Constant together

I solved this on Quora first:

Consider the function $\displaystyle f(x) = e^{e^{x}}$ Then, exploiting Euler’s formula: $\displaystyle\begin{aligned}f(e^{it}) &= e^{e^{e^{it}}} = e^{e^{\cos t + i\sin t}} = e^{e^{\cos t}e^{i\sin t}}\\&=e^{e^{\cos t}(\cos(\sin t) + i\sin(\sin t))}\\&=e^{e^{\cos t}\cos(\sin t)}(\cos (e^{\cos t}\sin(\sin t) + i\sin(e^{\cos t}\sin(\sin t))\end{aligned}$ Also since $f(x)$ is analytic around $x_{0} = 0$ , the following series converges to $f(x)$ in the neighborhood of zero: $\displaystyle f(x) = \sum_{n = 0}^{\infty}a_{n}x^{n}$ Then we have, $\displaystyle f(e^{it}) = \sum_{n = 0}^{\infty}a_{n}e^{int}$ Equating real parts, $\displaystyle \Re\{f(e^{it})\} = \Re\left\{\sum_{n = 0}^{\infty}a_{n}e^{int}\right\}$ Now integrate both sides from $0$ to $\pi$ and use the fact that $span\{1, \cos(nx)\}$ is an orthogonal set of functions for $(0,\pi) \to \mathbb{R}$ i.e. $\forall n \in \mathbb{Z}$ $\int_{0}^{\pi}\cos(nt)\,dt = 0.$ $\displaystyle \begin{aligned}\int_{0}^{\pi}e^{e^{\cos t}\cos(\sin t)}(\cos (e^{\cos t}\sin(\sin t)) &= a_{0}\\&= \int_{0}^{\pi}e^{e^{0}}\,dt\\&= \pi e\end{aligned}$