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Convergence Comparisons · Tutorial 346 of 998

Proving Divergence With Lower Bounds

Build useful lower bounds from dominant powers and logarithmic factors, then use direct comparison to establish divergence.

Intermediate 9 min read

What You'll Learn

  • State when a lower-bound comparison proves divergence of an improper integral.
  • Use dominant powers to find divergent lower bounds for rational integrands.
  • Compare logarithmic tails with the critical reciprocal-logarithm model.
  • Handle lower bounds near finite singular endpoints.
  • Identify why an upper bound by a divergent function does not prove divergence.

Why Lower Bounds Prove Divergence

To prove that an improper integral diverges, it is often enough to show that its integrand is at least as large as a simpler function whose integral is known to diverge. The comparison must point in the right direction: a large accumulated area under the simpler function forces at least as much accumulated area under the original one.

This is especially useful when an integrand contains several terms. Rather than finding an antiderivative, you can estimate the numerator from below and the denominator from above, or simplify logarithmic factors on the interval where they are being considered. The resulting bound may expose a familiar power or logarithmic divergence.

Definition: An improper integral of a nonnegative function diverges to \(+\infty\) if its truncated integrals grow without bound as the truncation approaches the improper endpoint. For example, \(\int_A^\infty f(x)\,dx=+\infty\) means \(\lim_{R\to\infty}\int_A^R f(x)\,dx=+\infty\).

The Direct Comparison Principle for Divergence

For an infinite interval, suppose \(f\) and \(g\) are continuous and nonnegative for \(x\ge A\), and \(f(x)\ge g(x)\) there. If \(\int_A^\infty g(x)\,dx\) diverges to \(+\infty\), then \(\int_A^\infty f(x)\,dx\) also diverges to \(+\infty\). The corresponding statement holds near a finite endpoint: if \(f(x)\ge g(x)\ge0\) near \(a\), and the improper integral of \(g\) there diverges, then so does the integral of \(f\).

1
Choose a comparison function.
Find a nonnegative \(g\) whose improper integral is known to diverge.
2
Prove the pointwise inequality.
Establish \(f(x)\ge g(x)\) throughout a whole tail or a whole neighborhood of the singular endpoint.
3
Compare truncated integrals.
Integrate the inequality over truncated intervals and pass to the improper limit.

The reason this works is contained in the inequality between truncated integrals. For \(R>A\), nonnegativity and \(f\ge g\) give:

$$ \int_A^R f(x)\,dx\ge \int_A^R g(x)\,dx. $$

If the right-hand side grows without bound as \(R\to\infty\), the left-hand side must also grow without bound. Near a finite endpoint \(a\), the same argument applies to \(\int_{a+\varepsilon}^c f(x)\,dx\) and \(\int_{a+\varepsilon}^c g(x)\,dx\) as \(\varepsilon\to0^+\). This is the direct comparison argument for divergence.

Theorem: If \(f\) and \(g\) are nonnegative on a tail (or sufficiently near a finite improper endpoint), \(f\ge g\), and the improper integral of \(g\) diverges to \(+\infty\), then the improper integral of \(f\) diverges to \(+\infty\). A finite initial part of the interval does not affect this conclusion, provided its integral is finite.

Useful Divergent Comparison Functions

Two power benchmarks are especially useful. For \(p\le1\), the tail integral \(\int_1^\infty x^{-p}\,dx\) diverges. For \(p\ge1\), the endpoint integral \(\int_0^1 x^{-p}\,dx\) diverges. These facts follow by integrating powers when \(p\ne1\), and by integrating \(1/x\) when \(p=1\). In particular, the logarithmic boundary cases are:

$$ \int_1^\infty \frac{dx}{x}=+\infty, \qquad \int_0^1 \frac{dx}{x}=+\infty. $$

Logarithmic factors can also produce divergence at a threshold. For \(x>1\), substitution \(u=\ln x\) gives the model tail \(\int dx/(x\ln x)\), whose truncated integral is \(\ln(\ln R)\) up to a constant and therefore grows without bound. Near zero, the substitution \(u=\ln(1/x)\) gives the corresponding model \(\int dx/(x\ln(1/x))\). The direction of the comparison is still decisive: the original integrand must be bounded below by the divergent model.

Worked Example: A Rational Integrand on an Infinite Interval

Worked Example: A Rational Integrand on an Infinite Interval

Determine whether the following integral diverges:

$$ \int_2^\infty \frac{3x^2+1}{x^3+4x}\,dx. $$

For \(x\ge2\), we have \(x^2\ge4\), so \(4x\le x^3\). Therefore the denominator satisfies \(x^3+4x\le2x^3\), while the numerator satisfies \(3x^2+1\ge3x^2\). Both denominators are positive, so these bounds imply:

$$ \frac{3x^2+1}{x^3+4x} \ge \frac{3x^2}{2x^3} = \frac{3}{2x} \qquad (x\ge2). $$

The comparison integral diverges:

$$ \int_2^R \frac{3}{2x}\,dx = \frac{3}{2}\bigl(\ln R-\ln2\bigr) = \frac{3}{2}\ln\!\left(\frac{R}{2}\right) \longrightarrow+\infty \quad\text{as }R\to\infty. $$

By direct comparison, the original integral diverges to \(+\infty\). The bound also shows why comparing dominant powers is useful here: the leading terms in the numerator and denominator leave a reciprocal-power lower bound.

Check: The derived inequality holds on the entire interval of integration, and its right-hand side has truncated integral tending to \(+\infty\). It therefore proves divergence in the required direction.

Worked Example: A Logarithmic Factor in the Tail

Worked Example: A Logarithmic Factor in the Tail

Consider:

$$ \int_e^\infty \frac{2x+5}{x^2(\ln x+1)}\,dx. $$

For \(x\ge e\), \(\ln x\ge1\), so \(\ln x+1\le2\ln x\). Also \(2x+5\ge2x\). Thus:

$$ \frac{2x+5}{x^2(\ln x+1)} \ge \frac{2x}{2x^2\ln x} = \frac{1}{x\ln x}. $$

For \(R>e\), the comparison integral is evaluated by the substitution \(u=\ln x\), \(du=dx/x\):

$$ \int_e^R \frac{dx}{x\ln x} = \int_1^{\ln R}\frac{du}{u} = \ln(\ln R)-\ln1 = \ln(\ln R). $$

As \(R\to\infty\), \(\ln(\ln R)\to+\infty\). The original integral therefore diverges by the lower-bound comparison.

Check: The inequalities use \(\ln x\ge1\), which is valid for every \(x\ge e\), and the comparison function has divergent tail integral. Both conditions needed for the argument hold.

Worked Example: A Logarithmic Singularity at Zero

Worked Example: A Logarithmic Singularity at Zero

Determine whether:

$$ \int_0^{e^{-1}}\frac{dx}{x\bigl(\ln(1/x)+1\bigr)} $$

diverges. For \(0<x\le e^{-1}\), \(\ln(1/x)\ge1\), so \(\ln(1/x)+1\le2\ln(1/x)\). Hence:

$$ \frac{1}{x\bigl(\ln(1/x)+1\bigr)} \ge \frac{1}{2x\ln(1/x)}. $$

To evaluate the truncated comparison integral, take \(0<\varepsilon<e^{-1}\) and set \(u=\ln(1/x)\). Since \(du=-dx/x\), the bounds \(x=\varepsilon\) and \(x=e^{-1}\) correspond to \(u=\ln(1/\varepsilon)\) and \(u=1\), respectively. Thus:

$$ \int_\varepsilon^{e^{-1}}\frac{dx}{2x\ln(1/x)} = \frac12\int_1^{\ln(1/\varepsilon)}\frac{du}{u} = \frac12\ln\!\bigl(\ln(1/\varepsilon)\bigr). $$

As \(\varepsilon\to0^+\), \(\ln(1/\varepsilon)\to+\infty\), and this expression tends to \(+\infty\). The original integral diverges by comparison.

Check: The comparison inequality holds throughout the interval \(0<x\le e^{-1}\), and the truncated comparison integral tends to \(+\infty\) at the same endpoint.

Worked Example: A Power Lower Bound Near an Endpoint

Worked Example: A Power Lower Bound Near an Endpoint

Consider the finite-endpoint integral:

$$ \int_0^1\frac{dx}{x^2+x^{3/2}}. $$

For \(0<x\le1\), \(x^2\le x^{3/2}\), since \(x^{1/2}\le1\). It follows that \(x^2+x^{3/2}\le2x^{3/2}\), so:

$$ \frac{1}{x^2+x^{3/2}} \ge \frac{1}{2x^{3/2}}. $$

For \(0<\varepsilon<1\), integrate the lower bound:

$$ \int_\varepsilon^1\frac{dx}{2x^{3/2}} = \frac12\left[-2x^{-1/2}\right]_\varepsilon^1 = \varepsilon^{-1/2}-1. $$

This tends to \(+\infty\) as \(\varepsilon\to0^+\), so the original integral diverges to \(+\infty\). The power \(x^{-3/2}\) is a useful lower comparison because its exponent is at least \(1\), the divergent threshold at a zero endpoint.

Check: For every \(0<\varepsilon<1\), the truncated original integral is at least \(\varepsilon^{-1/2}-1\), which grows without bound as the endpoint is approached.

A Common Mistake: Using an Upper Bound to Claim Divergence

A comparison with a divergent function proves divergence only when the original nonnegative integrand is bounded below by that function. An upper bound by a divergent function gives no such conclusion. For example, when \(x\ge1\), \(x^{-2}\le x^{-1}\), and \(\int_1^\infty x^{-1}\,dx\) diverges. It would be incorrect to conclude from this inequality that \(\int_1^\infty x^{-2}\,dx\) diverges.

In fact, direct calculation shows the opposite:

$$ \int_1^R \frac{dx}{x^2} = \left[-\frac{1}{x}\right]_1^R = 1-\frac{1}{R} \longrightarrow1 \quad\text{as }R\to\infty. $$

The integral converges. The correction is to check the direction before applying comparison: to prove divergence, establish a lower bound by a function with a divergent improper integral. An upper bound by a divergent function may still hold for an integrable function, as this example demonstrates.

Choosing and Checking a Lower Bound

A practical lower-bound argument usually involves two choices: where the comparison should hold and which divergent model to use. The estimate need not be sharp; it only needs to be valid on a full tail or sufficiently close to the singular endpoint. A bound that holds at just a few points, or only for a finite portion of the interval, cannot establish divergence of the improper integral.

  • For a rational expression, bound the numerator below and the positive denominator above. Keep track of positivity before dividing.
  • For a power near infinity, aim for a lower bound proportional to \(x^{-p}\) with \(p\le1\); near zero, aim for one with \(p\ge1\).
  • For a logarithmic tail, check whether a lower bound by \(1/(x\ln x)\) is available. Its integral diverges.
  • For a finite logarithmic singularity, use \(1/(x\ln(1/x))\) as a comparison where the logarithm is positive.
  • State the interval on which every estimate holds, then evaluate the truncated comparison integral and take the relevant limit.

These comparisons matter because divergence is sometimes easier to prove than to compute. A lower bound can establish that the accumulated area is unbounded even when the original integrand has no convenient antiderivative. The argument is complete once the inequality, the divergent comparison integral, and the improper limit have all been checked.

Check Your Understanding

For each question, identify the comparison direction and the improper limit involved.

  1. What pointwise inequality is needed to prove that a nonnegative improper integral diverges by direct comparison?
  2. For \(x\ge2\), why can \(x^3+4x\) be bounded above by \(2x^3\)?
  3. Evaluate \(\int_e^R dx/(x\ln x)\), and state its limit as \(R\to\infty\).
  4. Why does \(x^{-2}\le x^{-1}\) for \(x\ge1\) fail to prove that \(\int_1^\infty x^{-2}\,dx\) diverges?
  5. For an integral near zero, what happens to \(\int_\varepsilon^1 x^{-3/2}\,dx\) as \(\varepsilon\to0^+\)?

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