\( \int _ { \pi } ^ { \infty } \frac { \sin x } { x ^ { 2 } } d x \)
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Bonus Knowledge
To solve the integral \( \int _ { \pi } ^ { \infty } \frac { \sin x } { x ^ { 2 } } d x \), we can use integration by parts. Let: - \( u = \frac{\sin x}{x} \) (which we'll differentiate) - \( dv = \frac{1}{x} \, dx \) (which we'll integrate) This gives us: - \( du = \left(\frac{\cos x}{x} - \frac{\sin x}{x^2}\right) \, dx \) - \( v = \ln x \) Now, applying integration by parts yields: \[ \int u \, dv = uv - \int v \, du \] Evaluating the boundary terms at \( \pi \) and \( \infty \): 1. As \( x \to \infty \): \( \frac{\sin x}{x} \ln x \to 0 \) because \( \frac{\sin x}{x} \) oscillates between -1 and 1, while \( \ln x \) grows, leading to a net product approaching 0. 2. At \( x = \pi \): \( \frac{\sin \pi}{\pi} \ln \pi = 0 \). Thus, the boundary terms contribute 0. This allows us to focus on the remaining integral: \[ -\int \ln x \left( \frac{\cos x}{x} - \frac{\sin x}{x^2} \right) dx \] This integral is still complex, but evaluating directly or using convergence tests shows this specific integral converges due to the oscillatory nature of the sine function and the \( x^{-2} \) decay. In conclusion: The integral converges, and its value can be evaluated numerically if required, but exact solvability may involve special functions or numerical approximations.