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.

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} $.
