I don't quite understand the jump in this inequality given in our notes (R>1) $\int_0^\pi \frac{e^{-R\sin t}}{R^2 - 1} dt \leq \frac{2\pi R}{R^2 - 1}$ surely if we were to use to the bounds for an integral then as $e^{-R\sin t} \leq 1$ for $0
$\int_0^\pi \frac{e^{-R\sin t}}{R^2 - 1} dt \leq (\pi - 0)\frac{1}{R^2 - 1} = \frac{\pi}{R^2 - 1} \quad ?$ Is 'my' bound just a better bound than the one above? Thanks!