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Convergence Comparisons · Tutorial 344 of 998

Oscillatory Factors and Absolute Integrability

Learn to decide absolute integrability by combining a simple upper bound with lower bounds on repeated intervals where the oscillatory factor remains large.

Intermediate 10 min read

What You'll Learn

  • Define absolute integrability on an infinite interval using an improper integral.
  • Use |sin x| ≤ 1 to prove convergence when p > 1.
  • Construct repeated intervals on which |sin x| ≥ 1/2.
  • Prove divergence for p ≤ 1, including the critical case p = 1.
  • Explain why oscillations do not make the absolute integral converge below the threshold.
  • Apply the classification to specific exponents with explicit bounds.

Why Oscillation Needs Careful Comparison

The function \(\sin x\) repeatedly changes sign and has infinitely many zeros as \(x\) increases. Those features can make an integral involving \(\sin x\) behave differently from an integral involving a positive function. For absolute integrability, however, the relevant integrand is \(|\sin x|\): every positive and negative oscillation contributes nonnegative area.

We will classify the improper integral of \(|\sin x|/x^p\) over \([1,\infty)\). Starting at \(1\) isolates the behavior at infinity; any separate finite-endpoint issue would need its own check. The key is to use two complementary estimates: an upper bound for convergence, and lower bounds on repeated intervals for divergence.

Definition: The function \(f\) is absolutely integrable on \([1,\infty)\) if the improper integral \(\int_1^\infty |f(x)|\,dx\) is finite, meaning that \(\lim_{R\to\infty}\int_1^R |f(x)|\,dx\) exists as a finite number.

The Classification

For \(x\ge 1\), \(x^p\) is positive for every real \(p\), so the integrand \(|\sin x|/x^p\) is well-defined and nonnegative. Its absolute integrability has a sharp threshold at \(p=1\).

Theorem: For real \(p\), the improper integral $$ \int_1^\infty \frac{|\sin x|}{x^p}\,dx $$ converges if and only if \(p>1\). It diverges for every \(p\le 1\), including the critical case \(p=1\).

The proof of convergence is direct: \(|\sin x|\le 1\), so the integrand is bounded above by \(x^{-p}\). Divergence needs more care. The factor \(|\sin x|\) is not bounded below by a positive number everywhere, because it vanishes repeatedly. Instead, we use intervals of fixed length on which it is at least \(1/2\), and show that the contributions from those intervals have a divergent sum.

Convergence from an Upper Bound

For every real \(p\) and \(x\ge 1\), the elementary bound \(|\sin x|\le 1\) gives

$$ 0\le \frac{|\sin x|}{x^p}\le \frac{1}{x^p}. $$

If \(p>1\), the comparison integral is finite:

$$ \int_1^\infty \frac{1}{x^p}\,dx = \lim_{R\to\infty}\left[\frac{x^{1-p}}{1-p}\right]_1^R = \frac{1}{p-1}. $$

Here \(R^{1-p}\to 0\) because \(1-p<0\). The direct comparison test for nonnegative functions therefore proves convergence. In general, this test says that if \(0\le f(x)\le g(x)\) on the tail and \(\int g\) converges, then \(\int f\) converges. The inequality must hold throughout the tail, not just at selected points.

Repeated Intervals for a Lower Bound

For each integer \(k\ge 1\), consider the interval

$$ I_k=\left[2\pi k+\frac{\pi}{6},\,2\pi k+\frac{5\pi}{6}\right]. $$

These intervals are disjoint. On each one, the angle \(x-2\pi k\) lies between \(\pi/6\) and \(5\pi/6\), where the sine is at least \(1/2\). Since shifting by \(2\pi k\) does not change sine, \(|\sin x|\ge 1/2\) throughout \(I_k\). Each interval has length \(2\pi/3\).

This gives a definite amount of oscillatory mass on every interval. To compare the denominator with the interval index, note that for \(x\in I_k\) and \(k\ge 1\),

$$ \frac{13\pi}{6}k \le 2\pi k+\frac{\pi}{6} \le x \le 2\pi k+\frac{5\pi}{6} \le \frac{17\pi}{6}k. $$

The last inequality follows from \(5\le 5k\). Thus \(x/k\) lies between the two positive constants \(13\pi/6\) and \(17\pi/6\). For any fixed real \(p\), define

$$ c_p=\min\left\{\left(\frac{13\pi}{6}\right)^{-p}, \left(\frac{17\pi}{6}\right)^{-p}\right\}>0. $$

The function \(t\mapsto t^{-p}\) is monotone or constant on the positive real numbers, so its minimum on this range occurs at an endpoint. Consequently, \((x/k)^{-p}\ge c_p\), or \(x^{-p}\ge c_p k^{-p}\), throughout \(I_k\). Combining this with the sine bound gives

$$ \frac{|\sin x|}{x^p} \ge \frac{c_p}{2}k^{-p} \qquad (x\in I_k). $$

Why the Lower Bounds Force Divergence

Integrating the preceding inequality over each \(I_k\), whose length is \(2\pi/3\), yields

$$ \int_{I_k}\frac{|\sin x|}{x^p}\,dx \ge \frac{\pi c_p}{3}k^{-p}. $$

When \(p\le 1\), the series \(\sum_{k=1}^{\infty} k^{-p}\) diverges. Indeed, if \(p\le 0\), then \(k^{-p}\ge 1\) for every \(k\). If \(0<p\le 1\), then \(k^{-p}\ge 1/k\), and the harmonic series diverges. One way to see the latter is to group its terms from \(2^j\) through \(2^{j+1}-1\): each group contains \(2^j\) terms, each at least \(1/2^{j+1}\), so each group has sum at least \(1/2\).

Because the \(I_k\) are disjoint and the integrand is nonnegative, the integral over \([1,\infty)\) is at least the sum of its integrals over any finite collection of these intervals. Those finite sums are bounded below by \((\pi c_p/3)\sum k^{-p}\), which grows without bound when \(p\le 1\). Hence the improper integral diverges.

1
For convergence, bound from above.
Use \(|\sin x|\le 1\) and compare with the integrable power \(x^{-p}\) when \(p>1\).
2
For divergence, select intervals.
On each \(I_k\), the oscillatory factor is at least \(1/2\), despite having zeros elsewhere.
3
Compare the interval contributions.
Each is bounded below by a positive constant times \(k^{-p}\), whose sum diverges when \(p\le 1\).

Worked Examples

Worked Example: \(p=2\) gives a convergent integral

Consider \(\int_1^\infty |\sin x|/x^2\,dx\). For \(x\ge 1\), \(0\le|\sin x|/x^2\le1/x^2\). For a finite upper limit \(R>1\),

$$ 0\le \int_1^R\frac{|\sin x|}{x^2}\,dx \le \int_1^R\frac{1}{x^2}\,dx =\left[-\frac{1}{x}\right]_1^R =1-\frac{1}{R} \le 1. $$

As \(R\to\infty\), the comparison integral tends to \(1\). The nonnegative integrals on the left are increasing and bounded above by \(1\), so their limit is finite. Thus the absolute integral converges. The comparison checks directly: the integrand is nonnegative and never exceeds \(x^{-2}\), and \(\int_1^\infty x^{-2}\,dx=1\).

Worked Example: \(p=1\) is the divergent critical case

Let \(I_k=[2\pi k+\pi/6,\,2\pi k+5\pi/6]\), for \(k\ge1\). On \(I_k\), \(|\sin x|\ge1/2\), and \(x\le 2\pi k+5\pi/6\). Therefore,

$$ \int_{I_k}\frac{|\sin x|}{x}\,dx \ge \frac{1}{2(2\pi k+5\pi/6)}\cdot\frac{2\pi}{3} = \frac{2}{12k+5} \ge \frac{2}{17k}. $$

For the last inequality, \(12k+5\le17k\) when \(k\ge1\). Summing over the first \(N\) disjoint intervals gives

$$ \int_1^\infty\frac{|\sin x|}{x}\,dx \ge \sum_{k=1}^{N}\frac{2}{17k} =\frac{2}{17}\sum_{k=1}^{N}\frac{1}{k}. $$

The right side grows without bound as \(N\to\infty\), so the integral diverges. This verifies the critical case using an explicit lower bound on each interval, rather than treating the zeros of sine as if they controlled the total area.

Worked Example: \(p=1/2\) gives divergence

For \(k\ge1\), every \(x\in I_k\) satisfies \(x\le(17\pi/6)k\). Since \(x\mapsto x^{-1/2}\) decreases for positive \(x\), on \(I_k\) we have

$$ \frac{|\sin x|}{\sqrt{x}} \ge \frac{1}{2\sqrt{(17\pi/6)k}}. $$

Integrating this constant lower bound over an interval of length \(2\pi/3\) gives

$$ \int_{I_k}\frac{|\sin x|}{\sqrt{x}}\,dx \ge \frac{2\pi/3}{2\sqrt{17\pi/6}}\frac{1}{\sqrt{k}} = \frac{\pi}{3\sqrt{17\pi/6}}\frac{1}{\sqrt{k}}. $$

The coefficient is positive and independent of \(k\). Since \(\sum_{k=1}^{N}1/\sqrt{k}\ge\sum_{k=1}^{N}1/k\), these lower bounds have unbounded partial sums. The improper integral therefore diverges. The final comparison is valid because the integrand is nonnegative and the selected intervals are disjoint.

Common Pitfall: Using Zeros to Infer Convergence

A tempting but incorrect argument is: “\(\sin x\) has infinitely many zeros, so \(|\sin x|/x^p\) is often small and the integral should converge.” Zeros are only isolated points; they do not erase the area on the surrounding intervals. For example, at \(p=1\), the incorrect calculation might replace the oscillatory factor by its zeros and claim

$$ |\sin(2\pi k)|=0 \quad\Longrightarrow\quad \int_1^\infty\frac{|\sin x|}{x}\,dx=0. $$

This does not follow: evaluating a function at isolated points gives no value for its integral over intervals. On the intervals \(I_k\), the correct bound is \(|\sin x|\ge1/2\), and the integral over each is at least \(2/(17k)\). Their sum diverges. A valid comparison must hold across intervals of positive length, not merely at a list of points.

Why the Threshold Matters

The numerator oscillates, but its absolute value has a uniformly bounded upper size and repeatedly reaches a fixed positive size over intervals of fixed length. The upper estimate determines that decay faster than \(1/x\) is enough for convergence. The interval lower estimates show that decay like \(1/x\), or slower, leaves a divergent accumulation of area. This is why the critical exponent is \(p=1\), not an exponent inferred from the locations of the zeros.

The conclusion concerns absolute integrability: it classifies the integral with \(|\sin x|\) in the numerator. For this question, the sign changes of \(\sin x\) cannot cancel area, so the interval-based lower bounds are essential.

Check Your Understanding

Use the upper-bound and interval lower-bound ideas to answer these questions.

  1. For which real values of \(p\) does \(\int_1^\infty |\sin x|/x^p\,dx\) converge?
  2. Why is \(|\sin x|\le1\) sufficient to prove convergence when \(p>1\), but not to prove divergence when \(p\le1\)?
  3. What is the length of each interval \(I_k\), and what lower bound for \(|\sin x|\) holds throughout it?
  4. Explain why the lower bounds on \(I_k\) imply divergence at \(p=1\).
  5. Why can the zeros of \(\sin x\) not be used by themselves to establish convergence of the absolute integral?

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