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Inspired by a question I saw these days, I try to calculate in closed form

$$\int_0^{\pi/2}\frac{\sin(x)\log{\sin{(x)}}}{x}\,dx$$

So far no fruitful idea that is worth sharing. What way would you propose? Note I prefer ways suggested, not necessarily solutions, but I have nothing against any of the options you prefer.

user 1591719
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    What way would you propose? - In accordance with the ancient Chinese wisdom of the Tao Te Ching, I would propose the way of non-action. :-$)~\dfrac{\sin x}x$ doesn't make any sense on $\bigg(0,\dfrac\pi2\bigg)$; it only makes sense on $(0,\infty)$. So, at best, you might want to try $\displaystyle\int_0^\infty\frac{\sin(x)}x\cdot\ln\big(\sin^2x\big)~dx$. – Lucian Mar 03 '15 at 23:16
  • @Lucian Chinese wisdom might be right to a certain extent, but not sure if I wanted to try the integral you proposed. – user 1591719 Mar 03 '15 at 23:28
  • After working on this for a while, I only managed to get $$\int^\frac{\pi}{2}_0\frac{\sin{x}\ln(\sin{x})}{x}dx=\frac{1}{2}\int^1_0\frac{(x+1)\ln{x}\ln(1+x)}{x^{3/2}\left(\ln^2{x}+\pi^2\right)}dx-\mathrm{Si}\left(\frac{\pi}{2}\right)\ln{2}$$ – M.N.C.E. Apr 03 '15 at 13:50

1 Answers1

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It's not a closed form, but I hope can be useful. Using $$\log\left(\sin\left(x\right)\right)=-\log\left(2\right)-\sum_{n\geq1}\frac{\cos\left(2nx\right)}{n}$$ we have$$\int_{0}^{\pi/2}\frac{\sin\left(x\right)\log\left(\sin\left(x\right)\right)}{x}=-\log\left(2\right)\textrm{Si}\left(\frac{\pi}{2}\right)-\sum_{n\geq1}\frac{1}{n}\int_{0}^{\pi/2}\frac{\sin\left(x\right)\cos\left(2nx\right)}{x}dx.$$ Now we use the identity $$\sin\left(x\right)\cos\left(2nx\right)=\frac{1}{2}\left(\sin\left(x-2nx\right)+\sin\left(2nx+x\right)\right)$$ to obtain $$\int_{0}^{\pi/2}\frac{\sin\left(x\right)\log\left(\sin\left(x\right)\right)}{x}=-\log\left(2\right)\textrm{Si}\left(\frac{\pi}{2}\right)+\frac{1}{2}\sum_{n\geq1}\frac{\textrm{Si}\left(\frac{\pi}{2}\left(2n+1\right)\right)-\textrm{Si}\left(\frac{\pi}{2}\left(2n-1\right)\right)}{n}.$$ Now, we have that $$\frac{\textrm{Si}\left(\frac{\pi}{2}\left(2n+1\right)\right)-\textrm{Si}\left(\frac{\pi}{2}\left(2n-1\right)\right)}{n}=O\left(\frac{1}{\pi n^{2}}\right)$$ at $n\rightarrow\infty$ so we have the approximation $$\int_{0}^{\pi/2}\frac{\sin\left(x\right)\log\left(\sin\left(x\right)\right)}{x}\simeq-\log\left(2\right)\textrm{Si}\left(\frac{\pi}{2}\right)-\frac{\zeta\left(2\right)}{2\pi}.$$ Note that numerically the integral is $-1.05585$ and my result is $-1.21193...$

Marco Cantarini
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