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i've tried to Evaluate $$\int_{0}^{\frac{\pi}{6}}x\ln^2(2\sin(x))dx$$ without using Contour integral first i changed $2\sin(x)$ into polar form ,and i got $$\int_{0}^{\frac{\pi}{6}}x\ln^2(2\sin(x))dx = \Re(\int_{0}^{\frac{\pi}{6}}x\ln^2(1-e^{2ix})dx)+\int_{0}^{\frac{\pi}{6}} x(x-\frac{\pi}{2})^2 dx $$ then to find $$\Re(\int_{0}^{\frac{\pi}{6}}x\ln^2(1-e^{2ix})dx)$$ So i used Harmonic Number identities $$\sum_{n=1}^{\infty} \frac{H_{n}x^{n+1}}{(n+1)} = \frac{\ln^2(1-x)}{2}$$ then substituted $x$ with $e^{2ix}$ and multiply x on both sides and take an integral $$\int_{0}^{\frac{\pi}{6}}\sum_{n=1}^{\infty} \frac{e^{2i(n+1)x}xH_n}{n+1} = \frac{1}{2}\int_{0}^{\frac{\pi}{6}}x\ln^2(1-e^{2ix})dx $$ and considered only real part $$ \Re(\frac{1}{2}\int_{0}^{\frac{\pi}{6}}x\ln^2(1-e^{2ix})dx) = \Re(\sum_{n=1}^{\infty} \frac{H_{n}}{n+1}\bigg[\frac{e^{\frac{i\pi(n+1)}{3}}(2i\pi(n+1))-6}{-24(n+1)^2} \bigg])$$ So i got $$\Re(\sum_{n=1}^{\infty} \frac{H_{n}}{n+1}\bigg[\frac{e^{\frac{i\pi(n+1)}{3}}(2i\pi(n+1))-6}{-24(n+1)^2} \bigg]) = \frac{1}{4}\bigg[\sum_{n=1}^{\infty}\frac{H_{n}\cos(\frac{\pi(n+1)}{3})}{(n+1)^3}-\sum_{n=1}^{\infty}\frac{H_{n}}{(n+1)^3}\bigg]+\sum_{n=1}^{\infty}\frac{H_{n}\sin(\frac{\pi(n+1)}{3})}{12(n+1)^2}$$ last but not least,So i change $\cos(\frac{\pi(n+1)}{3})$ into a polar form again then i've no idea How to evaluate it properly with out any huge expansion So i think i might need to use $$\sum_{n=1}^{\infty} \frac{{H_n}x^{n+1}}{(n+1)^3}$$ to complete the whole progress ,but i'm stuck at this point Could someone give me a hand please.

Unik Sillavich
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  • I didn't look at everything but I think you need to be careful with the complex logarithm, I get that ,$$2\sin(x)=\frac{e^{ix}-e^{-ix}}{i}=\frac{e^{2ix}-1}{ie^{ix}}=\frac{1-e^{2ix}}{-ie^{ix}}=\frac{1-e^{2ix}}{e^{ix+\frac{3\pi}{2}+2n\pi}}\rightarrow\ln(2\sin(x))=\ln(1-e^{2ix})-\ln(e^{ix+\frac{3\pi}{2}+2n\pi})=\ln(2\sin(x))=\ln(1-e^{2ix})-ix+\frac{3\pi}{2}+2n\pi$$ Also I think you have to be consistent with your variables, if you replace $x$ with $e^{ix}$ then you should do it everywhere doesn't seems fine to write $x^{n+1}=xe^{nix}$. Maybe someone can answer how to calculate this integral. – Dabed Jul 06 '19 at 17:55
  • Here there is a similar integral $I=\int_{0}^{\pi/4}x\ln^{2}(\sin(x))dx$ – Dabed Jul 06 '19 at 18:01
  • Oh,thanks you so much i finally did it yesterday,here https://www.facebook.com/photo.php?fbid=547414695791491&set=pb.100015688965158.-2207520000.1562596979.&type=3&theater i posted it on facebook – Unik Sillavich Jul 08 '19 at 14:45
  • I'm bad at integrating so I suppose what was more spot on was my second comment with the link but glad it helped anyway, I didn't went trough what you did in the photo but if your results agree with the numerical value you should definitely post your solution and accept it. – Dabed Jul 08 '19 at 22:59

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