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?

• This question should actually get more attention... I already grabbed my popcorn to see how real methods come in action. If this does not get attention I'm considering to put a bounty on it (if that is possible, I don't know how that works). – Shashi Jan 7 '18 at 21:14
• I have put a bounty on this question. @Nilknarf Are there some updates concerning this question? – Shashi Jan 8 '18 at 20:47
• @Shashi No... Do you suggest that I add anything in particular to the question? – Frpzzd Jan 8 '18 at 23:55
• I was curious whether you have solved it in the mean time – Shashi Jan 9 '18 at 8:00

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

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

• Not thought about a single contour and yet so simple! Conclusion: the author of the book lied lol (jk I know one does not have full vision on everything) – Shashi Jan 9 '18 at 8:03

$$e^{\cos x}\sin(\sin x) = \text{Im}\, e^{\cos x+i\sin x} = \text{Im}\exp\left(e^{ix}\right) = \text{Im}\sum_{n\geq 0}\frac{e^{nix}}{n!}=\sum_{n\geq 1}\frac{\sin(nx)}{n!}$$ and since $\int_{0}^{+\infty}\frac{\sin(nx)}{x}\,dx = \frac{\pi}{2}$ for any $n>0$, we have $$\int_{0}^{+\infty}\frac{e^{\cos x}\sin(\sin x)}{x}\,dx = \frac{\pi}{2}\sum_{n\geq 1}\frac{1}{n!}=\color{red}{\frac{\pi}{2}(e-1)}.$$

• Is it true that interchanging series and integral is allowed since the series with sine is uniformly convergent? The answer is so neat. I guess it is even easier with Real methods after seeing the answers. – Shashi Jan 9 '18 at 7:55
• "Real methods=without Residue Theorem " in the last comment – Shashi Jan 9 '18 at 8:06
• @Shashi: indeed, this is a small variation on the improper Riemann-integrability of $\frac{\sin x}{x}$ on $\mathbb{R}^+$, which can be proved by real (Fourier-analytic), almost-real (Laplace transform) or complex techniques (residue theorem) in few steps, anyway. – Jack D'Aurizio Jan 9 '18 at 14:45