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Apply the Cauchy integral theorem to $F(z)=f(z)(L(z)-i\pi )$ on the slit plane, which is star-shaped about $-1$. For fixed $\epsilon >0$, first let $\delta \to 0$. The upper bank is traversed from $\epsilon$ to one and the lower bank from one to $\epsilon$, so their limiting sum isThe outer arc tends to the positively oriented unit-circle integral, interpreting the logarithm by its one-sided boundary values at the cut. On the inner arc, $|L-i\pi |\le |\log \epsilon |+\pi$ and $f$ is bounded near zero. Its integral is therefore $O(\epsilon |\log \epsilon |)$ and tends to zero. The sum of all four contour pieces is zero. Letting $\epsilon \to 0$ givesThe integral around the full circle is a piecewise integral of the branch's boundary values; $L$ is not a single holomorphic function across the cut-crossing point. This is the centred logarithmic keyhole identity.

For real $a_r$, put $h(z)={\sum }_{r=1}^n{a}_r{z}^r$. The quadratic form equals ${\int }_0^{1}h(x{)}^{2}dx$, proving its nonnegativity. Apply the centred logarithmic keyhole identity bound to the entire function $h^2$. Since $|h(e^{i\theta }{)}^2|=|h(e^{i\theta }){|}^2$, orthogonality of the finite Fourier sum gives ${\int }_0^{2\pi }|h(e^{i\theta }){|}^{2}d\theta =2\pi {\sum }_r{a}_r^2$. Multiplication by the contour estimate's factor $1/2$ proves the upper bound. The matrix $(1/(r+s+1))$ is the principal submatrix of a Hilbert matrix obtained by deleting its first row and column.