integral inequality

[Grigorios Kostakos]

Prove that \(\displaystyle\int_{0}^{\frac{\pi}{2}}{e^{-k\sin{x}}\,dx}<\frac{\pi}{2k}\,(1-e^{-k})\,,\quad k>0\,.\)

[achilleas]

Let \(f(x)=\sin x-\dfrac{2x}{\pi}\) for \(x\in [0,\pi/2]\). Then \(f'(x)=\cos x-\dfrac{2}{\pi}\). Clearly, \( f'\) is strictly decreasing in \([0,\pi/2]\) and since
\[
f'(\pi/2)>0>f'(0),\]
there is a unique \(\xi\in (0,\pi/2)\) such that \(f'(\xi)=0\).
Hence \(f'(x)>0\) for all \(x\in (\xi,\pi/2]\) and \(f'(x)<0\) for all \(x\in [0,\xi)\). Accordingly, we can easily see that
\[
f(x)\geq 0
\]
for all \(x\in [0,\pi/2]\), with equality if and only if \(x=0\) or \(x=\pi/2\).
Hence
\[
e^{-k\sin x}\leq e^{-\frac{2kx}{\pi}}
\]
for all \(x\in [0,\pi/2]\), with equality if and only if \(x=0\) or \(x=\pi/2\).
Integrating the above inequality from \(0\) to \(\pi/2\) we obtain
\[
\int_0^{\pi/2} e^{-k\sin x}\,dx< \int_0^{\pi/2} e^{-\frac{2kx}{\pi}}\,dx=\dfrac{\pi}{2k}\left(1-e^{-k}\right),
\]
as desired.

The proof is complete