Evaluate the following integral or state that it diverges. \( \int_{3}^{\infty} \frac{8}{x^{2}} \sin \left(\frac{\pi}{x}\right) d x \) Select the correct choice and, if necessary, fill in the answer box to complete your choice. A. The improper integral converges and \( \int_{3}^{\infty} \frac{8}{x^{2}} \sin \left(\frac{\pi}{x}\right) d x=\square \). (Type an exact answer, using \( \pi \) and radicals as needed.)

Let \[ I = \int_{3}^{\infty} \frac{8}{x^2} \sin\left(\frac{\pi}{x}\right) dx. \] We use the substitution \[ u = \frac{\pi}{x}, \] so that \[ \frac{du}{dx} = -\frac{\pi}{x^2} \quad \Longrightarrow \quad du = -\frac{\pi}{x^2}\,dx. \] Solving for \( dx \) gives \[ dx = -\frac{x^2}{\pi}\, du. \] Substitute into the integral: \[ I = \int_{x=3}^{x\to\infty} \frac{8}{x^2} \sin\left(\frac{\pi}{x}\right) \left(-\frac{x^2}{\pi}\, du\right) = -\frac{8}{\pi} \int_{x=3}^{x\to\infty} \sin(u)\, du. \] Next, we change the limits of integration. When \( x = 3 \), \[ u = \frac{\pi}{3}. \] When \( x \to \infty \), \[ u = \frac{\pi}{\infty} = 0. \] Thus, the integral becomes \[ I = -\frac{8}{\pi} \int_{u=\pi/3}^{0} \sin(u)\, du. \] Reversing the limits of integration, we obtain: \[ I = \frac{8}{\pi} \int_{0}^{\pi/3} \sin(u)\, du. \] Now, evaluate the integral: \[ \int_{0}^{\pi/3} \sin(u)\, du = \left[-\cos(u)\right]_{0}^{\pi/3} = \left[-\cos\left(\frac{\pi}{3}\right)+\cos(0)\right]. \] Since \[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \quad \text{and} \quad \cos(0)= 1, \] we have \[ \int_{0}^{\pi/3} \sin(u)\, du = -\frac{1}{2} + 1 = \frac{1}{2}. \] Substitute back into the expression for \( I \): \[ I = \frac{8}{\pi} \cdot \frac{1}{2} = \frac{4}{\pi}. \] Thus, the improper integral converges and its value is \[ \frac{4}{\pi}. \]