= Solution
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 $\varepsilon>0$, first let $\delta\to0$. The upper bank is traversed from $\varepsilon$ to one and the lower bank from one to $\varepsilon$, so their limiting sum is
$$
\int_\varepsilon^1f(x)(\log x-i\pi)\,dx+\int_1^\varepsilon f(x)(\log x+i\pi)\,dx=-2\pi i\int_\varepsilon^1f(x)\,dx.
$$
The 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, $|L-i\pi|\leq|\log\varepsilon|+\pi$ and $f$ is bounded near zero. Its integral is therefore $O(\varepsilon|\log\varepsilon|)$ and tends to zero. The sum of all four contour pieces is zero. Letting $\varepsilon\to0$ gives
$$
\boxed{\int_0^1f(x)\,dx=\frac1{2\pi i}\int_\phi f(z)(L(z)-i\pi)\,dz.}
$$
The 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>.
\Image[/past-exam-of-the-metric-spaces-and-complex-analysis-course-of-the-university-of-oxford/2017-logarithmic-keyhole.png]
{title=The logarithmic jump on a keyhole contour recovers a real-segment integral}
{description=The upper and lower banks have opposite traversal directions and logarithmic boundary values differing by $2\pi i$. The small circle is traversed clockwise. The slit width is illustrative; its limit is taken before the inner radius tends to zero.}
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