What is $\int_0^\infty \ln \left(1-\frac{\sin x}{x} \right) dx $? I'm looking for closed-form expressions for the integral $$I:=\int_0^\infty \ln \left(1-\frac{\sin x}{x} \right) dx .$$
Some related integrals that I've found include: $$J:=-\int_0^\infty \log(1-\cosh(x))\frac{x^2}{e^x}\,dx = 6 + 2\ln(2) - {2\pi^{2} \over 3} - 4\zeta(3) - 2\pi i,$$
and this one: $$K:= \int_0^{\frac{\pi}{2}}\log\Big{(}\cos(x)\Big{)}dx = - \frac{\pi}{2} \log(2) ,$$ and (p. 530) $$L:=\int_{0}^{\infty} \big{(} \ln(1-e^{-x}) \big{)}  dx = - \frac{\pi^{2}}{6} .$$
Moreover, I've encountered various integrals of the sinc function, including: $$M:= \int_{0}^{\pi/2} \ln \bigg{(} \frac{\sin(x)}{x} \bigg{)} dx = \frac{\pi}{2} \Big{(} 1 - \ln(\pi) \Big{)} .$$ However, I haven't found any closed-form expressions for $I$ yet, only an approximation: $$I \approx -5.75555891011162780816$$ and I'm not sure how to proceed with the integral.
 A: More precise computation of the given integral: $$I=-5.031379591902842520548271636746403412607399342991051\cdots$$
This is computed using PARI/GP. The integrand is oscillating, so intnum cannot handle it directly, and PARI/GP user's guide "suggests" expanding it into Fourier series. Instead, using $$\int_0^\infty f(x)\,dx+\sum_{n=1}^\infty\int_0^\pi\big(f(2n\pi-x)+f(2n\pi+x)\big)\,dx$$ for our integrand $f(x)=\log(1-\frac{\sin x}{x})$, we have $$f(2n\pi-x)+f(2n\pi-x)=\log\left[1-\left(\frac{x-\sin x}{2n\pi}\right)^2\right]-\log\left[1-\left(\frac{x}{2n\pi}\right)^2\right]$$ and, recognizing the infinite product for the sine function, we obtain $$I=\int_0^\pi\log\frac{\sin\big((x-\sin x)/2\big)}{\sin(x/2)}\,dx=\color{blue}{\int_0^\pi\log\left(2\sin\frac{x-\sin x}{2}\right)dx}.$$ I compute this as follows.
sinctail(x)=if(x>1e-3,(1-sinc(x))/x^2,suminf(n=0,(-x^2)^n/(2*n+3)!));
kernel(x)=log(sinctail(x))+log(sinc((x-sin(x))/2));
answer()=3*Pi*(log(Pi)-1)+intnum(x=0,Pi,kernel(x));

An amusing consequence: using Bessel functions, we have $\color{blue}{I=-\pi\sum_{n=1}^\infty J_n(n)/n}$.

Another line of thoughts, thanks to the observation by @Tyma Gaidash: $$y=y(\lambda,x)=2\sum_{n=1}^\infty J_n(n\lambda)\frac{\sin nx}{n}\qquad(|\lambda|\leqslant 1)$$ is the solution of $y=\lambda\sin(x+y)$. Writing $y(x):=y(1,x)$, we get $$-I=\int_0^\infty\frac{y(x)}{x}\,dx=\frac12\int_0^\pi y(x)\cot\frac x2\,dx$$ similarly to the above. Now let's get rid of the implicitly defined $y(x)$ by an inverse substitution. Namely, for $0<x<\pi/2-1$, $y(x)$ grows from $0$ to $1$, and we have $x=\arcsin y-y$; similarly, for $\pi/2-1<x<\pi$, $y(x)$ goes from $1$ to $0$, and we have $x=\pi-\arcsin y-y$. Thus
\begin{align*}
-I&=\frac12\int_0^1y\left(\frac1{\sqrt{1-y^2}}-1\right)\cot\frac{\arcsin y-y}2\,dy\\&+\frac12\int_0^1y\left(\frac1{\sqrt{1-y^2}}+1\right)\tan\frac{\arcsin y+y}2\,dy\\\implies I&=\color{blue}{\int_0^1\frac{(1-y^2-\cos y)\ y}{(y-\sin y)\sqrt{1-y^2}}\,dy}.
\end{align*}
This gives an alternative way(s) of computation of $I$.
A: Expand as follows
$$\int_0^\infty \ln \left(1-\frac{\sin x}{x} \right) dx 
=-\sum_{n\ge1}\frac1n \int_0^\infty \left( \frac{\sin x}{x} \right)^n dx  $$
where $\int_0^\infty \left( \frac{\sin x}{x} \right)^n dx = 
 \frac{\pi}{2^n (n-1)!}  
    \sum_{k=0}^{\lfloor n/2\rfloor} (-1)^k {n \choose k} (n-2k)^{n-1} $.
A: The final sum looks like a Kapteyn Series. Maybe it can be simplified or is the solution to a transcendental equation. From @metamorphy’s answer with Bessel J:
$$-\frac I\pi=\sum_{n=1}^\infty\frac{\text J_n(n)}n= \sum_{n=1}^\infty\frac{\text J_n(n)}n \cos\left(2\pi n\right)$$
Inverse Beta Regularized $\text I^{-1}_s(a,b)$ we have a partial closed form and a Kepler equation series solution:
$$x-\sin(x)=y\implies x=\boxed{\text{hav}^{-1}\left(\text I^{-1}_\frac y\pi\left(\frac32,\frac12\right)\right)=y+2\sum_{n=1}^\infty \frac{\text J_n(n)}n\sin(y n)}\tag{1}$$
which is shown here with the inverse haversine.
$$\sum_{n=1}^\infty\frac{\text J_n(n)}n \sin\left(\frac\pi2n\right)= \sum_{n=1}^\infty\frac{\text J_{4n-3}(4n-3)}{4n-3}-\sum_{n=1}^\infty\frac{\text J_{4n-1}(4n-1)}{4n-1} $$
$$\frac 2{\sqrt3}\sum_{n=1}^\infty \frac{\text J_n(n)}n \sin\left(\frac\pi3n\right)= \sum_{n=1}^\infty\frac{\text J_{6n-5}(6n-5)}{6n-5}+ \sum_{n=1}^\infty\frac{\text J_{6n-4}(6n-4)}{6n-4}-\sum_{n=1}^\infty\frac{\text J_{6n-2}(6n-2)}{6n-2}-\sum_{n=1}^\infty\frac{\text J_{6n-1}(6n-1)}{6n-1}$$
$$\vdots$$
All we need to do now is find something like $\sin(yn)=1,n\in\Bbb N$, but no value of $y$ satisfies this condition, so we must add multiple sines together or manipulate the sum. Then we remember the Fourier series:
$$\sum_{n=1}^\infty\frac{\text J_n(n)}n \cos\left(2\pi n\right)=\sum_{n=1}^\infty \sum_{m=1}^\infty a_m \frac{\text J_n(n)}n\sin\left(\frac{\pi m n }L\right)\mathop=^? \sum_{m=1}^\infty a_m \sum_{n=1}^\infty \frac{\text J_n(n)}n\sin\left(\frac{\pi m n }L\right),a_m= \frac 2L \int _0^L \cos(2\pi x)\sin\left(\frac{\pi m x}L\right)dx=\frac{2m((-1)^m\cos(2\pi L)-1)}{\pi(4L^2-m^2)}$$
where $a_m$ are series coefficients. If the series converges on $[-L,L]$, then we get a sum made of $\text{hav}^{-1}\left(\text I^{-1}_\frac mL\left(\frac32,\frac12\right)\right)$ terms using $(1)$
$$\text{hav}^{-1}\left(\text I^{-1}_\frac mL\left(\frac32,\frac12\right)\right),0\le \frac mL\le \pi$$
$$\text{hav}^{-1}\left(\text I^{-1}_{\frac m L-1}\left(\frac12,\frac32\right)\right)+\pi,\pi\le \frac mL\le 2\pi$$
$$\text{hav}^{-1}\left(\text I^{-1}_{\frac m L-2}\left(\frac32,\frac12\right)\right)+2\pi,2\pi\le \frac mL\le 3\pi$$
$$\text{hav}^{-1}\left(\text I^{-1}_{\frac m L-3}\left(\frac12,\frac32\right)\right)+3\pi,3\pi\le \frac mL\le 4\pi$$
$$\text{hav}^{-1}\left(\text I^{-1}_{\frac mL-4}\left(\frac32,\frac12\right)\right)+4\pi,4\pi\le \frac mL\le 5\pi$$
$$\vdots$$
Does this method work?
