Calculate the improper integral (or establish its

Calculate the improper integral (or establish its divergence): $$ \int_2^\infty \cos x \, dx $$

This is an improper integral at infinity. Unlike the previous examples, the integrand does not decay to zero as $x \to \infty$, which suggests potential divergence.

Step 1: Express as a limit

$$\int_2^\infty \cos x \, dx = \lim_{R \to \infty} \int_2^R \cos x \, dx$$

Step 2: Find the antiderivative

The antiderivative of $\cos x$ is:

$$\int \cos x \, dx = \sin x + C$$

Step 3: Evaluate the definite integral

$$\int_2^R \cos x \, dx = [\sin x]_2^R = \sin R - \sin 2$$

Step 4: Analyze the limit

$$\int_2^\infty \cos x \, dx = \lim_{R \to \infty} (\sin R - \sin 2)$$

As $R \to \infty$, we need to evaluate $\lim_{R \to \infty} \sin R$.

Key observation: The function $\sin R$ oscillates between $-1$ and $1$ for all $R$. It does not approach any single value as $R \to \infty$.

Therefore:

$$\lim_{R \to \infty} \sin R \quad \text{does not exist}$$

Since the limit does not exist, the integral diverges.

Final Answer

$$\boxed{\int_2^\infty \cos x \, dx = \text{divergent}}$$

The integral diverges (by oscillation).

Detailed explanation of divergence

The integral diverges because:

• The antiderivative is $\sin R - \sin 2$
• As $R$ increases, $\sin R$ oscillates between $-1$ and $1$
• The values $\sin R - \sin 2$ oscillate between approximately $-1 - 0.909 = -1.909$ and $1 - 0.909 = 0.091$
• Since the limit doesn't settle to a single value, the integral diverges

This is a different type of divergence than we've seen before - it's divergence by oscillation rather than divergence to $\pm\infty$.

Contrast with convergent integrals

Compare with integrals that converge:

Example 1: $\int_1^\infty \frac{\cos x}{x^2} \, dx$ converges

This integral converges because the factor $\frac{1}{x^2}$ causes the integrand to decay to zero fast enough, even though $\cos x$ oscillates.

Example 2: $\int_0^\infty e^{-x}\cos x \, dx$ converges

This converges because $e^{-x}$ decays exponentially, overwhelming the oscillation of $\cos x$.

Our integral: $\int_2^\infty \cos x \, dx$ diverges

This diverges because $\cos x$ doesn't decay at all - it just oscillates.

Graphical interpretation

Consider the graph of $y = \cos x$:

• The function oscillates between $-1$ and $1$ forever
• Each "positive hump" adds area above the $x$-axis
• Each "negative hump" subtracts area below the $x$-axis
• The accumulated signed area doesn't converge to a fixed value
• Instead, it oscillates as we extend the upper limit

The partial integral $\int_2^R \cos x \, dx = \sin R - \sin 2$ oscillates as $R$ increases, never settling down.

Behavior of partial integrals

Let's look at some specific values:

For $R = 2$: $\sin 2 - \sin 2 = 0$

For $R = \frac{\pi}{2}$: $\sin\frac{\pi}{2} - \sin 2 = 1 - 0.909 \approx 0.091$

For $R = \pi$: $\sin \pi - \sin 2 = 0 - 0.909 \approx -0.909$

For $R = \frac{3\pi}{2}$: $\sin\frac{3\pi}{2} - \sin 2 = -1 - 0.909 \approx -1.909$

For $R = 2\pi$: $\sin 2\pi - \sin 2 = 0 - 0.909 \approx -0.909$

The values oscillate between approximately $-1.909$ and $0.091$, never converging.

Comparison test failure

The comparison test, which works for positive functions, doesn't directly apply here because $\cos x$ changes sign.

However, we can note that:

$$|\cos x| \leq 1$$

If we consider $\int_2^\infty 1 \, dx$, this clearly diverges to $+\infty$.

But this doesn't tell us about $\int_2^\infty \cos x \, dx$ because of cancellation from sign changes.

Necessary condition for convergence

For $\int_a^\infty f(x) \, dx$ to converge, a necessary condition is:

$$\lim_{x \to \infty} f(x) = 0$$

In our case:

$$\lim_{x \to \infty} \cos x \quad \text{does not exist}$$

Since $\cos x$ oscillates between $-1$ and $1$, it doesn't approach 0 (or any value).

Therefore, the integral cannot converge.

Important note: This is only a necessary condition, not sufficient! Even if $f(x) \to 0$, the integral may still diverge (e.g., $\int_1^\infty \frac{1}{x} \, dx$).

Related integrals

Integral 1: $\int_0^\infty \sin x \, dx$

$$= \lim_{R \to \infty} [-\cos x]_0^R = \lim_{R \to \infty} (-\cos R + 1)$$

Since $\lim_{R \to \infty} \cos R$ doesn't exist, this also diverges.

Integral 2: $\int_1^\infty \frac{\sin x}{x} \, dx$

This integral actually converges! (It converges to approximately 0.504)

The factor $\frac{1}{x}$ causes enough decay to ensure convergence despite the oscillation.

Integral 3: $\int_1^\infty \frac{\cos x}{x} \, dx$

This also converges (conditionally).

Absolute vs. conditional convergence

An integral is said to be:

• Absolutely convergent if $\int_a^\infty |f(x)| \, dx$ converges
• Conditionally convergent if $\int_a^\infty f(x) \, dx$ converges but $\int_a^\infty |f(x)| \, dx$ diverges

For $\int_2^\infty \cos x \, dx$:

• The integral itself diverges
• The absolute value integral $\int_2^\infty |\cos x| \, dx$ also diverges (to $+\infty$)

So this integral is neither absolutely nor conditionally convergent - it simply diverges.

Summary

1. The integral $\int_2^\infty \cos x \, dx$ is improper at infinity
2. The antiderivative is $\sin x$
3. Evaluating gives: $\lim_{R \to \infty} (\sin R - \sin 2)$
4. Since $\sin R$ oscillates and doesn't approach a limit, the integral diverges
5. This is divergence by oscillation, not divergence to infinity
6. Necessary condition violated: $\cos x$ doesn't approach 0 as $x \to \infty$

Key lessons:

• Not all divergent integrals go to $\pm\infty$ - some diverge by oscillation
• For convergence, the integrand must approach 0 at infinity (necessary but not sufficient)
• Oscillating functions like $\sin x$ and $\cos x$ typically lead to divergent integrals at infinity
• A decay factor (like $\frac{1}{x}$ or $e^{-x}$) is needed to "tame" oscillating integrands
• The behavior of $\int_a^\infty f(x) \, dx$ depends critically on how fast $f(x) \to 0$
• Trigonometric integrals to infinity require special care