1. The problem states that we need to evaluate or analyze the given improper integral: $$\int_1^\infty \frac{1}{x} e^{2 + \sin(1/x)} \, dx$$ and it is given that this integral diverges to infinity. 2. Let's analyze the behavior of the integrand: The function is $$f(x) = \frac{1}{x} e^{2 + \sin(1/x)}$$. Since $$\sin(1/x)$$ oscillates between -1 and 1 as $$x \to \infty$$, we know $$\sin(1/x)$$ approaches 0 because $$1/x \to 0$$. Therefore, $$e^{2 + \sin(1/x)} \approx e^2 e^{\sin(1/x)}$$, and $$e^{\sin(1/x)}$$ is close to $$e^0 = 1$$ for large $$x$$. 3. Hence, for large $$x$$, the integrand behaves roughly like: $$\frac{1}{x} e^2 = e^2 \frac{1}{x}$$. 4. We know the integral of $$1/x$$ from 1 to infinity diverges: $$\int_1^\infty \frac{1}{x} dx = \infty$$. 5. Since $$e^2 > 0$$ is a positive constant multiplier, the integral $$\int_1^\infty e^2 \frac{1}{x} dx$$ diverges to infinity. 6. Given that $$e^{2 + \sin(1/x)} \geq e^{2-1} = e^1$$ (since $$\sin(1/x) \geq -1$$), the integrand is bounded below by a positive constant times $$1/x$$, guaranteeing divergence. Hence, the integral diverges to infinity as stated. Final answer: $$\int_1^\infty \frac{1}{x} e^{2 + \sin(1/x)} \, dx = \infty$$ due to the asymptotic behavior like $$\frac{e^2}{x}$$, which is non-integrable over $$[1, \infty)$$.