I found some scholars clarified that the relationship between the Shannon transform and Stieltjes transform with
\begin{equation}
\frac{\gamma}{\log e}\frac{d}{d\gamma}\mathcal{V}_N(\gamma) = 1 - \frac{1}{\gamma}\mathcal{S}_N(-\frac{1}{\gamma})
\end{equation}
However, other scholars claimed that this equality was modified as
\begin{equation}
\mathcal{V}_N(x) = \frac{1}{N}\log \det (\mathbf{I}_N + \frac{1}{x}\mathbf{B}_N) = \int_{0}^{+\infty}\log (1+\frac{\lambda}{x})dF_N(\lambda) = \int_{x}^{+\infty}(\frac{1}{\omega}-\mathcal{S}_N(-\omega))d\omega
\end{equation}
Here $\mathcal{S}_X(.)$ denotes the stieltjes transform, whereas $\mathcal{V}_X(.)$ is the shannon transform. So how can I obtain the last equation from the first equation?
***I mean, given that***
\begin{equation}
\frac{\gamma}{\log e}\frac{d}{d\gamma}\mathcal{V}_N(\gamma) = 1 - \frac{1}{\gamma}\mathcal{S}_N(-\frac{1}{\gamma})
\end{equation}
***How can I obtain this***
\begin{equation}
\mathcal{V}_N(x) = \int_{x}^{+\infty}(\frac{1}{\omega}-\mathcal{S}_N(-\omega))d\omega
\end{equation}
I did some calculations, but not sure that they are equal.
Try this PDF on Stieljest and related transform. If people are interested in pusuing this further, please let me know, [email protected]. I find yahoo easier to handle.
Yang