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Prove that:
\[\displaystyle \int _{0}^{1}{\frac {\log^2{x}}{x^2-x\sqrt{2}+1}dx}=\frac{7\sqrt{2}}{128}\pi^3\,.\]

The integral can be re-written as: $$\int_{0}^{1}\frac{\ln^2 x}{x^2-\sqrt{2}x+1}\,dx=\frac{1}{2}\int_{0}^{\infty}\frac{\ln^2 x}{x^2-\sqrt{2}x+1}\,dx$$ Considering \( \displaystyle \oint _{\gamma}\frac{\ln^3 z}{z^2-\sqrt{2}z+1}\,dz \) whereas \( \gamma \) is a circle contour with a branch along the positive \( x \) axis we have that:

The function \( \displaystyle f(z)=\frac{\ln^3 z}{z^2-\sqrt{2}z+1} \) has poles at \( e^{i\pi/4 }, e^{7i\pi/4} \) and the residues at the poles are: $$\mathfrak{Res}\left ( f; e^{i\pi/4} \right )=-\frac{\pi^3}{64\sqrt{2}}, \;\; \mathfrak{Res}\left ( f; e^{7i\pi/4} \right )=\frac{343\pi^3}{64\sqrt{2}}$$ Expanding we get that: $$\oint_{\gamma}\frac{\ln^3 z}{z^2-\sqrt{2}z+1}\,dz=i\left ( -6\pi\int_{0}^{\infty}\frac{\ln^2 x}{x^2-\sqrt{2}x+1}\,dx+8\pi^3 \int_{0}^{\infty}\frac{dx}{x^2-\sqrt{2}x+1} \right )+12\pi^2 \int_{0}^{\infty}\frac{\ln x}{x^2-\sqrt{2}x+1}\,dx$$ But the contour integral is equal to \( \displaystyle \frac{171\pi^4\sqrt{2}}{32}i \) since \( \displaystyle \oint_{\gamma} \frac{\ln^3 z}{z^2-\sqrt{2}z+1}\,dz =2\pi i \sum residues \)

Setting the imaginary part of the integral expansion equal to this and solving for the integral in question (the squared one) we get: $$-6\pi \int_{0}^{\infty}\frac{\ln^2 x}{x^2-\sqrt{2}x+1}\,dx+8\pi^3 \int_{0}^{\infty}\frac{dx}{x^2-\sqrt{2}x+1}=\frac{171\pi^4 \sqrt{2}}{32}$$ The \( \displaystyle 8\pi^3\int_{0}^{\infty}\frac{dx}{x^2-\sqrt{2}x+1} \) integral is an old chestnut and we'll take it as a lemma. It equals to $$8\pi^3\int_{0}^{\infty}\frac{dx}{x^2-\sqrt{2}x+1}=\frac{12\pi^4}{\sqrt{2}}$$ Then the integral in question (after we solve for it and divide by two) equals to: $$\int_{0}^{1}\frac{\ln^2 x}{x^2-\sqrt{2}x+1}\,dx=\frac{7\pi^3 \sqrt{2}}{128}$$

galactus wrote:The \( \displaystyle 8\pi^3\int_{0}^{\infty}\frac{dx}{x^2-\sqrt{2}x+1} \) integral is an old chestnut and we'll take it as a lemma. It equals to $$8\pi^3\int_{0}^{\infty}\frac{dx}{x^2-\sqrt{2}x+1}=\frac{12\pi^4}{\sqrt{2}}$$

\(\displaystyle{\begin{aligned}\int_{0}^{\infty}\dfrac{1}{x^2-\sqrt{2}\,x+1}\,\mathrm{d}x&=\int_{0}^{\infty}\dfrac{1}{\displaystyle{\dfrac{1}{2}+\left(x-\dfrac{1}{\sqrt{2}} \right )^2}}\,\mathrm{d}x\\&=\int_{0}^{\infty}\dfrac{2}{1+\left(\sqrt{2}\,x-1 \right )^2}\,\mathrm{d}x\\&=\left[\sqrt{2}\,\arctan\,\left(\sqrt{2}\,x-1 \right ) \right ]_{0}^{\infty}\\&=\sqrt{2}\,\dfrac{\pi}{2}-\sqrt{2}\,\arctan\,(-1)\\&=\dfrac{\pi}{\sqrt{2}}+\dfrac{\pi\,\sqrt{2}}{4}\\&=\dfrac{6\,\pi}{4\,\sqrt{2}}\end{aligned}}\)

so :

\(\displaystyle{8\,\pi^3\,\int_{0}^{\infty}\dfrac{1}{x^2-\sqrt{2}\,x+1}\,\mathrm{d}x=8\,\pi^3\cdot \dfrac{6\,\pi}{4\,\sqrt{2}}=\dfrac{12\,\pi^4}{\sqrt{2}}}\) .