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"]

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.


$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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.

  • $\begingroup$ Very nice solution ! $\endgroup$ – Claude Leibovici Oct 24 '20 at 5:11
  • $\begingroup$ @ClaudeLeibovici Thanks. $\endgroup$ – Felix Marin Oct 24 '20 at 5:14
  • $\begingroup$ Unrelated: I've seen Gradshteyn & Ryzhik mentioned several times on this website. In your opinion, is it worth buying a copy? $\endgroup$ – FearfulSymmetry Oct 24 '20 at 19:07
  • 1
    $\begingroup$ @Integrand There's a $\displaystyle{\tt PDF}$ copy in Internet Archive. $\endgroup$ – Felix Marin Oct 24 '20 at 19:15

This is not an answer.

To me, this is one more mistery of CAS (I had a few of these in the last thirty years which I still do not understand).

May be, a part of possible explanation.

Using $$\frac{1}{y \sqrt{1-y^2}}=\sum_{n=0}^\infty (-1)^n \binom{-\frac{1}{2}}{n} y^{2 n-1}$$ we face the problem of $$I_n=\int_0^1 y^{2n-1}\sin(y)\,dy=\frac{\, _1F_2\left(n+\frac{1}{2};\frac{3}{2},n+\frac{3}{2};-\frac{1}{4}\right)}{2n+1}$$ the first expansions of the hypergeometric functions are given below as linear combinations of Bessel functions of the first kind $$\left( \begin{array}{cc} n & \sqrt{\frac 2{\pi }}\, I_n \\ 0 & \sqrt{\frac{2}{\pi }} \text{Si}(1) \\ 1 & J_{\frac{3}{2}}(1) \\ 2 & 3 J_{\frac{5}{2}}(1)-J_{\frac{7}{2}}(1) \\ 3 & 14 J_{\frac{7}{2}}(1)-J_{\frac{9}{2}}(1) \\ 4 & 97 J_{\frac{9}{2}}(1)-16 J_{\frac{11}{2}}(1) \\ 5 & 853 J_{\frac{11}{2}}(1)-45 J_{\frac{13}{2}}(1) \\ 6 & 9330 J_{\frac{13}{2}}(1)-1007 J_{\frac{15}{2}}(1) \end{array} \right)$$ that is to say $$I_n=\sqrt{\frac{\pi }{2}}\left(a_n J_{\frac{2n+1}{2}}(1)-b_n J_{\frac{2n+3}{2}}(1) \right)$$ But the $I_n$ simplify in terms of linear combinations of $\sin(1)$ and $\cos(1)$ $$\left( \begin{array}{cc} n & I_n \\ 1 & -\cos (1)+\sin (1) \\ 2 & 5 \cos (1)-3 \sin (1) \\ 3 & -101 \cos (1)+65 \sin (1) \\ 4 & 4241 \cos (1)-2723 \sin (1) \\ 5 & -305353 \cos (1)+196065 \sin (1) \\ 6 & 33588829 \cos (1)-21567139 \sin (1) \end{array} \right)$$

Considering now $$S_p=\text{Si}(1)+\sum_{n=1}^p (-1)^n \binom{-\frac{1}{2}}{n}\int_0^1 y^{2 n-1}\sin(y)\,dy$$ $$S_6=\text{Si}(1)+$$ $$\sqrt{\frac{\pi }{2}}\left(\frac{J_{\frac{3}{2}}(1)}{2}+\frac{9 J_{\frac{5}{2}}(1)}{8}+4 J_{\frac{7}{2}}(1)+\frac{3355 J_{\frac{9}{2}}(1)}{128}+\frac{52619 J_{\frac{11}{2}}(1)}{256}+\frac{1071945 J_{\frac{13}{2}}(1)}{512}-\frac{232617 J_{\frac{15}{2}}(1)}{1024} \right)$$ that is to say $$S_6=\text{Si}(1)+\frac{7 (1097603873 \cos (1)-704763287 \sin (1))}{1024}$$

  • $\begingroup$ The mystery (misery?) of CAS is one I know all too well... $\endgroup$ – FearfulSymmetry Oct 24 '20 at 19:08
  • 1
    $\begingroup$ @Integrand. The word I created on purpose in the combination of both misery and mystery and not a typo. Cheers :-) $\endgroup$ – Claude Leibovici Oct 26 '20 at 9:29

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