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Infinite-Interval Integrals · Tutorial 297 of 998

Positive Periodic Tails Such as Sine Squared

Learn how to prove that sine squared has an infinite integral on an infinite half-line even though its values remain bounded.

Beginner 9 min read

What You'll Learn

  • Define convergence for an improper integral over an infinite half-line
  • Use a trigonometric identity to find an antiderivative of sine squared
  • Prove that the integral of sine squared grows without bound
  • Evaluate shifted and rescaled sine-squared tails
  • Use positive area over each period to test other nonnegative periodic functions
  • Avoid inferring integral convergence from boundedness of the integrand

Introduction

A function can remain bounded forever and still contribute an unlimited total area. The key is that bounded height does not imply bounded accumulated area: if a function keeps returning to positive values, those contributions can add up over infinitely many cycles.

The function \(\sin^2 x\) makes this distinction especially clear. It never exceeds \(1\), and it repeatedly reaches \(0\). Nevertheless, its integral over an infinite half-line diverges. We will see this both from an antiderivative and by adding the same positive area over successive periods.

What an Infinite-Half-Line Integral Means

An integral with an infinite upper endpoint is defined by a limit of ordinary integrals over finite intervals. For a continuous function \(f\) on \([a,\infty)\), the improper integral \(\int_a^\infty f(x)\,dx\) converges to a finite value \(L\) if the limit of \(\int_a^T f(x)\,dx\) as \(T\) tends to infinity exists and equals \(L\). If the finite-interval integrals grow without bound, the improper integral diverges to infinity.

$$ \int_a^\infty f(x)\,dx = \lim_{T\to\infty}\int_a^T f(x)\,dx, $$

The limit is part of the definition; the symbol \(\infty\) is not an ordinary endpoint at which an antiderivative can simply be evaluated. In this tutorial the integrands are nonnegative, so their accumulated integrals cannot decrease as \(T\) increases. To prove divergence, it is enough to show that these accumulated areas become arbitrarily large.

Finding an Antiderivative of Sine Squared

The identity \(\cos(2x)=1-2\sin^2 x\) gives a useful expression for the square:

$$ \sin^2 x=\frac{1-\cos(2x)}{2}. $$

Integrating the constant term and the cosine term separately gives an antiderivative. The factor \(2\) inside the cosine must be accounted for: an antiderivative of \(\cos(2x)\) is \(\frac12\sin(2x)\).

$$ \int \sin^2 x\,dx = \int \left(\frac12-\frac12\cos(2x)\right)\,dx = \frac{x}{2}-\frac{\sin(2x)}{4}+C. $$

This formula separates the accumulated integral into a term that grows linearly with \(x\) and an oscillating term that stays bounded. That contrast is the reason the integral diverges.

Key formula: For any finite \(T\geq a\), $$ \int_a^T \sin^2 x\,dx = \frac{T-a}{2}-\frac{\sin(2T)-\sin(2a)}{4}. $$ The sine contribution is bounded, while the term \((T-a)/2\) grows without bound as \(T\) increases.

Worked Example: The Integral of Sine Squared from Zero

Worked Example: The Integral of Sine Squared from Zero

Determine whether \(\int_0^\infty \sin^2 x\,dx\) converges.

First use the identity \(\sin^2 x=(1-\cos(2x))/2\). On a finite interval from \(0\) to \(T\), this gives

$$ \int_0^T\sin^2 x\,dx = \int_0^T\left(\frac12-\frac12\cos(2x)\right)\,dx = \left[\frac{x}{2}-\frac{\sin(2x)}{4}\right]_0^T. $$

Substitute the upper and lower endpoints:

$$ \int_0^T\sin^2 x\,dx = \left(\frac{T}{2}-\frac{\sin(2T)}{4}\right) - \left(0-\frac{\sin(0)}{4}\right) = \frac{T}{2}-\frac{\sin(2T)}{4}. $$

Since \(-1\leq\sin(2T)\leq 1\), the oscillating term \(-\sin(2T)/4\) stays between \(-1/4\) and \(1/4\). Therefore,

$$ \frac{T}{2}-\frac14 \leq \int_0^T\sin^2 x\,dx \leq \frac{T}{2}+\frac14. $$

The lower bound tends to infinity as \(T\) tends to infinity, so the finite-interval integrals grow without bound. Thus \(\int_0^\infty\sin^2 x\,dx\) diverges to infinity.

As an exact check along whole periods, take \(T=n\pi\), where \(n\) is a positive integer. Then \(\sin(2n\pi)=0\), and

$$ \int_0^{n\pi}\sin^2 x\,dx = \frac{n\pi}{2}. $$

These values increase without bound as \(n\) increases. Differentiating the antiderivative also checks the calculation:

$$ \frac{d}{dx}\left(\frac{x}{2}-\frac{\sin(2x)}{4}\right) = \frac12-\frac{2\cos(2x)}{4} = \frac{1-\cos(2x)}{2} = \sin^2 x. $$

Why the Oscillation Cannot Cancel the Growth

The antiderivative limit gives a direct proof for any finite lower endpoint \(a\). For \(T\geq a\), the formula above and the bound \(|\sin u|\leq 1\) imply

$$ \int_a^T\sin^2 x\,dx = \frac{T-a}{2}-\frac{\sin(2T)-\sin(2a)}{4} \geq \frac{T-a}{2}-\frac12. $$

The last term is a fixed constant, while \((T-a)/2\) tends to infinity. Thus the integral from \(a\) to \(T\) tends to infinity for every real \(a\). The oscillating part can move the accumulated area up or down by only a bounded amount; it cannot offset a term that grows without limit.

Theorem: For every real number \(a\), the improper integral \(\int_a^\infty \sin^2 x\,dx\) diverges to infinity. In particular, the fact that \(0\leq\sin^2 x\leq 1\) does not make its integral over an infinite interval finite.

Worked Example: A Shifted Starting Point

Worked Example: A Shifted Starting Point

Determine whether \(\int_{\pi/4}^\infty\sin^2 x\,dx\) converges. Apply the finite-interval formula with \(a=\pi/4\) and upper endpoint \(T\):

$$ \int_{\pi/4}^T\sin^2 x\,dx = \frac{T-\pi/4}{2} - \frac{\sin(2T)-\sin(\pi/2)}{4}. $$

Since \(\sin(\pi/2)=1\), simplify:

$$ \int_{\pi/4}^T\sin^2 x\,dx = \frac{T}{2}-\frac{\pi}{8}-\frac{\sin(2T)}{4}+\frac14. $$

The sine term stays between \(-1/4\) and \(1/4\). Hence this expression is at least \(T/2-\pi/8\), which tends to infinity. The improper integral therefore diverges to infinity.

For an exact sequence of endpoints, let \(T=n\pi\). Then \(\sin(2T)=0\), so

$$ \int_{\pi/4}^{n\pi}\sin^2 x\,dx = \frac{n\pi}{2}-\frac{\pi}{8}+\frac14. $$

This grows without bound as \(n\) increases. The endpoint substitution can be checked by setting \(T=\pi/4\), which makes the expression equal to \(0\), as an integral over an interval of zero length must. The antiderivative check is the same derivative calculation as in the preceding example.

Area Repeated Over Each Period

There is also a useful period-by-period explanation. The function \(\sin^2 x\) has period \(\pi\): adding \(\pi\) to \(x\) changes the sign of \(\sin x\), but squaring removes that sign. Each period contributes the same area.

$$ \int_0^\pi \sin^2 x\,dx = \left[\frac{x}{2}-\frac{\sin(2x)}{4}\right]_0^\pi = \frac{\pi}{2}. $$

After \(n\) complete periods, the accumulated area is \(n\pi/2\). Every period adds a positive amount, so the total cannot settle at a finite value. This reasoning extends to other continuous, nonnegative periodic functions.

Theorem: Let \(f\) be continuous, nonnegative, and periodic with period \(P>0\) on \([a,\infty)\). If the area over one period, $$ A=\int_a^{a+P}f(x)\,dx, $$ is positive, then \(\int_a^\infty f(x)\,dx\) diverges to infinity.

To prove this, periodicity gives the same area on every interval of length \(P\). More explicitly, for each positive integer \(n\), split the integral from \(a\) to \(a+nP\) into \(n\) intervals of length \(P\):

$$ \int_a^{a+nP}f(x)\,dx = \sum_{k=0}^{n-1}\int_{a+kP}^{a+(k+1)P}f(x)\,dx = \sum_{k=0}^{n-1}A = nA. $$

Because \(A>0\), the values \(nA\) tend to infinity. These are finite-interval integrals with upper endpoints tending to infinity, so the improper integral cannot converge to a finite value. For sine squared, \(P=\pi\) and \(A=\pi/2>0\), recovering the divergence just established.

Worked Example: A Change in Frequency

Worked Example: A Change in Frequency

Determine whether \(\int_0^\infty \sin^2(2x)\,dx\) converges. The identity for the squared sine gives \(\sin^2(2x)=(1-\cos(4x))/2\). Thus, for finite \(T\),

$$ \int_0^T\sin^2(2x)\,dx = \int_0^T\left(\frac12-\frac12\cos(4x)\right)\,dx = \left[\frac{x}{2}-\frac{\sin(4x)}{8}\right]_0^T = \frac{T}{2}-\frac{\sin(4T)}{8}. $$

The sine term has absolute value at most \(1/8\), so the integral is at least \(T/2-1/8\). This lower bound tends to infinity, and the improper integral diverges to infinity.

For a period-based check, \(\sin^2(2x)\) has period \(\pi/2\). At \(T=n\pi/2\), the sine term vanishes, giving

$$ \int_0^{n\pi/2}\sin^2(2x)\,dx = \frac{n\pi}{4}. $$

These values grow without bound. Finally, differentiating the antiderivative verifies it:

$$ \frac{d}{dx}\left(\frac{x}{2}-\frac{\sin(4x)}{8}\right) = \frac12-\frac{4\cos(4x)}{8} = \frac{1-\cos(4x)}{2} = \sin^2(2x). $$

A Common Mistake

A frequent error is to reason that a bounded function must have a finite integral over an infinite interval. Boundedness only limits the height of the graph; it does not limit the total area when positive contributions continue indefinitely. In this case, \(0\leq\sin^2 x\leq1\), but that upper bound does not establish convergence.

Another mistake is to compute the antiderivative and treat the bounded oscillating term as if it could cancel the growing term. For example, the incorrect conclusion might be: “\(\frac{T}{2}-\frac{\sin(2T)}{4}\) oscillates, so it has a finite limit.” It does not oscillate within a fixed range: its first term increases without bound, while the second term stays between \(-1/4\) and \(1/4\). The corrected estimate makes the distinction explicit:

$$ \frac{T}{2}-\frac14 \leq \frac{T}{2}-\frac{\sin(2T)}{4} \leq \frac{T}{2}+\frac14. $$

Both bounds grow without bound as \(T\) increases, so the accumulated integral does too. A bounded oscillation cannot cancel linear growth.

Check Your Understanding

Use an antiderivative, a period argument, or both to answer each question.

  1. Find an antiderivative of \(\sin^2 x\), and verify it by differentiation.
  2. Evaluate \(\int_0^{n\pi}\sin^2 x\,dx\) for a positive integer \(n\). What does this imply about the integral over the infinite half-line?
  3. Why does the bounded term involving \(\sin(2T)\) not make \(\int_0^T\sin^2 x\,dx\) converge to a finite limit?
  4. What two conditions in the periodic-function theorem ensure that each repeated period contributes a fixed positive area?
  5. Does the fact that \(0\leq\sin^2 x\leq1\) prove convergence of its integral over an infinite interval? Explain.

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