Investigate the convergence of the integral: integral

Investigate the convergence of the integral: $$ \int_0^\infty \frac{\sin x \cdot \arctan x}{1+x^2} \, dx $$

This integral is improper at infinity. We need to analyze the behavior of the integrand and determine whether the integral converges.

Step 1: Analyze the integrand

The integrand is:

$$f(x) = \frac{\sin x \cdot \arctan x}{1+x^2}$$

Let's examine each component:

• $\sin x$: oscillates between $-1$ and $1$
• $\arctan x$: increases from $0$ to $\frac{\pi}{2}$ as $x$ goes from $0$ to $\infty$
• $\frac{1}{1+x^2}$: decreases from $1$ to $0$ as $x$ goes from $0$ to $\infty$

Step 2: Check if the integral is improper at $x = 0$

As $x \to 0^+$:

$$\sin x \approx x, \quad \arctan x \approx x, \quad 1+x^2 \approx 1$$

So:

$$f(x) \approx \frac{x \cdot x}{1} = x^2 \to 0$$

The function is continuous at $x = 0$, so there's no issue at the lower limit.

Step 3: Behavior at infinity

As $x \to \infty$:

• $|\sin x| \leq 1$
• $\arctan x \to \frac{\pi}{2}$
• $\frac{1}{1+x^2} \sim \frac{1}{x^2}$

Therefore, for large $x$:

$$\left|f(x)\right| = \left|\frac{\sin x \cdot \arctan x}{1+x^2}\right| \leq \frac{|\sin x| \cdot |\arctan x|}{1+x^2} \leq \frac{1 \cdot \frac{\pi}{2}}{x^2} = \frac{\pi}{2x^2}$$

Step 4: Apply the comparison test

We have:

$$\left|\frac{\sin x \cdot \arctan x}{1+x^2}\right| \leq \frac{\pi}{2x^2} \quad \text{for large } x$$

We know that:

$$\int_1^\infty \frac{1}{x^2} \, dx = 1 \quad \text{(converges)}$$

Therefore:

$$\int_1^\infty \frac{\pi}{2x^2} \, dx = \frac{\pi}{2} \quad \text{(converges)}$$

By the comparison test, since $\int_1^\infty \left|f(x)\right| \, dx$ converges, we have:

$$\int_1^\infty \frac{\sin x \cdot \arctan x}{1+x^2} \, dx \quad \text{converges absolutely}$$

Step 5: Complete integral

Since:

• $\int_0^1 \frac{\sin x \cdot \arctan x}{1+x^2} \, dx$ is a proper integral (finite interval, continuous integrand)
• $\int_1^\infty \frac{\sin x \cdot \arctan x}{1+x^2} \, dx$ converges absolutely (by comparison test)

We conclude:

$$\int_0^\infty \frac{\sin x \cdot \arctan x}{1+x^2} \, dx \quad \text{converges absolutely}$$

Conclusion

$$\boxed{\text{The integral } \int_0^\infty \frac{\sin x \cdot \arctan x}{1+x^2} \, dx \text{ CONVERGES (absolutely)}}$$

Detailed analysis

Absolute convergence:

An integral $\int_a^\infty f(x) \, dx$ is said to converge absolutely if $\int_a^\infty |f(x)| \, dx$ converges.

If an integral converges absolutely, it also converges (but not vice versa).

In our case:

$$\int_0^\infty \left|\frac{\sin x \cdot \arctan x}{1+x^2}\right| \, dx \leq \int_0^\infty \frac{\frac{\pi}{2}}{1+x^2} \, dx = \frac{\pi}{2} \cdot \frac{\pi}{2} = \frac{\pi^2}{4}$$

Actually, let's be more precise. For all $x \geq 0$:

$$|\sin x| \leq 1, \quad |\arctan x| \leq \frac{\pi}{2}$$

So:

$$\left|\frac{\sin x \cdot \arctan x}{1+x^2}\right| \leq \frac{\frac{\pi}{2}}{1+x^2}$$

And we know:

$$\int_0^\infty \frac{1}{1+x^2} \, dx = [\arctan x]_0^\infty = \frac{\pi}{2} - 0 = \frac{\pi}{2}$$

Therefore:

$$\int_0^\infty \left|\frac{\sin x \cdot \arctan x}{1+x^2}\right| \, dx \leq \frac{\pi}{2} \cdot \frac{\pi}{2} = \frac{\pi^2}{4}$$

This confirms absolute convergence.

Comparison with related integrals

Integral 1: $\int_0^\infty \frac{\sin x}{1+x^2} \, dx$

This also converges absolutely by comparison: $\left|\frac{\sin x}{1+x^2}\right| \leq \frac{1}{1+x^2}$

Integral 2: $\int_0^\infty \frac{\arctan x}{1+x^2} \, dx$

This converges: $\frac{\arctan x}{1+x^2} \leq \frac{\pi/2}{1+x^2}$

Integral 3: $\int_0^\infty \sin x \, dx$

This diverges by oscillation (as we saw earlier).

Key observation: The factor $\frac{1}{1+x^2}$ provides enough decay at infinity to ensure convergence, even with the oscillating $\sin x$ and the growing $\arctan x$.

Why the decay factor is crucial

Without the decay factor $\frac{1}{1+x^2}$, consider:

$$\int_0^\infty \sin x \cdot \arctan x \, dx$$

For large $x$: $\sin x \cdot \arctan x \approx \frac{\pi}{2}\sin x$

This would be like $\int_0^\infty \sin x \, dx$, which diverges by oscillation.

The decay factor $\frac{1}{1+x^2} \sim \frac{1}{x^2}$ is "strong enough" (exponent 2 > 1) to dominate the oscillation and bounded growth, ensuring convergence.

General principle

For integrals of the form:

$$\int_0^\infty \text{(oscillating)} \cdot \text{(bounded)} \cdot \text{(decay)} \, dx$$

Convergence typically requires the decay factor to be at least $O(1/x^{1+\varepsilon})$ for some $\varepsilon > 0$.

In our case:

• Oscillating: $\sin x$ (bounded by 1)
• Bounded growth: $\arctan x$ (bounded by $\pi/2$)
• Decay: $\frac{1}{1+x^2} \sim \frac{1}{x^2}$ (strong enough)

Result: Absolute convergence ✓

Summary

1. The integrand is continuous on $[0, \infty)$
2. At infinity, $\left|f(x)\right| \leq \frac{\pi/2}{1+x^2}$
3. Since $\int_0^\infty \frac{1}{1+x^2} \, dx = \frac{\pi}{2}$ converges, the comparison test applies
4. The integral converges absolutely
5. Absolute convergence implies regular convergence

Key lessons:

• The comparison test is powerful for determining convergence
• Absolute convergence is a stronger condition than regular convergence
• Bounding oscillating/growing terms by constants simplifies analysis
• The decay rate $\frac{1}{x^2}$ is sufficient to overcome bounded oscillation
• The factor $\frac{1}{1+x^2}$ acts as a "damping" term ensuring convergence
• Even with oscillation and growth, strong enough decay ensures convergence