be the time of transmitting a pulse with eigenvalue $\lambda_k$. Let $r$ be the unique real positive root of the polynomial $\sum_{k=1}^Mx^{-T(\lambda_k)} - 1$. Then, in the noiseless case, the capacity is obtained as $${\mathsf{C}}= \log_2(r)$$ and the capacity-achieving distribution is given by $$\begin{aligned}
 P_X^\diamond(\lambda_k) &= {\operatorname{e}^{-\ln(r){T{\mathopen{}\mathclose\bgroup\originalleft}(\lambda_k{\aftergroup\egroup\originalright})}}}, & k&=1,\ldots,M.
\label{eqn:opt_dist_noiseless}\end{aligned}$$

Suppose that the $k$-th eigenvalue is transmitted with probability $P_k$. For any fixed average symbol interval $\bar{T}(X) = \sum_kP_kT(\lambda_k)$, where ${T{\mathopen{}\mathclose\bgroup\originalleft}(\lambda_M{\aftergroup\egroup\originalright})} \leq \bar{T}(X) \leq {T{\mathopen{}\mathclose\bgroup\originalleft}(\lambda_1{\aftergroup\egroup\originalright})}$, we are interested in the distribution that maximizes the entropy while leading to the average symbol duration $\bar{T}(X)$. It is known that this distribution takes the form [@cover_inf_theory Ch. 12] $$\begin{aligned}
    P_k &= \frac{{\operatorname{e}^{-\theta {T{\mathopen{}\mathclose\bgroup\originalleft}(\lambda_k{\aftergroup\egroup\originalright})}}}}{\xi(\theta)}
    \label{eqn:maximizer}\end{aligned}$$ where $\xi(\theta) = \sum_i{\operatorname{e}^{-\theta T(\lambda_i)}}$ ensures that $\sum_k P_k = 1$ and $\theta$ has to be selected such that $\sum_kP_kT(\lambda_k) = \bar{T}$. In the noiseless case, the MI is given by ${\mathbb{I}(X;Y)} = {\mathbb{H}(X)}$. The entropy ${\mathbb{H}(X)}$ then is $$\begin{aligned}
    {\mathbb{H}(X)} =: {\mathbb{H}(\theta)} &= -\sum_{k=1}^M P_k \log_2{\mathopen{}\mathclose\bgroup\originalleft}(\frac{{\operatorname{e}^{-\theta {T{\mathopen{}\mathclose\bgroup\originalleft}(\lambda_k{\aftergroup\egroup\originalright})}}}}{\xi(\theta)}{\aftergroup\egroup\originalright})\\
                                     &= \frac{1}{{\ln(2)}}\sum_{k=1}^M P_k (\theta {T{\mathopen{}\mathclose\bgroup\originalleft}(\lambda_k{\aftergroup\egroup\originalright})} + \ln(\xi(\theta)))\\
                                     &= \frac{\theta \bar{T}(X)}{{\ln(2)}} + \log_2(\xi(\theta)).\end{aligned}$$ The time-scaled MI hence takes the form $$\begin{aligned}
 {\mathsf{I}(X;Y)} &=  \frac{\theta}{\ln(2)} + \frac{\log_2(\xi(\theta))}{\bar{T}(X)} \\
 &= \frac{\theta}{\ln(2)} + \frac{\log_2(\xi(\theta))}{\sum_kP_kT(\lambda_k)} \\
 &= \frac{\theta}{\ln(2)} + \frac{\log_2(\sum_k{\operatorname{e}^{-\theta T(\lambda_k)}})\sum_k{\operatorname{e}^{-\theta T(\lambda_k)}}}{\sum_k{\operatorname{e}^{-\theta T(\lambda_k)}}T(\lambda_k)}\,. \end{aligned}$$ In order 