The pdf of the Normal distribution is
$$ f(x) = \frac{1}{\sigma \sqrt{2 \pi}} e^{-\frac{1}{2}\left(\frac{z-\mu}{\sigma}\right)^{2}} $$
The pdf of the Johnson-SU distribution is $$ f(x) = \frac{\delta}{\lambda \sqrt{2 \pi}} \frac{1}{\sqrt{1+\left(\frac{x-\xi}{\lambda}\right)^{2}}} e^{-\frac{1}{2}\left(\gamma+\delta \sinh ^{-1}\left(\frac{x-\xi}{\lambda}\right)\right)^{2}} $$
The differential entropy of the Normal pdf is derived as \begin{align}\label{equation:hN} h(X) = & -\int_{-\infty}^{\infty} f(x) \ln f(x) \mathrm{d} x\\ =&-\int_{-\infty}^{\infty} \left(2\pi \sigma^2\right)^{-\frac{1}{2}} e^{-(x-\mu)^2 / 2\sigma^2} \ln\left[ \left(2\pi \sigma^2\right)^{-\frac{1}{2}} e^{-(x-\mu)^2 / 2\sigma^2} \right] \mathrm{d} x\\ = &\frac{1}{2} \ln(2\pi \sigma^2) \int_{-\infty}^{\infty} (2\pi \sigma^2)^{-\frac{1}{2}} e^{-(x-\mu)^2 / 2\sigma^2} \mathrm{d} x\\ &+ \frac{1}{2\sigma^2} \left(2\pi \sigma^2\right)^{-\frac{1}{2}} (x-\mu)^2 e^{-(x-\mu)^2 / 2\sigma^2} \mathrm{d} x\\ =& \frac{1}{2} \ln (2\pi\sigma^2) + \frac{1}{2}\\ =& \frac12\ln(2\pi\sigma^2) + \frac12\ln e\\ =& \frac12\left(\ln(2\pi\sigma^2) + \ln e\right)\\ =& \frac{1}{2} \ln (2\pi e \sigma^2) \end{align}
What is the derivation of the Johnson-SU pdf's differential entropy?
First Attempt
\begin{align} h(X) = &-\int_{-\infty}^{\infty} f(x) \ln f(x) \mathrm{d} x\\ =& \mathbb{E}[ -\ln f(x)] \\ =& \mathbb{E} \left[ -\ln \left( \frac{\delta}{\lambda \sqrt{2 \pi}} \frac{1}{\sqrt{1+\left(\frac{x-\xi}{\lambda}\right)^{2}}} e^{-\frac{1}{2}\left(\gamma+\delta \sinh ^{-1}\left(\frac{x-\xi}{\lambda}\right)\right)^{2}} \right) \right] \\ =& \mathbb{E} \left[ -\ln \left( \frac{\delta}{\lambda \sqrt{2 \pi}} \right) -\ln \left(\frac{1}{\sqrt{1+\left(\frac{x-\xi}{\lambda}\right)^{2}}}\right) - \ln \left( e^{-\frac{1}{2}\left(\gamma+\delta \sinh ^{-1}\left(\frac{x-\xi}{\lambda}\right)\right)^{2}} \right) \right] \\ =& \delta \ln \left( \lambda \sqrt{2 \pi} \right) - \mathbb{E} \left[ -\ln \left(\sqrt{1+\left(\frac{x-\xi}{\lambda}\right)^{2}}\right) -\frac{1}{2}\left(\gamma+\delta \sinh ^{-1}\left(\frac{x-\xi}{\lambda}\right)\right)^{2}\right] \\ =& \delta \ln \left( \lambda \sqrt{2 \pi} \right) - \mathbb{E} \left[ -\frac{1}{2} \ln \left(1+\left(\frac{x-\xi}{\lambda}\right)^{2}\right) -\frac{1}{2}\left(\gamma+\delta \sinh ^{-1}\left(\frac{x-\xi}{\lambda}\right)\right)^{2}\right] \\ =& \delta \ln \left( \lambda \sqrt{2 \pi} \right) -\frac{1}{2} \mathbb{E} \left[ - \ln \left(1+\left(\frac{x-\xi}{\lambda}\right)^{2}\right) - \left(\gamma+\delta \log \left[\left(\frac{x-\xi}{\lambda}\right) + \sqrt{\left(\frac{x-\xi}{\lambda}\right)^2+1}\right] \right)^{2}\right] \\ = & ? \end{align} Note: $\sinh^{-1} z= \log (z + \sqrt{z^2+1})$.
Is the attempt so far correct? How to simplify and complete the derivation?