Definite integral - possible evaluation using real methods? The book "inside interesting integrals" gives the following exercise for the chapter about contour integration and the residue theorem:
$$\int_{0}^\infty \frac{e^{\cos x}\sin(\sin x)}{x}dx=\space\space ?$$
This can be solved using the function
$$f(z)=\frac{\exp(e^{iz})}{z}$$
on a quarter-circular contour, and is pretty straightforward. The answer turns out to be
$$\frac{\pi}{2}(e-1)$$
However, in the book, the author makes the following comment:

Edward Copson (1901-1980), who was professor of mathematics at the University of St. Andrews in Scotland, wrote "A definite integral which can be evaluated using Cauchy's method of residues can always be evaluated by other means, though generally not so simply." Here's an example of what Copson meant, an integral attributed to the great Cauchy himself. It is easily done with contour integration, but would (I think) otherwise be pretty darn tough.

Does anyone know how to evaluate this integral using real methods?
 A: Perhaps surprisingly, a straightforward trick works. To this end we refer to the following easy-to-prove lemma.

Lemma. Define $\operatorname{Si}(x) = \int_{0}^{x} \frac{\sin t}{t} \, dt$. Then 
  
  
*
  
*$ \int_{0}^{x} \frac{\sin(yt)}{t} \, dt = \operatorname{Si}(xy)$, and
  
*$ \operatorname{Si}(x) = \frac{\pi}{2} + \mathcal{O}(x^{-1})$ as $x \to \infty$.
  

Then for $R > 0$,
\begin{align*}
\int_{0}^{R} \frac{e^{\cos x}\sin(\sin x)}{x} \, dx
&= \int_{0}^{R} \frac{1}{x}\operatorname{Im}(e^{e^{ix}}) \, dx \\
&= \int_{0}^{R} \frac{1}{x}\sum_{n=1}^{\infty} \frac{\sin(nx)}{n!} \, dx \\
&= \sum_{n=1}^{\infty} \frac{1}{n!} \int_{0}^{R} \frac{\sin(nx)}{x} \, dx \\
&= \sum_{n=1}^{\infty} \frac{1}{n!} \operatorname{Si}(nR) \\
&= \sum_{n=1}^{\infty} \frac{1}{n!} \left( \frac{\pi}{2} + \mathcal{O}\left( (nR)^{-1} \right) \right) \\
&= \frac{\pi}{2}(e - 1) + \mathcal{O}(R^{-1})
\end{align*}
Letting $R \to \infty$ proves the claim.
A: $$e^{\cos x}\sin(\sin x) = \text{Im}\, e^{\cos x+i\sin x} = \text{Im}\exp\left(e^{ix}\right) = \text{Im}\sum_{n\geq 0}\frac{e^{nix}}{n!}=\sum_{n\geq 1}\frac{\sin(nx)}{n!}$$
and since $\int_{0}^{+\infty}\frac{\sin(nx)}{x}\,dx = \frac{\pi}{2}$ for any $n>0$, we have
$$ \int_{0}^{+\infty}\frac{e^{\cos x}\sin(\sin x)}{x}\,dx = \frac{\pi}{2}\sum_{n\geq 1}\frac{1}{n!}=\color{red}{\frac{\pi}{2}(e-1)}.$$
