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I need to show by contour integration that $$\int_0^\infty \frac{(\log x)^2}{x^2+x+1} dx = \frac{16}{81}\sqrt 3 \pi^2$$

I usually approach a contour integral involving $\log x$ by considering a semi-circular contour indented at $0$. But here the other straight arm doesn't simplify due to odd $x$ in denominator.

Can somebody help me how to proceed?

utkarshk5
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    Check this: https://math.stackexchange.com/a/294566/42969 – Martin R Apr 10 '21 at 11:05
  • @Martin R, that's precisely what I needed. Thank you so much. – utkarshk5 Apr 10 '21 at 11:23
  • Using integration in the complex plane ("keyhole") a general expression for $I(k)=\int_0^\infty \frac{(\log x)^k}{x^2+x+1} dx$ can be easily obtained: $I(k)=\frac{2\pi^{k+1}}{\sqrt3}\frac{d^k}{ds^k}\Bigl(\frac{\sin\frac{s}{3}}{\sin s}\Bigr)|_{s=0}$. It gives $I(0)=\frac{2\pi}{\sqrt3}$, $I(1)=0$, $I(2)=\frac{16\pi^3}{81\sqrt3 }$, etc. – Svyatoslav Apr 10 '21 at 13:52

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