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\( \int _ { \pi } ^ { \infty } \frac { \sin x } { x ^ { 2 } } d x \)

Ask by Fowler Li. in Iran
Jan 14,2025

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Tutor-Verified Answer

Answer

The integral evaluates to approximately \(-0.073667\).

Solution

To evaluate the improper integral \[ \int_{\pi}^{\infty} \frac{\sin x}{x^2} \, dx, \] we can use integration by parts and properties of special functions. ### Step 1: Integration by Parts Let's set: - \( u = \sin x \) ⇒ \( du = \cos x \, dx \) - \( dv = \frac{dx}{x^2} \) ⇒ \( v = -\frac{1}{x} \) Applying integration by parts: \[ \int \frac{\sin x}{x^2} \, dx = -\frac{\sin x}{x} + \int \frac{\cos x}{x} \, dx. \] ### Step 2: Evaluate the Limits Now, evaluate the integral from \( \pi \) to \( \infty \): \[ \int_{\pi}^{\infty} \frac{\sin x}{x^2} \, dx = \left[ -\frac{\sin x}{x} \right]_{\pi}^{\infty} + \int_{\pi}^{\infty} \frac{\cos x}{x} \, dx. \] As \( x \to \infty \), \( \frac{\sin x}{x} \) oscillates between \(-\frac{1}{x}\) and \(\frac{1}{x}\), both of which approach 0. At \( x = \pi \), \( \sin \pi = 0 \). Therefore, the boundary term is 0: \[ \left[ -\frac{\sin x}{x} \right]_{\pi}^{\infty} = 0 - 0 = 0. \] Thus, \[ \int_{\pi}^{\infty} \frac{\sin x}{x^2} \, dx = \int_{\pi}^{\infty} \frac{\cos x}{x} \, dx. \] ### Step 3: Express in Terms of the Cosine Integral The remaining integral is known as the **Cosine Integral**, denoted by \( \text{Ci}(x) \), which is defined as: \[ \text{Ci}(x) = -\int_{x}^{\infty} \frac{\cos t}{t} \, dt. \] Thus, \[ \int_{\pi}^{\infty} \frac{\cos x}{x} \, dx = -\text{Ci}(\pi). \] ### Conclusion Putting it all together, the original integral evaluates to: \[ \int_{\pi}^{\infty} \frac{\sin x}{x^2} \, dx = -\text{Ci}(\pi). \] ### Numerical Approximation The cosine integral \( \text{Ci}(\pi) \) has a numerical value approximately: \[ \text{Ci}(\pi) \approx 0.073667. \] Therefore, \[ \int_{\pi}^{\infty} \frac{\sin x}{x^2} \, dx \approx -0.073667. \] ### Final Answer \[ \boxed{\, -\,\text{Ci}(\pi) \approx -0.073667\,} \]

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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.

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