M Cauchy Présente

For any $0 < \epsilon < 1 < R$ , let $\Gamma_{\epsilon,R}$ be the contour $\Gamma_{\epsilon,R} \,=\, \gamma_1 + \gamma_2 - \gamma_3 - \gamma_4$ where

• $\gamma_1$ is the straight line segment from $\epsilon$ to $R$ ,
• $\gamma_2$ is the semicircular arc $z = Re^{i\theta}$ for $0 \le \theta \le \pi$ ,
• $\gamma_3$ is the straight line segment from $-\epsilon$ to $-R$ ,
• $\gamma_4$ is the semicircular arc $z = \epsilon e^{i\theta}$ for $0 \le \theta \le \pi$ .

Since the function $f(z) \; = \; \frac{e^{e^{iz}} - 1}{z}$ is analytic on the punctured disk $\mathbb{C} \backslash \{0\}$ , we deduce that $\int_{\Gamma_{\epsilon,R}} f(z) \,dz \; = \; 0$ for all $0 < \epsilon < 1 < R$ . Now $\int_{\gamma_1} f(z)\,dz \; = \; \int_{\epsilon}^R \frac{e^{e^{ix}} - 1}{x}\,dx \qquad \qquad \int_{\gamma_3} f(z)\,dz \; = \; \int_{\epsilon}^R \frac{e^{e^{-ix}} - 1}{x}\,dx$ and hence $\left(\int_{\gamma_1} - \int_{\gamma_3}\right) f(z)\,dz \; =\; \int_\epsilon^R \frac{e^{e^{ix}} - e^{e^{-ix}}}{x}\,dx \; = \; 2i\int_\epsilon^R e^{\cos x}\sin(\sin x)\,\frac{dx}{x}$ Since $f(z)$ has a simple pole at $z=0$ , we deduce that $\lim_{\epsilon \to 0} \int_{\gamma_4} f(z)\,dz \; = \; \pi i \mathrm{Res}_{z=0}f(z) \; = \; \pi i (e-1)$ and so we deduce, letting $\epsilon \to 0$ , that $2i\int_0^R e^{\cos x}\sin(\sin x)\,\frac{dx}{x} + \int_{\gamma_2} f(z)\,dz - \pi i(e-1) \; = \; 0$ If $z \in \gamma_2$ and $w = e^{iz}$ , then $z = Re^{i\theta}$ for $0 \le \theta \le \pi$ , and so $|w| = e^{-R\sin\theta}$ . Thus $|f(z)| \; \le \; \displaystyle \left|\frac{e^w - 1}{z}\right| \; = \; \frac{1}{R}\left|\sum_{n=1}^\infty \frac{w^n}{n!}\right| \; \le \; \frac{1}{R}\sum_{n=1}^\infty \frac{|w|^n}{n!} \; \le \; \frac{|w|}{R}\sum_{n=1}^\infty \frac{1}{n!} \; \le \; \frac{e}{R}e^{-R\sin\theta}$ so that $\left|\int_{\gamma_2} f(z)\,dz\right| \; \le \; e\int_0^\pi e^{-R\sin\theta}\,d\theta \; = \; 2e\int_0^{\frac12\pi}e^{-R\sin\theta}\,d\theta$ Jordan's Inequality tells us that $\sin\theta \,\ge\, \tfrac{2}{\pi}\theta$ for $0 \le \theta \le \tfrac12\pi$ , so it follows that $\int_{\gamma_2} f(z)\,dz \; = \; O\big(R^{-1}\big) \qquad \qquad R \to \infty$ which implies, letting $R \to \infty$ , that $\int_0^\infty e^{\cos x}{\sin(\sin x)} \frac{dx}{x} \; = \; \tfrac12\pi(e-1)$ so that $P=2$ , $Q=1$ , $R=1$ and $S=1$ , making the answer $2111$ .

Let $\displaystyle \int_Cf(z)dz=\int_C \frac{e^{e^{iz}}}{z} dz$

where $C$ is the contour as shown in figure

Using the Cauchy's theorem :

$\displaystyle \int\limits_{r}^{R}f(x)dx + \int_{\gamma}f(z)dz+\int\limits_{-R}^{-r}f(x)dx + \int_{\Gamma}f(z)dz = 0$

Now, consider $\displaystyle \int_{\Gamma}f(z)dz = \int\limits_{0}^{\pi}\frac{e^{e^{iRe^{i\theta}}}}{Re^{i\theta}}Rie^{i\theta} d\theta$

As $R \to \infty$

$\displaystyle \int\limits_{0}^{\pi} ie^{e^{iRe^{i\theta}}}d\theta \to i\pi$

Similarly, as $r \to 0$

$\displaystyle \int\limits_{-\pi}^{0} ie^{e^{ire^{i\theta}}}d\theta \to -ei\pi$

Substituting these back into Cauchy's theorem, we get

$\displaystyle \int\limits_{-\infty}^{\infty} f(x)dx = i\pi(e-1)$

or $\displaystyle \int\limits_{0}^{\infty} \frac{e^{e^{ix}}}{x} dx = i\frac{\pi}{2}(e-1)$

On expanding left hand side , and comparing imaginary parts, we can easily get

$\displaystyle \int\limits_{0}^{\infty} \frac{e^{\cos x}\sin(\sin x)}{x} dx = \frac{\pi}{2}(e-1)$

Hence , the answer is $\boxed{2111}$

A solution using the "Feynmann trick":

$\forall a> 0$ , let us define the real valued function $f(a)=\int_{0}^{\infty}e^{-a x}e^{\cos x}\frac{\sin (\sin x)}{x} dx$ Then we can write $\frac{df}{da}=-\int_0^{\infty}e^{-a x}e^{\cos x}{\sin (\sin x)} dx$ Now note that we can write, $\begin{aligned} -\frac{df(a)}{da}=& \Im\left(\int_0^\infty {e^{(-a x +e^{ikx})}} dx\right)=\Im(g(a)) \end{aligned}$ Furthermore, $\begin{aligned} g(a)=& \int_{0}^{\infty}\sum_{k\ge 0}\frac{e^{(-a+ik)x}}{k!} dx\\ \ =&\sum_{k\ge 0}\frac{1}{k!}\int_0^\infty {e^{(-a+ik)x}} \end{aligned}$ The second step can be justified using Fubini's theorem . Now, note that $\begin{aligned} h(a): &= \int_0^\infty {e^{(-a+ik)x}}dx\\ \ &=\frac{1}{a-ik} \end{aligned}$ which follows since the integral converges for $a>0$ . Then, it follows, after some algebraic manipulation, and an application of the Monotone Convergence Theorem $\begin{aligned} -\frac{df(a)}{da}= & \sum_{k\ge 0}\frac{1}{k!((k+1)^2+a^2)}\\ \implies f(a)= & -\sum_{k\ge 0}\frac{1}{(k+1)!}\arctan\left(\frac{a}{k+1}\right)+C \end{aligned}$ where $C$ is the integration constant. Now note that $\lim_{a\to \infty}f(a)=0=-\frac{\pi}{2}(e-1)+C\implies C=\frac{\pi(e-1)}{2}$ . Finally, note that $f(a)$ is a monotonically decreasing function of $a$ and is lower bounded by the desired expression. Thus using Bounded Convergence Theorem , we get the evaluation of the expression as $\frac{\pi(e-1)}{2}$ , which yields the answer $\boxed{2111}$ .

Maybe the best way of solving this integral is by complex analysis; I chose the elementary method: Let:

$\displaystyle I= \int_0^{\infty} e^{\cos x} \sin (\sin x) \frac{dx}{x}$

Recalling the definitions and properties of hyperbolic sine and hyperbolic cosine :

$\displaystyle \cosh x + \sinh x = e^x \;\;\;\; => \;\; \cosh (\cos x ) + \sinh (\cos x) = e^{\cos x}$

Therefore:

$\displaystyle I = \int_0^{\infty} \cosh (\cos x ) \sin (\sin x) \frac{dx}{x} + \int_0^{\infty} \sinh (\cos x) \sin (\sin x) \frac{dx}{x}$

Now consider ( For real numbers $a$ and $b$ ) :

$\displaystyle \sin (a+ib) = \sin a \cos ib + \cos a \sin ib = \sin (a) \cosh (b) + i\cos (a) \sinh (b)$

$\displaystyle \cos (a+ib) = \cos a \cos ib - \sin a \sin ib = \cos (a) \cosh (b) - i\sin (a) \sinh (b)$

Therefore:

$\displaystyle \sin (a) \cosh (b) = \Re \sin (a+ib) \;\;\;\;\;\text{and}\;\;\;\;\;\; \sin (a) \sinh (b) = - \Im \cos (a+ib)$

Hence :

$\displaystyle \sin (\sin x) \cosh (\cos x) = \Re \sin (\sin x+i\cos x) = \Re \sin ( i e^{-ix} )$

And,

$\displaystyle \sin (\sin x) \sinh (\cos x) = \Im \cos (\sin x+i\cos x) = \Im \cos ( i e^{-ix} )$

Now substitute these in the expression of $I$ :

$\displaystyle I = \Re \int_0^{\infty} \sin ( i e^{-ix} ) \frac{dx}{x} + \Im \int_0^{\infty} \cos ( i e^{-ix} ) \frac{dx}{x}$

Using Taylor series expansion of the sine and the cosine functions:

$\displaystyle I = \Re \int_0^{\infty} \sum_{n=0}^{\infty} (-1)^n \frac{(i e^{-ix})^{2n+1}}{(2n+1)!} \frac{dx}{x} + \Im \int_0^{\infty} \sum_{m=0}^{\infty} (-1)^m \frac{(i e^{-ix})^{2m}}{(2m)!} \frac{dx}{x}$

Note that $i^{2m} = (-1)^m$ and $i=e^{i\frac{\pi}{2}}$ :

$\displaystyle I = \Re \int_0^{\infty} \sum_{n=0}^{\infty} \frac{e^{i(\frac{\pi}{2} -x(2n+1)})}{(2n+1)!} \frac{dx}{x} + \Im \int_0^{\infty} \sum_{m=0}^{\infty} \frac{e^{-ix(2m)}}{(2m)!} \frac{dx}{x}$

taking the real and the imaginary parts :

$\displaystyle I = \int_0^{\infty} \sum_{n=0}^{\infty} \frac{\sin ((2n+1)x)}{(2n+1)!} \frac{dx}{x} + \int_0^{\infty} \sum_{m=0}^{\infty} \frac{\sin (2mx)}{(2m)!} \frac{dx}{x}$

Now using Fubini's Theorem:

$\displaystyle I = \sum_{n=0}^{\infty} \frac{1}{(2n+1)!} \int_0^{\infty} \frac{\sin ((2n+1)x)}{x} dx + \sum_{m=0}^{\infty} \frac{1}{(2m)!} \int_0^{\infty} \frac{\sin (2mx)}{x} dx$

And use the well-known Dirichlet Integral :

$\displaystyle I = \sum_{n=0}^{\infty} \frac{1}{(2n+1)!} (\frac{\pi}{2} ) + \sum_{m=1}^{\infty} \frac{1}{(2m)!} (\frac{\pi}{2} )$

Note that we omitted the $m=0$ term of the second summation , because the term will disappear before performing the Dirichlet Integral. Now it is only the Taylor series of hyperbolic sine and hyperbolic cosine:

$\displaystyle I= \frac{\pi}{2} (\sinh (1) + \cosh (1) - 1) = \boxed{\frac{1}{2} \pi ^1 (1e -1) }$