Confusing sign error in proof of Jensen's formula

Question

I'm trying to prove Jensen's formula relating the number of zeros of an entire function to its logarithmic averages on circles. A step in the proof is to show that $$\frac{1}{2\pi} \int_0^{2\pi} \log \vert Re^{i\theta} - \alpha \vert d\theta = \log R$$ when $\alpha \in \mathbb{C}^{\times}$ and $R > \vert \alpha \vert$. Since $\log \vert e^{-i\theta} \vert = 0$, one can rewrite this as $$\frac{1}{2\pi} \int_0^{2\pi} \log \vert R - \alpha e^{-i\theta} \vert d\theta.$$ Here's what I am confused about. We are integrating in the clockwise direction over the unit circle, so this should be equal to minus the integral over the circle in the counterclockwise direction, i.e. $$-\frac{1}{2\pi} \int_0^{2\pi} \log \vert R - \alpha e^{i\theta} \vert d\theta.$$ But the above is equal to $-\log R$.

Answer 1

Reversing the direction of integration along the unit circle is done via the substitution $\theta = 2\pi - \phi$: $$ \int_0^{2 \pi} f(e^{i\theta}) \, d\theta = \int_{2 \pi}^0 f(e^{-i\phi}) \, (-1)d\phi = \int_0^{2 \pi}f(e^{-i\phi}) \, d\phi \, . $$ Both the substitution and the swapping of the integral bounds introduce a factor $(-1)$. In your case: $$ \frac{1}{2\pi} \int_0^{2\pi} \log \vert R - \alpha e^{-i\theta} \vert d\theta = \frac{1}{2\pi} \int_0^{2\pi} \log \vert R - \alpha e^{i\theta} \vert d\theta = \log R \, . $$