(H. Priestley complex Analysis Chapter 7 Exercise 9)

Suppose $f$ is holomorphic inside and on $\gamma(0,1)$. By integration around the usual keyhole like this one : Integration of $\ln $ around a keyhole contour , prove that:

$$\int_0^1 f(x)\,dx=\frac{1}{2i \pi} \int_{\gamma(0,1)} f(z) (\ln(z) - i \pi)\, dz$$

where $\ln(z)$ denotes the branch of the logarithm with imaginary part between $[0,2\pi]$

I am not sure how to tackle this question :(

the branch of the $\ln$ will be $\ln(z) = \ln (|z|) + i \theta$

I thought that maybe if I integrate $f(z) (\ln(z) - i \pi)$ along the entire contour I will get $0$ by Cauchy's theorem as it is holomorphic inside by design:

$$ \int_{\gamma(0,1)} f(z) (\ln(z) - i \pi)\, dz + \int_{\text{top edge}} f(z) (\ln(z) - i \pi)\, dz + \int_{\text{bottom edge}} f(z) (\ln(z) - i \pi)\, dz - \int_{\gamma(0,\epsilon)} f(z) (\ln(z) - i \pi)\, dz = 0,$$


$$ \int_{\gamma(0,1)} f(z) (\ln(z) - i \pi) i\, dz + \int_{0}^{1} f(x) (\ln(x) - i \pi)\, dx + \int_{1}^{0} f(x) (\ln(x)+2i \pi - i \pi)\, dx - \int_{\gamma(0,\epsilon)} f(\epsilon e^{i\theta}) (\ln(\epsilon) +i\theta - i \pi)\, dz = 0.$$

That is, with a fortunate cancellation:

$$ \int_{\gamma(0,1)} f(z) (\ln(z) - i \pi)\, dz -2i \pi \int_{0}^{1} f(x)\, dx - \int_{\gamma(0,\epsilon)} f(\epsilon e^{i\theta}) (\ln(\epsilon) +i\theta - i \pi) \epsilon i e^{i\theta}\, d\theta = 0.$$

however I'm not sure about what happens with the very last integral, if the integrand were holomorphic apart from a simple pole then it would just be $2\pi \times\text{residue}$, i.e. $0$ since $f$ is holomorphic inside but with the $\ln()$ it makes it complicated...


You have tackled it correctly, all that is missing is that you see

$$\lim_{\epsilon \to 0} \int_{\gamma(0,\epsilon)} f(z)(\ln z - i\pi)\,dz = 0.$$

The standard estimate

$$\biggl\lvert \int_\Gamma g(z)\,dz\biggr\rvert \leqslant L(\Gamma)\cdot \sup \{ \lvert g(\zeta)\rvert : \zeta \in \operatorname{Tr}(\Gamma)\}$$

yields that, since $f$ is bounded, and

$$\lim_{\epsilon \to 0} \epsilon\cdot \lvert\ln\epsilon\rvert = 0.$$

The length of $\gamma(0,\epsilon)$ is bounded by $2\pi\epsilon$, and the integrand is bounded by $\lVert f\rVert_\infty(\pi + \lvert\ln\epsilon\rvert)$.

  • $\begingroup$ Of course ! Many thanks $\endgroup$ – user3203476 Jul 5 '15 at 13:57

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