The problem requires examining whether the given improper integral converges and, if it does, calculating its value. The integral is:

$J = \int_{0}^{\infty} x \cdot \cos(x^2) \, dx$

Step 1: Determine if the integral converges

The convergence of the improper integral can be examined using the following observations:

1. The integrand $x \cos(x^2)$ oscillates due to the cosine term, and the amplitude of the function depends on $x$. However, its behavior as $x \to \infty$ suggests we need to analyze the integral further.

2. The substitution $u = x^2$ (with $du = 2x dx$) transforms the integral into a simpler form: $J = \int_{0}^{\infty} x \cos(x^2) \, dx \quad \Rightarrow \quad J = \frac{1}{2} \int_{0}^{\infty} \cos(u) \, du$ The term $\int_{0}^{\infty} \cos(u) \, du$ does not converge, as cosine oscillates indefinitely without settling to a limit. Hence, the integral diverges.

Step 2: Conclusion

The improper integral $J = \int_{0}^{\infty} x \cos(x^2) \, dx$ is not defined, as it diverges due to the oscillatory nature of the integrand $\cos(x^2)$.

Do you need any additional explanations or a different approach? Let me know!

5 Related Questions:

1. How do we determine the convergence of oscillatory integrals in general?
2. What happens if the upper limit of the integral is finite instead of infinite?
3. What is the result of integrating $\int_{0}^{\infty} e^{-x^2} \cos(x^2) dx$?
4. How does substitution help in evaluating improper integrals?
5. What techniques are available to approximate the value of an oscillatory integral?

Tip:

For oscillatory integrals, always consider transformations or bounds that can simplify convergence analysis.