Cauchy integral formula

Triple exponential integral

We show the proof of the following integral: $$\int_{0}^{\pi} e^{e^{\cos x}\cos\sin x}\cos(e^{cos x} \sin \sin x) dx = \pi e$$ On the way we will find this beautiful result: $$\int_{0}^{\pi } e^{e^{e^{ix}}} + e^{e^{e^{-ix}}}dx =2\pi e$$ Proof $$\begin{align*} \int_{0}^{\pi} e^{e^{\cos x}\cos\sin x}\cos(e^{cos x} \sin \sin x) dx = &\int_{0}^{\pi} e^{e^{\cos x}\cos\sin x} \left[ \frac{ e^{ie^{\cos x}\sin\sin x} + e^{-ie^{\cos x}\sin\sin x}}{2}\right]dx \\ =& \frac{1}{2}\int_{0}^{\pi } e^{e^{\cos x}\cos\sin x+ie^{\cos x}\sin\sin x} + e^{e^{\cos x}\cos\sin x-ie^{\cos x}\sin\sin x}dx\\ =& \frac{1}{2}\int_{0}^{\pi } e^{e^{\cos x}(\cos\sin x+i\sin\sin x)} + e^{e^{\cos x}(\cos\sin x-i\sin\sin x)}dx\\ =& \frac{1}{2}\int_{0}^{\pi } e^{e^{\cos x}e^{i\sin x}} + e^{e^{\cos x}e^{-i\sin x}}dx\\ =& \frac{1}{2}\int_{0}^{\pi } e^{e^{\cos x + i\sin x}} + e^{e^{\cos x-i\sin x}}dx\\ =& \frac{1}{2}\int_{0}^{\pi } e^{e^{e^{ix}}} + e^{e^{e^{-ix}}}dx\\ =& \frac{1}{2}\int_{0}^{\pi } \left[ \sum_{n=0}^{\infty} \frac{(e^{e^{ix}})^n}{n!} + \sum_{0}^{\infty} \frac{(e^{e^{-ix}})^{n}}{n!} \right]dx\\ =& \frac{1}{2}\int_{0}^{\pi } \sum_{n=0}^{\infty} \frac{(e^{e^{ix}})^n+(e^{e^{-ix}})^{n}}{n!} dx\\ =& \frac{1}{2}\sum_{n=0}^{\infty} \frac{1}{n!}\int_{0}^{\pi } (e^{e^{ix}})^n+(e^{e^{-ix}})^{n}dx\\ =& \frac{1}{2}\sum_{n=0}^{\infty} \frac{1}{n!} \left[\int_{0}^{\pi } (e^{e^{ix}})^ndx+ \int_{0}^{\pi }(e^{e^{-ix}})^{n}dx\right]\\ =& \frac{1}{2}\sum_{n=0}^{\infty} \frac{1}{n!}\left[\int_{0}^{\pi } (e^{e^{ix}})^ndx- \int_{\pi}^{0 }(e^{e^{-ix}})^{n}dx\right]\\ =& \frac{1}{2}\sum_{n=0}^{\infty} \frac{1}{n!}\left[\int_{0}^{\pi } (e^{e^{ix}})^ndx+ \int_{-\pi}^{0 }(e^{e^{ix}})^{n}dx\right]\\ =& \frac{1}{2}\sum_{n=0}^{\infty} \frac{1}{n!}\int_{-\pi}^{\pi } (e^{e^{ix}})^ndx \\ =& \frac{1}{2}\sum_{n=0}^{\infty} \frac{1}{n!}\oint_{\gamma} \frac{e^{zn}}{zi} dz \quad \left( \textrm{where } \gamma(x) = e^{ix} \quad x\in [-\pi,\pi]\right)\\ =& \frac{1}{2}\sum_{n=0}^{\infty} \frac{1}{n!} 2\pi \quad (\textrm{Cauchy integral formula}) \\ =& \pi e \end{align*}$$ $$\boxed{\int_{0}^{\pi} e^{e^{\cos x}\cos\sin x}\cos(e^{cos x} \sin \sin x) dx = \pi e }$$ Corollary $$\boxed{ \int_{0}^{\pi } e^{e^{e^{ix}}} + e^{e^{e^{-ix}}}dx =2\pi e}$$