There is a very short and elementary way to calculate this integral I. First write sin[x] as 2*sin[x / 2]*sin[x / 2+pi/2] Then log[sin[x]= log[2]+log[sin[x] /2] +log[sin[x/2+pi/2] . The substitutions t= x/2 and t=x/2+pi/2] immediately lead to the equation I+ log[2]*pi = 2*I ,i.e. I = - pi*log[2] .
6:10 it depends on the branch cut of the logarithm, I'll pick it to be the negative real axis, then sin(x+iy)=sin(x)cosh(y)+icos(x)sinh(x), the imaginary part has to be equal to 0, and the real part has to be less than 0, so sin(x)
30:24 use the taylor expansion of e^x, then you have x ln( x(2ie^itheta+(2ie^itheta)^2 x/2! +...) = xlnx+xln(2ie^itheta+....), the limit as x goes to 0 is then 0.
If you decompose log(sin(x))=iz-log(2)-i*Pi/2+log(1-exp(-2i*x)) and integrate from 0 to Pi, then Int_0^Pi log(sin(x)) dx = -Pi*log(2) + Int_0^Pi log(1-exp(-2i*x)) dx. You can then use complex analysis to show that the last integral is 0. Substitute z=exp(-2i*x), then this integral becomes Int_{|z|=1} log(1-z) dz/(-2i*z) where the contour is going clockwise. log(1-z)/z has a branch-cut on (1,infinity) and is holomorphic inside the disk of radius 1 about z=0. Hence this integral is zero by Cauchy. You may want to check the integral over the semi-circle of radius epsilon about z=1, but it is not difficult to see that it is zero as epsilon -> 0. Btw: What do you mean by standard solution? I once saw a solution using integration by parts twice (using the symmetry about Pi/2) and you end up with Int_0^{Pi/2} (x/sin(x))^2 dx and then there was a trick, but I don't remember.
I know the solution you're thinking of. It uses integration by parts once and then Feynman's trick after substituting y = tan(x). So you were right about there being a trick! It's a nice alternative to the normal solution that exploits symmetry within the integrand.
This is amazing! Made me sub as primitive functions of logarithms of trigonometric functions are currently in my scope. As much as I enjoy a purely real proof... :)
