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.

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 $I$.
