I saw this problem on a Facebook post. I tried to look for a solution but it was really a mess. The only thing I could see is that even though $\frac{\sin(\sin(x))}{x} e^{\cos(x)}$ is not defined for $0$, $\lim_{x\to0}\frac{\sin(\sin(x))}{x} e^{\cos(x)}$ is equal to e but nothing more. I'd like to see how to tackle something like this.
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6Integrate $f(z) = e^{e^{iz}}/z$ around semicircle above real axis, show integral around big circle $\to 0$. So $\int_\mathbb{R} f(x) dx$ is related to half of residue of $f(z)$ at $z=0$. Convergence (in Riemann's sense) is proved along the way. – pisco Dec 30 '19 at 10:22
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Correction: the integral around big circle does not tend to $0$, but to a non-zero value. The value of your integral is $\pi(e-1)/2$. – pisco Dec 30 '19 at 15:46