What is $\int_{0}^1\ln(\sin(x)) dx$? [duplicate]

Let me introduce the polylogarithm$$\operatorname{Li}_s(z):=\sum_{n\ge1}z^nn^{-s}$$(technically we analytically continue the RHS, but I don't think you want that kind of complex-analytic pedantry right now). In particular, $\frac{d}{dz}\operatorname{Li}_2(z)=-z^{-1}\ln(1-z)$ so$$\frac{d}{dx}\operatorname{Li}_2(e^{2ix})=-2i\ln(1-e^{2ix})=-2i\ln\sin x+2x-\color{blue}{2i\ln(-2i)}.$$Applying $\frac{-1}{2i}\int dx$,$$\int\ln\sin xdx=\frac{i}{2}\left(\operatorname{Li}_2(e^{2ix})-x^2\right)+\color{blue}{2i\ln(-2i)}x+C.$$I'll leave you to ponder what the blue coefficient should be, i.e. which branch of the complex logarithm it needs.

As @jjagmath noted, it's highly unlikely the $\int_0^1$ result has a closed form, except for what the above indefinite integral implies. Wolfram Alpha can do no better, although at least it will help you test your thoughts about the blue value. The result is in line with @MariuszIwaniuk comment.

Edit: we can prove $\int_0^1(-\ln\sin x)dx<1-\sin1$ or equivalently $\int_0^1\left(-\ln\frac{\sin x}{x}\right)dx<-\sin1$ by noting the latter integrand increases from $0$ at $x=0$ to $-\ln\sin1$ at $x=1$, so the latter integral has upper bound $\int_0^1(-\ln\sin1)dx=-\ln\sin1$.