Evaluating $\int_0^1 \frac{\sin(y)}{y\sqrt{1-y^2}}\,dy$ (a step of evaluating $\int_0^{\infty} \frac{\sin(\sin(x))}{x}\,dx$)

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.


I "guess" those integrations use somehow the generating functions.

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

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 $$S_6=\text{Si}(1)+\frac{7 (1097603873 \cos (1)-704763287 \sin (1))}{1024}$$