How do I compute the indicator function of an entire function? Let $F(z)$ be an entire function of finite exponential type. The indicator function of $F$ is defined as $$h_F(\theta)={\overline{\displaystyle\lim}_{r\rightarrow\infty}}\frac{\ln|F(re^{i\theta})|}{r}\;,~~\theta\in\mathbb{R}\;.$$ Suppose that $F$ is given by $$F(z)=(a_3z^3+a_2z^2+a_1z+a_0)e^{z\tau}+(b_2z^2+b_1z+b_0)\;,$$ where $\tau>0$ and the coefficients $a_j,b_j\in\mathbb{R}$. I know what $h_F(\theta)$ is in this case. What I am looking for is a step-by-step explanation of how one gets $h_F(\theta)$. I'm not interested in the answer, but rather how to get the answer. This type of calculation does involve asymptotic analysis, so please explain this part well.
 A: It's a little messy, because many variables are playing. If you look at it carefully though, there are only three important variables: $\tau$, $\rho$ and $\theta$.
In this case, an "asymptotic analysis" is just the analysis of the limit for various values of the variables noted above.
First you need to restrict your variables, so wlog assume $\tau>0$ and $-\pi\lt \theta\lt\pi$. The last restriction specifies the principal branch of the complex logarithm, so we can work without the $|.|$.
Now define:
$\begin{align}
h_F^*(\theta)&=\lim_{\rho\to +\infty}\frac{\ln F(\rho\cdot e^{i\theta})}{\rho}\\
&=\lim_{\rho\to+\infty}\ln F\left(\rho\cdot e^{i\theta}\right)^{\frac{1}{\rho}}\\
\end{align}
$
Now we set:
$$R(\rho,\theta)=F\left(\rho\cdot e^{i\theta}\right)^{\frac{1}{\rho}}$$
and look at:
$\begin{align}
\lim_{\rho\to+\infty}R(\rho,\theta)=
\lim_{\rho\to+\infty}e^{\ln F\left(\rho\cdot e^{i\theta}\right)^{\frac{1}{\rho}}}=\lim_{\rho\to+\infty}F\left(\rho\cdot e^{i\theta}\right)^{\frac{1}{\rho}}
\end{align}$
We now consider cases, based on the phase angles $\theta$:


*

*$\theta=0$: In this case,
$$R(\rho,0)=((\alpha_3 \rho^3+\alpha_2 \rho^2+\alpha_1 \rho+\alpha_0)e^{\tau\rho}+\beta_2\rho^2+\beta_1\rho+\beta_0)^{\frac{1}{\rho}}$$


Using calculus and assuming $\tau>0$, the above has limit:
$$\lim_{\rho\to+\infty}R(\rho,0)=e^\tau\Rightarrow$$
$$h^*_F(0)=\tau$$


*

*$\theta=\frac{\pi}{2}$: In this case (modulo typos),
$$R(\rho,\pi/2)=((-\alpha_3 i \rho^3-\alpha_2 \rho^2+\alpha_1 i \rho-\alpha_0)e^{i\tau\rho}-\beta_2\rho^2+\beta_1 i\rho-\beta_0)^{\frac{1}{\rho}}$$


Again using calculus and assuming $\tau>0$, the above has limit:
$$\lim_{\rho\to+\infty}R(\rho,\pi/2)=1=e^{\tau\cdot\cos\left(\frac{\pi}{2}\right)}\Rightarrow$$
$$h^*_F(\pi/2)=0$$
Repeating the process and omitting the details of the limits, we find for the other phase angles,


*

*$\theta=\frac{\pi}{n}$, $n\in \mathbb{N}$:


$$\lim_{\rho\to+\infty}R(\rho,\pi/n)=e^{\tau\cdot\cos\left(\frac{\pi}{n}\right)}\Rightarrow$$
$$h^*_F(\pi/n)=\tau\cdot \cos\left(\frac{\pi}{n}\right)$$
The above, with the restriction $\tau>0$. The assumption $\tau\le 0$ warrants an additional analysis, because the exponent of the exponential becomes negative.
