= Centred logarithmic keyhole identity
{title2=$\int_0^1f(x)dx=\frac1{2\pi i}\oint_{|z|=1}f(z)(L(z)-i\pi)dz$}
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 $-2\pi i\int_\varepsilon^1f(x)dx$. The inner circle contributes $O(\varepsilon|\log\varepsilon|)$, which vanishes. The <Cauchy integral theorem> on the slit plane therefore gives the identity.
On $z=e^{i\theta}$, $|L-i\pi|=|\theta-\pi|\leq\pi$. The <estimation lemma> consequently gives
$$
\left|\int_0^1f(x)dx\right|\leq\frac1{2\pi}\int_0^{2\pi}|f(e^{i\theta})||\theta-\pi|d\theta\leq\frac12\int_0^{2\pi}|f(e^{i\theta})|d\theta.
$$
Subtracting 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.
\Image[/past-exam-of-the-metric-spaces-and-complex-analysis-course-of-the-university-of-oxford/2017-logarithmic-keyhole.png]
{title=Opposite bank orientations turn a logarithmic jump into a real-segment integral}
Back to article page
