How did this integral inequality come to be?

Question

I've seen this result $$ \int\limits_0^\pi e^{-R\sin\theta}d\theta \le \frac{\pi}{R}\left(1-e^{-R}\right) $$

with $R>0$, where the source said to have derived this inequality using Jordan's inequality, which is that for $\theta\in [0,\pi/2]$, $\frac{2}{\pi}\theta\le\sin\theta\le \theta$.

Here are my comments regarding this inequality:

1. Not so while ago, I derived this inequality, but now I cannot remember how (I probably made an error somewhere at that time, that's why things don't seem to add up now).

2. How would you use Jordan's inequality to derive the inequality above?

Answer 1

Since \begin{align} -\frac{2R}{\pi} x\geq -R\sin x \geq -Rx \end{align} for $x\in [0, \pi/2]$, then we see that \begin{align} \int^{\pi/2}_0 e^{-R\sin x}\ dx \leq \int^{\pi/2}_0 e^{-\frac{2R}{\pi} x}\ dx = \frac{\pi}{2R}\left(1-e^{-R} \right). \end{align} Do the same type of estimate for $x \in [\pi/2, \pi]$.

Answer 2

Jordan's inequality can be obtained by noting the values of the corresponding differentials and the initial values of the functions.

For the second part, note that the integral is same in [0,90] and [90,180]. Split the interval and use the left part of Jordan's Inequality.