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In evaluating this integral:

$$\int_0^\infty \frac{\Im{\left(e^{e^{ix}} \right)}}{x}\text{d}x$$

My means of evaluation was to expand the numerator of the integrand as a fourier series (a.k.a. Taylor series of $e^u$ where $u=e^{ix}$) and then exchange the order of integration and summation.

But what theorems can be used to justify this interchange?

The decay of the integrand is not fast enough for absolute convergence, so nothing in that direction looks promising.

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The usual trick is to improve convergence by introducing an exponential smoothing factor $e^{-\lambda x}$, then let $\lambda\to 0^+$. Have a look at the zeta regularization technique, too.

$e^{z}$ is an entire function, hence $$ e^{e^{ix}} = \sum_{n\geq 0}\frac{e^{nix}}{n!} \tag{1}$$ holds as an identity for every $x\in\mathbb{C}$, so $$ \text{Im}\left(e^{e^{ix}}\right) = \sum_{n\geq 0}\frac{\sin(nx)}{n!}\tag{2} $$ holds as an identity for every $x\in\mathbb{R}$. We may write the original integral as $$ I=\lim_{\lambda\to 0^+}\int_{0}^{+\infty}\frac{\text{Im}\left(e^{e^{ix}}\right)e^{-\lambda x}}{x}\,dx=\lim_{\lambda\to 0^+}\sum_{n\geq 1}\int_{0}^{+\infty}\frac{\sin(nx)}{n! x} e^{-\lambda x}\,\,dx\tag{3} $$ and the $\int-\sum$ exchange is justified by the dominated convergence theorem. $(3)$ leads to: $$ I = \lim_{\lambda\to 0^+}\sum_{n\geq 1}\frac{\arctan\frac{1}{\lambda}}{n!}=(e-1)\lim_{\lambda\to 0^+}\arctan\frac{1}{\lambda}=\color{red}{\frac{\pi(e-1)}{2}}. \tag{4}$$

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  • $\begingroup$ Justification for the assertion that $ I = \lim_{\lambda\to 0^+}\int_{0}^{+\infty}\frac{\text{Im}\left(e^{e^{ix}}\right)e^{-\lambda x}}{x}\,dx $ follows from Daniel Fischer's answer to this question. $\endgroup$ – Random Variable Jul 3 '16 at 16:57
  • $\begingroup$ @RandomVariable: thanks for the improvement. $\endgroup$ – Jack D'Aurizio Jul 3 '16 at 16:57

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