Question

Determine whether the following integrals converge or diverge. $${\int }_1^{\mathrm{\infty }}\frac{{\sin }^{2}x}{x^2}dx$$

Short answer

Answer: The integral $\int_{1}^{\infty} \frac{\sin^2 x}{x^2} dx$ converges.

Step-by-step solution

Identify the improper integral

The given integral is an improper integral because the upper limit of integration is infinity: $${\int }_1^{\mathrm{\infty }}\frac{{\sin }^{2}x}{x^2}dx$$

Apply the comparison test

Perform a comparison test by considering a known integral for comparison. For this exercise, we can use the following integral for comparison: $${\int }_1^{\mathrm{\infty }}\frac{1}{x^2}dx$$ This integral converges because it is of the form ${\int }_1^{\mathrm{\infty }}\frac{1}{x^p}dx$ where $p>1$. Now, we need to show that the following inequality holds: $$0\le \frac{{\sin }^{2}x}{x^2}\le \frac{1}{x^2}$$ It is clear that $\frac{{\sin }^{2}x}{x^2}\ge 0$ for all $x$. To show the other part of the inequality, note that $0\le {\sin }^{2}x\le 1$ for all $x$. Therefore, we have: $$\frac{{\sin }^{2}x}{x^2}\le \frac{1}{x^2}$$

Determine if the given integral converges or diverges

Since both inequalities hold: $$0\le \frac{{\sin }^{2}x}{x^2}\le \frac{1}{x^2}$$ And, we know that the integral ${\int }_1^{\mathrm{\infty }}\frac{1}{x^2}dx$ converges, it follows by the comparison test that the given integral also converges: $${\int }_1^{\mathrm{\infty }}\frac{{\sin }^{2}x}{x^2}dx$$ Thus, the given integral converges.

Key concepts

Comparison Test

The comparison test is a powerful tool for determining the convergence or divergence of an improper integral.
The basic idea is to compare the given integral with another integral whose convergence is already known.

• If the given integral is less than or equal to a known convergent integral, the given integral will also converge.
• If the given integral is greater than or equal to a known divergent integral, the given integral will also diverge.

This is particularly useful when dealing with complex functions that are difficult to integrate directly. For example, in our exercise, we used the integral ${\int }_1^{\mathrm{\infty }}\frac{1}{x^2}dx$ which is known to converge. Since $0\le \frac{{\sin }^{2}x}{x^2}\le \frac{1}{x^2}$, we can conclude that our original integral converges as well.

Convergent Integrals

Convergent integrals are integrals whose value approaches a finite number as the upper limit of integration tends to infinity.
The concept is crucial when evaluating integrals over an infinite interval.
Generally, integrals of the form ${\int }_1^{\mathrm{\infty }}\frac{1}{x^p}dx$ converge if $p>1$. This criterion helps in determining whether an improper integral is convergent or divergent. In our case, the integral ${\int }_1^{\mathrm{\infty }}\frac{1}{x^2}dx$ is known to converge since $p=2>1$.Additionally, convergent integrals often serve as benchmarks when applying comparison tests, as their behavior is already understood, simplifying the process of analyzing more complicated functions.

Trigonometric Integrals

Trigonometric integrals involve integrals of functions that include trigonometric functions like sine and cosine.
These integrals may look intimidating at first, but they follow specific rules that make them manageable.
In our exercise, we dealt with the integral ${\int }_1^{\mathrm{\infty }}\frac{{\sin }^{2}x}{x^2}dx$.
Some key aspects to consider:

• Trigonometric functions like $\sin x$ and $\cos x$ are bounded between -1 and 1, which makes them suitable for comparison tests. For instance, ${\sin }^{2}x$ lies between 0 and 1, simplifying the upper-bound comparison.
• Utilizing known properties of trigonometric functions, such as periodicity and symmetry, can help simplify the analysis and computation of integrals.

Handling trigonometric components carefully allows for effective application of other calculus strategies, like the comparison test, to study the nature of these integrals.