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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 $0<t<\pi$ then the bound for the integral would be

$$\int_0^\pi \frac{e^{-R\sin t}}{R^2 - 1} dt \leq (\pi - 0)\frac{1}{R^2 - 1} = \frac{\pi}{R^2 - 1} \quad ?$$ Is 'my' bound just a better bound than the one above? Thanks!

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For which $R$ do you need such an inequality? – Davide Giraudo Mar 27 '12 at 17:26
Ah $R>1$, I'll add that in – user26069 Mar 27 '12 at 17:28
Yes, it seems your bound is sharper. – David Mitra Mar 27 '12 at 17:36
Well, then we have $e^{-R\sin t}\le 1 < 2R$, and the author of the math that is in your notes wants a $2\pi R$ in the numerator for some ulterior purpose. – anon Mar 27 '12 at 17:37

1 Answer 1

up vote 0 down vote accepted

If $R > 1$ your inequality is correct, but then the notes are also right.
If $0 < R < 1$ your inequality is in the wrong direction, because $R^2-1$ is negative. However, the notes are also wrong for approximately $ .3923824170 < R < 1$.

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