Inspired by this post where the value of $\int_0^{\infty}\frac{\sin(\tan(x))}{x}\,dx$ was found to be $\frac{\pi}{2}(1-e^{-1})$, I set out to do the same thing with $\int_0^{\infty}\frac{\sin(\sin(x))}{x}\,dx$. Convergence is slow, which makes numerical estimation difficult, but after coaxing Mathematica for a while, I got:
NIntegrate[Sin[Sin[x]]/x, {x, 0, 20000 Pi}, MaxRecursion -> 20, WorkingPrecision -> 20, Method -> "DoubleExponential"]
1.4446949333948902084
My method is largely similar and currently I have achieved a Pyrrhic victory: I got down to an integral Mathematica was able to evaluate, but I don't see how to evaluate the integral myself.
My approach was largely similar to the linked post: use periodicity and a series expansion using reciprocals to rewrite the integrand.
$$
\int_0^{\infty} \frac{\sin(\sin(x))}{x}\,dx = \frac{1}{2}\int_{-\infty}^{\infty} \frac{\sin(\sin(x))}{x}\,dx
$$
$$
=\frac{1}{2}\sum_{n=-\infty}^{\infty} \int_{n\pi}^{(n+1)\pi} \frac{\sin(\sin(x))}{x}\,dx
$$Now substitute $x=z+n\pi$:
$$
=\frac{1}{2}\sum_{n=-\infty}^{\infty} \int_{0}^{\pi} \frac{\sin(\sin(z+n\pi))}{z+n\pi}\,dz
$$
$$
=\frac{1}{2}\sum_{n=-\infty}^{\infty} (-1)^n \int_{0}^{\pi} \frac{\sin(\sin(z))}{z+n\pi}\,dz
$$Swap the sum and integral and use the series representation for cosecant:
$$
=\frac{1}{2} \int_{0}^{\pi} \sin(\sin(z))\sum_{n=-\infty}^{\infty} \frac{(-1)^n} {z+n\pi}\,dz
$$
$$
=\frac{1}{2} \int_{0}^{\pi} \sin(\sin(z))\csc(z)\,dz=\int_{0}^{\pi/2} \sin(\sin(z))\csc(z)\,dz,
$$where the last inequality is by symmetry. Now I substituted $\sin(z)=y$ which leads to the integral in the title:
$$
= \int_0^1 \frac{\sin(y)}{y\sqrt{1-y^2}}\,dy
$$(Note: at this point in the first linked post, the substitution is much nicer because the Pythagorean identity gives us a plus instead of a minus.) Now Mathematica cooperated: it tells me this integral is equal to
$$
\frac{1}{4} \pi ^2 \pmb{H}_0(1) J_1(1)-\frac{1}{4} \pi (\pi
\pmb{H}_1(1)-2) J_0(1) \approx 1.4447091498105593077;
$$here $J_a$ and $\pmb{H}_a$ are the Bessel and Struve functions, respectively.
My question: I would appreciate if someone could explain how this last integral was evaluated (it was 'known' in a way the original wasn't). I tried a series expansion using the Cauchy product for $\sin(y)/y$ and $(1-y^2)^{-1/2}$ but couldn't quite get a hold of the coefficients. If by some miracle the closed-form could be simplified a bit, that would be good as well.
Best Answer
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\int_{0}^{1}{\sin\pars{y} \over y\root{1 - y^{2}}}\,\dd y} \\ = &\ \int_{0}^{1}{1 \over \root{1 - y^{2}}}\ \overbrace{\pars{\int_{0}^{1} \cos\pars{ky}\,\dd k}}^{\ds{\sin\pars{y} \over y}}\ \,\dd y \\[5mm] = &\ \int_{0}^{1}\int_{0}^{1} {\cos\pars{ky} \over \root{1 - y^{2}}}\,\dd y\,\dd k = {\pi \over 2}\int_{0}^{1}\on{J}_{0}\pars{k}\,\dd k \end{align} where I used a Bessel $\ds{\on{J}_{\nu}}$ Integral Representation. $\ds{\on{\bf H}_{\nu}}$ is a Struve Function. \begin{align} &\bbox[5px,#ffd]{\int_{0}^{1}{\sin\pars{y} \over y\root{1 - y^{2}}}\,\dd y} \\[5mm] = &\ {\pi \over 2}\,\on{J}_{0}\pars{1} + {\pi^{2} \over 4}\on{J}_{1}\pars{1} \on{\bf H}_{0}\pars{1} - {\pi^{2} \over 4} \on{J}_{0}\pars{1}\on{\bf H}_{1}\pars{1} \\[5mm] = &\ 1.44470914981055930772056106554\ldots \end{align} The last result is given by Formula $\ds{{\bf 8}.}$, page $660$, of Table of Integrals, Series and Products ( seventh edition ) by I. S. Gradshteyn and I. M. Ryzhik.
I "guess" those integrations use somehow the generating functions.