For an entire function $$f(z)=\sum_{n=0}^\infty c_n z^n \, ,$$ we simply have that $$\frac{1}{2\pi} \int_0^{2\pi} f(Re^{it}) \, {\rm d}t = c_0$$ $\forall R>0$. The same is true if $R\rightarrow 0$, but we only consider a finite arc $\theta<2\pi$. Now I'm wondering if anything can be said about $$\lim_{R\rightarrow \infty} \int_0^\pi f(Re^{it}) \, {\rm d}t = \pi c_0 + 2\pi i \lim_{R\rightarrow \infty}\, \sum_{\substack{n=1 \\ n\text{ odd}}}^\infty \frac{c_n R^n}{n} $$ so for half a circle instead.
In particular I'm interested in the case for which this expression is equal to $0$ for $c_0=0$. Can you come up with a function $\neq 0$ for which this is the case? For example $$f(z) = \frac{e^z-1}{\log z}, \frac{e^z-1}{\sqrt{z}},...$$ would satisfy this property, but these, unfortunately, are not entire.
If instead I approach this from the RHS, i.e. $$\lim_{R\rightarrow \infty} \sum_{n=0}^\infty \frac{c_{2n+1}}{2n+1} \, R^{2n+1} = 0 \, ,$$ then $$c_{2n+1}=\frac{(-1)^n}{n!} (2n+1)$$ would give $$\sum_{n=0}^\infty \frac{c_{2n+1}}{2n+1} \, R^{2n+1} = Re^{-R^2} \, .$$