Upper bound for $\int_0 ^\pi e^{-A\cos t} dt$ In complex analysis I'm very often using inequality for $A >0$ that:
$$\int_0^\pi e^{-A\sin(t)} dt < \frac \pi A$$
But to be honest I'm not sure how to prove this inequality. Moreover, recently I faced a problem with bounding integral:
$$\int_0^\pi e^{-A\cos(t)} dt $$
and I'm not sure how can I obtain analogous inequality. I tried to use fact that $\sin(t + \frac \pi 2) = \cos(t)$:
$$\int_0^\pi e^{-A\cos(t)} dt = \int_0^\pi e^{-A\sin(t + \frac \pi 2)} dt$$
now by substitution $u = t + \frac \pi 2$ we have that:
$$\int_{\frac \pi 2}^ \frac{3\pi}{2}e^{-A\sin(u)}du$$
But now I cannot use my upper bound for sine integral, since the interval is now $[\frac \pi 2; \frac {3\pi}{2}]$ instead of $[0, \pi]$. Could you please give me a hand with deriving analogous inequality?
 A: Substituting $u=\pi-x$ gives
$$ \int_{\pi/2}^{\pi} e^{-A\sin(t)}dt=\int_0^{\pi/2}e^{-A\sin(u)}du $$
Moreover, $\forall t\in\left[0,\frac{\pi}{2}\right],\sin(t)\geqslant\frac{2}{\pi}t$ therefore
$$ \int_0^{\pi}e^{-A\sin(t)}dt=2\int_0^{\pi/2}e^{-A\sin(t)}dt\leqslant 2\int_0^{\pi/2} e^{-\frac{2A}{\pi}t}=\frac{\pi}{A}\left(1-e^{-A}\right)<\frac{\pi}{A} $$
As for the other integral,
$$ \int_0^{\pi}e^{-A\cos(t)}dt=\pi I_0(A)\underset{A\rightarrow +\infty}{\sim}\sqrt{\frac{\pi}{2A}}e^{A} $$
where $I_0$ is the modified Bessel function https://en.wikipedia.org/wiki/Bessel_function#Asymptotic_forms .
A: The usual way to derive inequalities is to use convexity arguments. THe function $t\mapsto \sin(t)$ has second derivative $-\sin t$, which is $\leq 0$ on the interval $[0,\frac{\pi}{2}]$. So, $\sin t$ is a concave function on this interval, meaning that the straight line segment joining $(0,\sin (0))$ to $(\frac{\pi}{2},\sin\left(\frac{\pi}{2}\right))$, i.e joining the points $(0,0)$ and $(\frac{\pi}{2},1)$, lies below the graph of $\sin t$ (draw a picture to convince yourself). Therefore, we have the inequality
\begin{align}
0\leq \frac{2t}{\pi}\leq\sin t \qquad (0\leq t\leq \frac{\pi}{2}).
\end{align}
By multiplying by $-A$ and integrating, we get
\begin{align}
\int_0^{\pi/2}e^{-A\sin t}\,dt&\leq \int_0^{\pi/2}e^{-2At/\pi}\,dt=\frac{\pi}{2A}(1-e^{-A})<\frac{\pi}{2A}.
\end{align}
If you consider the interval from $0$ to $\pi$ then because $\sin t$ is symmetric around $\pi/2$, we get the estimate $\int_0^{\pi}e^{-A\sin t}\,dt<\frac{\pi}{A}$.
For $\cos t$, play a similar game:
\begin{align}
\int_0^{\pi}e^{-A\cos t}\,dt&=\int_{\pi/2}^{3\pi/2}e^{-A\sin(u)}\,du\\
&=\int_{\pi/2}^{\pi}e^{-A\sin u}\,du+\int_{\pi}^{3\pi/2}e^{-A\sin u}\,du\\
&=\int_0^{\pi/2}e^{-A\sin u}\,du+\int_0^{\pi/2}e^{A\sin u}\,du
\end{align}
(do the obvious change of variables in the last step).
Can you continue?
