Conditional Convergence of Oscillatory Improper Integrals

Why Oscillation Changes the Convergence Question

The function \(\sin x\) changes sign repeatedly. When it is multiplied by a decreasing factor such as \(1/x^p\), positive and negative contributions can cancel as the interval grows. This means the integral may converge even when the integral of its absolute value does not.

For the family

\int_1^\infty \frac{\sin x}{x^p}\,dx,

the exponent \(p\) controls two distinct effects: how quickly the factor \(1/x^p\) decreases, and whether oscillatory cancellation is enough to produce a finite limit. The key is to test ordinary convergence and absolute convergence separately.

Definitions and the Classification to Be Proved

For a continuous function \(f\) on \([1,\infty)\), its improper integral converges if the finite-interval integrals have a finite limit:

\int_1^\infty f(x)\,dx = \lim_{R\to\infty}\int_1^R f(x)\,dx.

It converges absolutely if \(\int_1^\infty |f(x)|\,dx\) is finite. It converges conditionally if the original integral converges but the integral of the absolute value diverges. These definitions distinguish cancellation in the signed integral from the size of all the contributions without cancellation.

Theorem: For real \(p\), \(\int_1^\infty \sin x/x^p\,dx\) converges absolutely when \(p>1\), converges conditionally when \(0<p\leq 1\), and diverges when \(p\leq 0\).

The proof has two parts. First, determine exactly when the signed integral converges. Then determine when its absolute value has a finite integral.

When the Signed Integral Converges

Suppose \(p>0\). Integration by parts on a finite interval \([A,B]\), where \(1\leq A<B\), gives a useful formula. Take \(u=x^{-p}\) and \(dv=\sin x\,dx\), so \(du=-p x^{-p-1}\,dx\) and \(v=-\cos x\). Then

\begin{aligned} \int_A^B x^{-p}\sin x\,dx &=\left[-x^{-p}\cos x\right]_A^B -\int_A^B p x^{-p-1}\cos x\,dx\\ &=A^{-p}\cos A-B^{-p}\cos B -p\int_A^B x^{-p-1}\cos x\,dx. \end{aligned}

As \(B\to\infty\), the boundary term \(B^{-p}\cos B\) tends to zero because \(p>0\) and \(|\cos B|\leq 1\). Also, the remaining integral has an absolutely convergent bound:

\int_A^\infty \left|p x^{-p-1}\cos x\right|\,dx \leq p\int_A^\infty x^{-p-1}\,dx =A^{-p}.

The last equality follows by evaluating the power integral: an antiderivative of \(x^{-p-1}\) is \(-x^{-p}/p\), and \(B^{-p}\to0\). Thus the integration-by-parts expression has a finite limit as \(B\to\infty\). In particular, the original improper integral converges whenever \(p>0\).

The same calculation gives a useful bound on the tail. Taking \(B\to\infty\) in the formula and applying the triangle inequality yields

\left|\int_A^\infty x^{-p}\sin x\,dx\right| \leq A^{-p}+p\int_A^\infty x^{-p-1}\,dx =2A^{-p}.

This bound tends to zero as \(A\to\infty\). It makes the role of oscillation precise: integration by parts transfers the sine to a cosine and introduces the extra decreasing factor \(x^{-1}\).

Why the Signed Integral Diverges When \(p\leq 0\)

For \(p\leq0\), the weight \(x^{-p}\) is at least \(1\) for \(x\geq1\). Consider the intervals \([2\pi n,\,2\pi n+\pi/2]\), for positive integers \(n\). On each interval, \(\sin x\geq0\), and

\int_{2\pi n}^{2\pi n+\pi/2} x^{-p}\sin x\,dx \geq \int_{2\pi n}^{2\pi n+\pi/2}\sin x\,dx =1.

If the improper integral converged, its finite-interval integrals would approach a limit. The integral over any interval whose two endpoints tend to infinity would then have to tend to zero: it is the difference of two finite-interval integrals approaching the same limit. But the displayed integrals are at least \(1\) for every \(n\). This contradiction proves divergence for every \(p\leq0\), including \(p=0\).

When Convergence Is Absolute

For \(p>1\), the estimate \(|\sin x|\leq1\) gives

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

The power integral \(\int_1^\infty x^{-p}\,dx\) converges for \(p>1\). Comparison therefore proves that \(\int_1^\infty |\sin x|/x^p\,dx\) converges in this range. Absolute convergence implies ordinary convergence because the absolute value of any signed tail integral is no larger than the corresponding tail integral of the absolute value.

For the other direction, we show that the absolute-value integral diverges for \(p\leq1\). For each positive integer \(k\), use the interval

\left[k\pi+\frac{\pi}{6},\,k\pi+\frac{5\pi}{6}\right].

Throughout this interval, \(|\sin x|\geq 1/2\). Also \(x/(k\pi)\) lies between \(1\) and \(2\). For a fixed real \(p\), this implies

x^{-p}\geq C_p(k\pi)^{-p}, \qquad C_p=\min(1,2^{-p})>0.

The interval has length \(2\pi/3\), so its contribution to the absolute-value integral satisfies

\int_{k\pi+\pi/6}^{k\pi+5\pi/6} \frac{|\sin x|}{x^p}\,dx \geq \frac12 C_p(k\pi)^{-p}\frac{2\pi}{3} = \frac{C_p\pi^{1-p}}{3}k^{-p}.

These intervals are disjoint. When \(p\leq1\), the series \(\sum_{k=1}^\infty k^{-p}\) diverges, as follows from the power-integral convergence benchmark. Adding the lower bounds over disjoint intervals therefore proves that \(\int_1^\infty |\sin x|/x^p\,dx\) diverges for \(p\leq1\).

Worked Examples

Worked Example: The Case \(p=1\)

Classify \(\int_1^\infty \sin x/x\,dx\). Here \(p=1>0\), so the integration-by-parts result proves ordinary convergence. In detail, for finite \(R>1\),

\begin{aligned} \int_1^R \frac{\sin x}{x}\,dx &=\left[-\frac{\cos x}{x}\right]_1^R -\int_1^R \frac{\cos x}{x^2}\,dx\\ &=\cos 1-\frac{\cos R}{R} -\int_1^R \frac{\cos x}{x^2}\,dx. \end{aligned}

As \(R\to\infty\), \(\cos R/R\to0\). The remaining integral converges absolutely because

\int_1^\infty \frac{|\cos x|}{x^2}\,dx \leq \int_1^\infty \frac{1}{x^2}\,dx = \lim_{R\to\infty}\left(1-\frac{1}{R}\right) =1.

Thus the signed integral has a finite limit. But \(p=1\leq1\), so the absolute-value integral diverges by the interval lower bound. The integral is conditionally convergent. As a tail check, the general estimate gives a tail magnitude at most \(2/A\), which tends to zero.

Worked Example: The Case \(p=2\)

For \(p=2\), use \(0\leq|\sin x|/x^2\leq1/x^2\). The comparison integral is

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

Therefore the integral of \(|\sin x|/x^2\) is finite, and the original integral converges absolutely. To check the ordinary convergence conclusion directly, for every \(R>1\),

\left|\int_R^\infty \frac{\sin x}{x^2}\,dx\right| \leq \int_R^\infty \frac{1}{x^2}\,dx = \frac{1}{R}.

This tail bound tends to zero. The classification is absolutely convergent.

Worked Example: The Case \(p=0\)

When \(p=0\), the integrand is \(\sin x\). Its finite-interval integral is exactly

\int_1^R \sin x\,dx = [-\cos x]_1^R = \cos 1-\cos R.

Along \(R=2\pi n\), this equals \(\cos1-1\). Along \(R=(2n+1)\pi\), it equals \(\cos1+1\). These are different values, so the limit as \(R\to\infty\) does not exist. The integral diverges. This agrees with the interval test for \(p\leq0\), and is also an explicit check that the signed integral itself fails to converge.

Common Pitfall: Confusing a Small Amplitude with Absolute Convergence

A tempting but incorrect argument is: “When \(p=1\), the factor \(1/x\) tends to zero, so the integral of \(|\sin x|/x\) must converge.” A factor tending to zero does not by itself make an infinite integral finite. In fact, on each interval from the absolute-value estimate, the contribution for \(p=1\) is at least

\frac{C_1}{3}k^{-1}, \qquad C_1=\min(1,2^{-1})=\frac12.

The sum of these lower bounds is \(\frac16\sum_{k=1}^\infty 1/k\), which diverges. The signed integral nevertheless converges because its positive and negative parts cancel. Taking the absolute value removes that cancellation; it cannot be treated as though the signs were still alternating.

Complete Classification

Values of \(p\)	Signed integral	Integral of the absolute value	Classification
\(p>1\)	Converges	Converges	Absolutely convergent
\(0<p\leq1\)	Converges	Diverges	Conditionally convergent
\(p\leq0\)	Diverges	Diverges	Divergent

The two thresholds have different causes. The condition \(p>0\) is needed for the oscillatory signed integral to converge. The stronger condition \(p>1\) is needed for the absolute-value integral to converge. Thus oscillation extends ordinary convergence beyond the range of absolute convergence, but only when the amplitude decreases.

Check Your Understanding

Use the definitions and estimates in this tutorial to answer each question.

Classify \(\int_1^\infty \sin x/x^{3/4}\,dx\), including its absolute convergence.
Why does integration by parts establish ordinary convergence for \(p>0\)? State what happens to each term as the upper endpoint tends to infinity.
For \(p\leq0\), why do integrals over \([2\pi n,\,2\pi n+\pi/2]\) rule out convergence?
Which estimate proves absolute convergence for \(p>1\)?
For \(p=1\), explain why convergence of the signed integral does not imply convergence of the integral of its absolute value.