Centred logarithmic keyhole identity (${\int }_0^{1}f(x)dx=\frac{1}{2\pi i}{\oint }_{|z|=1}f(z)(L(z)-i\pi )dz$)

For an entire function $f$, use the complex logarithm $L$ with argument in $(0,2\pi )$. The unit-circle integral in the title uses the branch's one-sided boundary values at its crossing with the cut. On a keyhole contour, the outward upper bank has $L=\log x$, and the inward lower bank has $L=\log x+2\pi i$. Their combined integral of $f(L-i\pi )$ is $-2\pi i{\int }_{\epsilon }^{1}f(x)dx$. The inner circle contributes $O(\epsilon |\log \epsilon |)$, which vanishes. The Cauchy integral theorem on the slit plane therefore gives the identity.

On $z=e^{i\theta }$, $|L-i\pi |=|\theta -\pi |\le \pi$. The estimation lemma consequently givesSubtracting the middle branch value $i\pi$ centres the weight and produces this bound. The contour crosses the cut only in the limiting boundary-value interpretation, so it is not a closed contour inside the logarithm's domain.