To find the integral

$\int_{0}^{\infty} \frac{\sin^2(x)}{x^2} \, dx,$

we can use a known result from definite integrals involving trigonometric functions and their squares. Let's outline the steps to solve it.

1. Simplify the integrand:

   We start by using the trigonometric identity:

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

   Substituting this into the integral gives:

   $\int_{0}^{\infty} \frac{\sin^2(x)}{x^2} \, dx = \int_{0}^{\infty} \frac{1 - \cos(2x)}{2x^2} \, dx.$

2. Separate the integral:

   $\int_{0}^{\infty} \frac{1 - \cos(2x)}{2x^2} \, dx = \frac{1}{2} \left( \int_{0}^{\infty} \frac{1}{x^2} \, dx - \int_{0}^{\infty} \frac{\cos(2x)}{x^2} \, dx \right).$

3. Evaluate each integral separately:

   • The first integral $\int_{0}^{\infty} \frac{1}{x^2} \, dx$ diverges. This suggests we should consider the convergence properties of the integrand.

   • For the second integral, use the fact that:

     $\int_{0}^{\infty} \frac{\cos(2x)}{x^2} \, dx$

     This integral converges to $\frac{\pi}{2}$.

4. Combine the results:

   $\int_{0}^{\infty} \frac{\sin^2(x)}{x^2} \, dx = \frac{1}{2} \left( \infty - \frac{\pi}{2} \right).$

Since the first part diverges and the second part is finite, this integral does not converge in the usual sense. However, in the context of improper integrals and specific regularization techniques, one might handle such cases differently.

The integral:

$\int_{0}^{\infty} \frac{\sin^2(x)}{x^2} \, dx$

diverges.

Would you like further details or have any other questions?

Here are some related questions you might find interesting:

1. How do we handle divergent integrals in regularization techniques?
2. Can you explain the concept of improper integrals?
3. What are some common methods for evaluating trigonometric integrals?
4. How can we use contour integration to solve improper integrals?
5. What is the method of residue in complex analysis?
6. How can we apply the Fourier transform to evaluate integrals?
7. What is the Dirichlet integral, and how is it solved?
8. How does the Riemann-Lebesgue lemma relate to integrals involving trigonometric functions?

Tip: When faced with an integral that diverges, consider if the context allows for regularization techniques or if the integral can be reinterpreted in a different mathematical framework.