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I integrate by part I assume $dx=dv$ and $\ln(\sin x)= u$ or I compute by Maple but its answer wasn't clearly (Maple answer: xln(1-exp((2I)x))+xln(sin(x))+(1/2*I)x^2+(1/2*I)*polylog(2, e((2*I)*x))) thanks for any hints

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Your formatting has gone astray. Can you please correct? Regards –  Amzoti Jan 16 '13 at 20:05
    
Do you have bounds? –  NeverBeenHere Jan 16 '13 at 20:24
    
I pasted Integral(log(sin(x)),x) into wolframalpha.com to get a result. The solution contained something called polylogarithm. –  miracle173 Jan 16 '13 at 20:34
    
whats polylogarithm ? how wolframalpha.com compute it? –  Maisam Hedyelloo Jan 16 '13 at 20:41
    
wolframalpha.com –  miracle173 Jan 16 '13 at 20:48

1 Answer 1

$$ \text{Re}\log(1-e^{2ix}) = \text{Re}\log(1-\cos 2x - i \sin ) = \log\Big( (1- \cos 2x)^{2}+ \sin^2{2x} \Big)^{1/2}$$

$$ = \frac{1}{2} \log(2 - 2 \cos 2x) = \frac{1}{2} \log(4 \sin^{2} x) = \log(2 \sin x) = \log(2) + \log(\sin x)$$

So

$$ \int_{0}^{z} \log(\sin) \ dx = \text{Re}\int_{0}^{z} \log(1-e^{2ix}) \ dx - \int_{0}^{z}\log 2 \ dx$$

$$ = -\text{Re} \int_{0}^{z}\sum_{n=1}^{\infty} \frac{e^{2inx}}{n} \ dx - z\log 2$$

$$=- \text{Re} \sum_{n=1}^{\infty} \frac{1}{n} \int_{0}^{z} e^{2inx} \ dx - z \log(2) = - \text{Re} \sum_{n=1}^{\infty} \frac{1}{2in^{2}} \Big(e^{2inz}-1 \Big) - z \log 2$$

$$ =\text{Re} \frac{i}{2}\sum_{n=1}^{\infty} \frac{1}{n^{2}}\Big(\cos 2nx + i \sin 2nx -1 \Big) - z \log 2 $$

$$=- \frac{1}{2}\sum_{n=1}^{\infty} \frac{\sin 2nz}{n^{2}}- z \log 2 = - \frac{1}{2} \text{Im} \ \text{Li}_{2} (e^{2iz}) - z \log 2$$

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