91Ó°ÊÓ

First, we rewrite the improper integral as a limit of definite integral: $$\int_{1}^{\infty} \frac{1}{x^{2}} \sin \frac{\pi}{x} d x = \lim_{b\to\infty} \int_{1}^{b} \frac{1}{x^{2}} \sin \frac{\pi}{x} d x$$

In order to determine the convergence of the integral, we can apply the Comparison Test. We compare the given integral to the integral of the positive function \(\int_{1}^{\infty} \frac{1}{x^{2}} dx\), which we know converges (since the exponent is greater than 1). We observe that for all \(x \geq 1\), $$0\leq \frac{1}{x^{2}} \sin \frac{\pi}{x} \leq \frac{1}{x^{2}}$$ Since the integral of the function \(\frac{1}{x^{2}}\) converges, the integral of the function \(\frac{1}{x^{2}} \sin \frac{\pi}{x}\) also converges by the Comparison Test.

Since we've determined that the integral converges, we can now evaluate its value using Integration by Parts. Let: $$u = \frac{1}{x^2}, \quad d v = \sin \frac{\pi}{x} d x$$ Differentiate \(u\) and integrate \(d v\) to obtain: $$d u = -\frac{2}{x^3} d x, \quad v = -\int \sin \frac{\pi}{x} d x$$ The integral of \(v\) is not elementary and requires integration by parts: Let: $$w = \frac{\pi}{x}, \quad d z = \sin \frac{\pi}{x} d x$$ Differentiate \(w\), and integrate \(d z\): $$d w = -\frac{\pi}{x^2} d x, \quad z = -\frac{1}{\pi}\cos \frac{\pi}{x}$$ Now, substitute back into the expression for \(v\): $$v = -\int \sin \frac{\pi}{x} d x = wz - \int z d w = -\frac{\pi}{x}\left(-\frac{1}{\pi}\cos \frac{\pi}{x}\right) - \int -\frac{1}{\pi}\cos \frac{\pi}{x} \left(-\frac{\pi}{x^2}\right) d x$$ Simplifying, we get: $$v = \frac{1}{x}\cos \frac{\pi}{x} + \int \frac{1}{x^2}\cos \frac{\pi}{x} d x$$ Now, apply Integration by Parts to the original integral: $$\int_{1}^{b} \frac{1}{x^{2}} \sin \frac{\pi}{x} d x = uv - \int v d u = \left(\frac{1}{x}\cos \frac{\pi}{x} + \int \frac{1}{x^2}\cos \frac{\pi}{x} d x\right)\left(-\frac{2}{x^3}\right) d x$$ We differentiate \(v\) and integrate \(d u\) to get: $$d v = -\frac{2}{x^3} \cos \frac{\pi}{x} + \frac{2\pi}{x^4}\sin\frac{\pi}{x}, \quad u = -\frac{1}{x}$$ Now, apply the limit of integration to the resulting integral: $$\lim_{b\to\infty} -\frac{2}{x} \cos \frac{\pi}{x}\bigg|_1^b - \int_{1}^{b} \left(-\frac{2}{x^3} \cos \frac{\pi}{x} + \frac{2\pi}{x^4}\sin\frac{\pi}{x}\right) d x$$ It can be shown that both terms in this expression approach zero as \(b\to\infty\): $$\lim_{b\to\infty} \frac{2}{x} \cos \frac{\pi}{x}\bigg|_1^b = 0$$ Therefore, the value of the improper integral is: $$\int_{1}^{\infty} \frac{1}{x^{2}} \sin \frac{\pi}{x} d x = \boxed{0}$$