Why the Integral of Cosine Does Not Converge

Introduction

An antiderivative can evaluate an integral over every finite interval without difficulty. But an integral over an interval extending to infinity asks for something more: as the finite endpoint moves farther and farther out, do the accumulated values approach one finite number?

For cosine, the finite-cutoff calculation is especially simple. Its antiderivative is sine, so the answer at a cutoff \(R\) is \(\sin R\). The crucial point is that this answer depends on where the cutoff falls in the cosine cycle. We will use two sequences of cutoffs to show that the values cannot approach one finite limit.

Definition by a Cutoff Limit

The function \(\cos x\) is continuous on every finite interval \([0,R]\), so its ordinary definite integral on that interval is defined. The right-infinite integral is defined by taking a limit of these finite integrals.

Definition: The improper integral \(\int_0^\infty \cos x\,dx\) converges to a finite value if the limit \(\lim_{R\to\infty}\int_0^R\cos x\,dx\) exists as a real number. If that limit does not exist as a finite real number, the improper integral diverges.

Here \(R\) is a finite cutoff, and \(R\to\infty\) means that the cutoff can be taken arbitrarily large. The notation \(\int_0^\infty \cos x\,dx\) does not mean that infinity is substituted into an antiderivative. Instead, we first calculate the integral for a finite \(R\), then examine the limit of those answers.

The Finite-Cutoff Formula

Since \(\frac{d}{dx}\sin x=\cos x\), the Fundamental Theorem of Calculus applies on every finite interval \([0,R]\). It gives

\int_0^R \cos x\,dx = \sin R-\sin 0 = \sin R.

Thus the question of convergence is exactly the question of whether \(\sin R\) approaches a single finite value as \(R\) increases without bound. The sine function repeatedly takes the values \(1\) and \(-1\) at cutoffs that grow arbitrarily large. We will make that observation precise using sequences.

Worked Example: Cutoffs at Integer Multiples of \(\pi\)

Consider the sequence of cutoffs \(R_n=n\pi\), where \(n\in\mathbb{N}\). Since \(n\geq 1\), every cutoff is positive, and \(R_n\to\infty\) as \(n\to\infty\).

For each \(n\), apply the finite-cutoff formula and evaluate the sine:

\int_0^{R_n}\cos x\,dx = \int_0^{n\pi}\cos x\,dx = \sin(n\pi)-\sin 0 = 0-0 = 0.

The calculation is checked by differentiating the antiderivative: \(\frac{d}{dx}\sin x=\cos x\). The endpoint substitution is also exact because \(\sin(n\pi)=0\) for every integer \(n\), and \(\sin 0=0\).

Along these particular cutoffs, the integral values are all zero. This fact alone does not establish convergence of the improper integral: convergence requires the same limiting value for cutoffs generally, not only for one selected sequence.

Worked Example: Cutoffs Where the Integral Equals One

Now choose \(T_n=2\pi n+\frac{\pi}{2}\), for \(n\in\mathbb{N}\). These cutoffs are positive and tend to infinity because \(2\pi n\to\infty\).

Substituting each cutoff into the finite-cutoff formula gives

\begin{aligned} \int_0^{T_n}\cos x\,dx &=\sin\left(2\pi n+\frac{\pi}{2}\right)\\ &=\sin\left(\frac{\pi}{2}+2\pi n\right)\\ &=1. \end{aligned}

The last equality uses the \(2\pi\)-periodicity of sine and \(\sin(\pi/2)=1\). The antiderivative check remains \(\frac{d}{dx}\sin x=\cos x\), and the lower endpoint contributes \(\sin 0=0\), as in the finite-cutoff formula. Therefore the integral values along this sequence are exactly \(1,1,1,\ldots\).

Worked Example: Cutoffs Where the Integral Equals Negative One

Take a second sequence, \(U_n=2\pi n+\frac{3\pi}{2}\), for \(n\in\mathbb{N}\). Again, \(U_n>0\) and \(U_n\to\infty\).

The finite-cutoff formula gives

\begin{aligned} \int_0^{U_n}\cos x\,dx &=\sin\left(2\pi n+\frac{3\pi}{2}\right)\\ &=\sin\left(\frac{3\pi}{2}+2\pi n\right)\\ &=-1. \end{aligned}

Here periodicity reduces the sine value to \(\sin(3\pi/2)=-1\). As a check, differentiating the antiderivative still gives cosine, and evaluating at the lower endpoint gives \(\sin 0=0\). Thus every integral value in this sequence is exactly \(-1\).

Why the Improper Integral Diverges

If a function of the cutoff \(R\) has a finite limit as \(R\to\infty\), then its values along every sequence of cutoffs tending to infinity must approach that same limit. This is a consequence of the definition of a limit: once \(R\) is sufficiently large, the function value must be close to the limit for every such \(R\), including the terms of any sequence that eventually lie beyond that threshold.

Theorem: The improper integral \(\int_0^\infty\cos x\,dx\) does not converge to a finite real number.

We can prove the claim directly from the cutoff formula and the two sequences above.

Assume a finite limit exists.
Suppose that \(\lim_{R\to\infty}\int_0^R\cos x\,dx=L\) for some real number \(L\). By the cutoff formula, this means \(\sin R\to L\).

Use the first sequence of cutoffs.
The cutoffs \(T_n=2\pi n+\frac{\pi}{2}\) tend to infinity, and \(\sin T_n=1\) for every \(n\). If \(\sin R\to L\), its values along these cutoffs must tend to \(L\); since they are constantly \(1\), this forces \(L=1\).

Use the second sequence of cutoffs.
The cutoffs \(U_n=2\pi n+\frac{3\pi}{2}\) also tend to infinity, and \(\sin U_n=-1\) for every \(n\). The same proposed limit must therefore satisfy \(L=-1\).

Reach a contradiction.
A real number cannot equal both \(1\) and \(-1\). The assumption of a finite limit is impossible, so the improper integral diverges.

This proof does not depend on estimating the integral approximately. The cutoff values are computed exactly, and two sequences tending to infinity give incompatible values. That is enough to rule out a single finite limit.

A Common Mistake: Checking Only One Sequence

The cutoffs \(R_n=n\pi\) give integral values equal to zero. A tempting but incorrect conclusion is that \(\int_0^\infty\cos x\,dx=0\), because the values along those cutoffs approach zero. This uses only one sequence and ignores other cutoffs.

The correction is to test whether the proposed limit works for all cutoffs tending to infinity. For \(T_n=2\pi n+\frac{\pi}{2}\), the integral values are all \(1\); for \(U_n=2\pi n+\frac{3\pi}{2}\), they are all \(-1\). The zero values along \(n\pi\) therefore cannot represent a limit of the full family of cutoff integrals.

Another invalid shortcut is to write \(\int_0^\infty\cos x\,dx=\sin(\infty)-\sin(0)\). The expression \(\sin(\infty)\) is not a defined real number. The correct procedure is to keep a finite cutoff, obtain \(\sin R-\sin 0\), and then ask whether that expression has a finite limit as \(R\to\infty\).

Why the Cutoff Definition Matters

For an improper integral, a calculation at one endpoint, or along a specially chosen set of endpoints, is not by itself a convergence test. The definition asks for a limit as the real cutoff \(R\) grows without bound. If that limit exists, every sequence of cutoffs tending to infinity must produce integral values approaching the same number.

Cosine makes the failure visible in a particularly direct way. The antiderivative converts each cutoff into a sine value, and periodicity supplies arbitrarily large cutoffs with different exact values. The conclusion is not that the finite integrals cannot be evaluated; each one is \(\sin R\). The conclusion is that those evaluated values have no single finite limit as the cutoff increases.

Check Your Understanding

Use the cutoff definition and the finite-interval formula to answer these questions.

State precisely what it means for \(\int_0^\infty\cos x\,dx\) to converge to a finite value.
Evaluate \(\int_0^R\cos x\,dx\) for a finite \(R>0\), showing the antiderivative and both endpoint values.
Why do the cutoffs \(R_n=n\pi\) not prove that the improper integral converges to zero?
Give a sequence of cutoffs tending to infinity along which the integral values are constantly \(-1\).
Explain why two sequences of cutoffs with different constant integral values rule out a finite limit.