How to prove $\int_{0}^{\frac{\pi}{2}}e^{-R\sin x}dx$ < $\bigl(\frac{\pi}{2R}\bigr)(1-e^{-R})$, for all $ R > 0$.

Question

Hi sorry for bothering you, I encountered some problems while solving the following question, hope you can give me some advice of my method or any other ideas to figure out how to prove it. Thanks a lot.

$R$ is an arbitrary real number, prove that

$$\int_{0}^{\frac{\pi}{2}}e^{-R\sin x}dx < \frac{\pi}{2R}(1-e^{-R}),\quad\text{ for all }\:R > 0.$$

Here's my method.

Instead of integrate the left equation directly, I chose to change the form of the equation right hand side as following.

\begin{align}\frac{\pi}{2R}(1-e^{-R}) &v= -\frac{\pi}{2R}(e^{-R\sin x}|_{0}^{\frac{\pi}{2}}) \\ &= -\frac{\pi}{2R}\int_{0}^{\frac{\pi}{2}}-R\cos x\ e^{-R\sin x}\,dx \\ &= \int_{0}^{\frac{\pi}{2}}\frac{\pi}{2}\cos x\ e^{-R\sin x}dx \end{align} now let $f(x) = e^{-R\sin x}$ and $g(x) = \frac{\pi}{2}\cos x\ e^{-R\sin x}$

At first I tried to use the property "$0 < f(x) < g(x)\implies 0 < \int f(x) < \int g(x)$", however I found that between $0$ and $\frac{\pi}{2}$, $g(x)$ is not always greater than $f(x)$, since $e^{-R} \le e^{-R\sin x}$ (i.e. $f(x))\le e^{0}$ and $0 \le \frac{\pi}{2}\cos x\ e^{-R\sin x}$ (i.e. $\,g(x)) \le \frac{\pi}{2}$)

And then I stuck. Did I think wrong in any steps or what could I do next? Could you give me some clues?

Thanks for spending time watching this whole thing, I really appreciate that. And hope you have a great day.

Answer 1

The inequality comes from the fact that

• $\sin x > \frac{2}{\pi}x$ for $0<x<\frac{\pi}{2}$

Hence, you get for $R > 0$

$$\int_{0}^{\frac{\pi}{2}}e^{-R\sin x}dx < \int_{0}^{\frac{\pi}{2}}e^{-\frac{2R}{\pi}x}dx = \frac{\pi}{2R}(1-e^{-R})$$