval that also includes points below \(A\). Likewise, if the integrand changes sign, the nonnegative comparison theorem cannot be applied as stated; sign and absolute convergence require separate care.

Check Your Understanding

For each question, state both the inequality direction and the known behavior of the comparison integral that would justify the conclusion.

To prove that an improper integral of a nonnegative function \(f\) converges, should a convergent comparison function be above or below \(f\)?
Suppose \(0\le g\le f\) and \(\int g\) diverges. What follows about \(\int f\), and why?
If \(0\le f\le g\) and \(\int g\) diverges, can direct comparison determine whether \(\int f\) converges? Explain.
On \(0<x\le1\), the function \(1\) is below \(1/x\). Why does the divergence of \(\int_0^1 1/x\,dx\) not prove that \(\int_0^1 1\,dx\) diverges?
Why must an inequality be checked throughout the interval used for the improper limit?