91Ó°ÊÓ

To determine if the integral converges or not, we can use the Comparison Test. The Comparison Test states that if \(0\leq f(x) \leq g(x)\) and the integral \(\int g(x) \,dx\) converges, then the integral \(\int f(x)\, dx\) also converges. Let's choose a simple function \(g(x) = \frac{1}{x^2}\). $$ 0 \leq \frac{\cos(\pi / x)}{x^2} \leq \frac{1}{x^2} $$ Since \(-1 \leq \cos(x) \leq 1\), it follows that \(0 \leq \frac{\cos(\pi / x)}{x^2} \leq \frac{1}{x^2}\), and we can compare the integrals.

To establish the convergence of \(\int_{2}^{\infty} \frac{1}{x^2} dx\), we can use the P-Integral Test. The P-Integral states that if the function is of the form \(g(x) =\frac{1}{x^p}\) with \(p > 0\), the integral converges if \(p > 1\) and diverges when \(0 < p \leq 1\). In our case, \(g(x) = \frac{1}{x^2}\) and \(p=2\). Since \(p > 1\), this integral should converge.

Since the integral \(\int_{2}^{\infty} \frac{1}{x^2} dx\) converges (by the P-Integral Test), by the Comparison Test, the integral of our original function, \(\int_{2}^{\infty} \frac{\cos(\pi/x)}{x^2} dx\), will also converge.

To evaluate the integral, we can execute integration by substitution. Let \(u = \pi/x\), so then \(du = -\pi/x^2 dx\). Converting the limits of integration from \(x\) to \(u\), we find that when \(x=2\), \(u=\pi/2\), and when \(x\to\infty\), \(u\to 0\). Thus, we can rewrite our integral as $$ \int_{2}^{\infty} \frac{\cos(\pi / x)}{x^2} dx = -\int_{\pi/2}^{0} \frac{\cos(u)}{\pi} du $$ Now, integrating the function with respect to \(u\) and applying the Fundamental Theorem of Calculus, we get $$ -\frac{1}{\pi} \int_{\pi/2}^{0} \cos(u) du = -\frac{1}{\pi} [\sin(u)]_{\pi/2}^0 = -\frac{1}{\pi}(\sin(0) - \sin(\pi/2)) = -\frac{1}{\pi}(0-1) = \frac{1}{\pi} $$ Therefore, the value of the definite integral is \(\boxed{\frac{1}{\pi}}\).