Question

Let
\[ I(R) = \int_0^R e^{-R \sin x} \, dx, \quad R > 0. \]
Which of the following is correct?

• A. \(I(R)>\frac{\pi}{2R}(1 - e^{-R})\)
• B. \(I(R)<\frac{\pi}{2R}(1 - e^{-R})\)
• C. \(I(R) = \frac{\pi}{2R}(1 - e^{-R})\)
• D. \(I(R)\) and \(\frac{\pi}{2R}(1 - e^{-R})\) are not comparable

Hint:

When dealing with oscillatory integrals, consider numerical techniques or series approximations for better insight into the behavior of the function.

Answer 1

The Correct Option is D

1. The integral in question is:

\( I(R) = \int_{0}^{R} e^{-R \sin x} \, dx. \)

2. The presence of \( e^{-R \sin x} \) with the oscillating \( \sin x \) makes the integrand's behavior complex, hindering direct calculation.

3. Analytical evaluation strategies include:

• Approximations of \( e^{-R \sin x} \) for small or large \( R \).
• A simple, closed-form solution for \( I(R) \) is unavailable due to the interplay between the oscillations of \( \sin x \) and exponential decay.

4. The expression \( \frac{\pi}{2R}(1 - e^{-R}) \) is a common approximation for integrals with oscillatory terms, but it is not precise.

5. Because \( I(R) \) and \( \frac{\pi}{2R}(1 - e^{-R}) \) have varying characteristics based on \( R \), a direct comparison for all \( R > 0 \) is invalid.