Prove that $\exists\,! \,\lambda \in (1/5,1/4)$ such that $\frac{1}{2\pi}\int_0^{2\pi}e^{\sin x}\,\mathrm{d}x=e^{\lambda}$ The following question came up in chat

Prove that $\exists\,! \,\lambda \in (1/5,1/4)$ such that
  $\displaystyle\frac{1}{2\pi}\int_0^{2\pi}e^{\sin x}\,\mathrm{d}x=e^{\lambda}$

Now the integral can be written as 
$$
 \frac{1}{2\pi}\int_0^{2\pi}e^{\sin x}\,\mathrm{d}x
  = \int_0^{1}e^{\sin 2\pi x}\,\mathrm{d}x
  = \sum_{k\geq 0} \int_0^{1} \frac{\sin^k(2\pi x)}{k!} \,\mathrm{d}x
  = \sum_{k\geq 0} \frac{1}{4^k(k!)^2} 
$$
Which is the same as $J_0(1)$, where $J$ is the bessel function of the first kind. Now one can also rewrite the problem by using $e^x = \sum_{k=0} x^k/k!$. So 
\begin{array}{ccc}
  e^{1/5} & < & \int_0^{1}e^{\sin 2\pi x}\,\mathrm{d}x & < & e^{1/4}\\
  \sum_{k\geq0} \frac{1}{5^k k!} & < & \sum_{k\geq 0} \frac{1}{4^k(k!)^2} 
  & < & \sum_{k\geq0} \frac{1}{4^k k!}
\end{array}
Alas the latter sums seems just as hard to prove than the integral form. 
Any help or suggestions would be appreciated. 

Anticlimax ahead:
$$\sum_{k=0}^\infty \frac{1}{4^k(k!)^2} < \sum_{k=0}^\infty \frac{1}{4^k\,k!}$$
follows since $4^k(k!)^2 \geqslant 4^k\, k!$ for all $k$ and the inequality is strict for $k\geqslant 2$.
For the inequality
$$\sum_{k=0}^\infty \frac{1}{5^k\,k!} < \sum_{k=0}^\infty \frac{1}{4^k(k!)^2},$$
we note that the terms for $k = 0$ are equal in both series, so it suffices to see
$$\sum_{k=2}^\infty \frac{1}{k!}\left(\frac{1}{5^k} - \frac{1}{4^k\,k!}\right) \leqslant \sum_{k=2}^\infty \frac{1}{5^k\,k!} \leqslant \frac{1}{50}\sum_{\nu=0}^\infty \frac{1}{15^\nu} = \frac{15}{14\cdot 50} = \frac{3}{140} < \frac{1}{20} = \frac{1}{4^1(1!)^2} - \frac{1}{5^1\,1!}.$$
