Entropy of a non-central chi-squared distribution with one degree of freedom

Question

Let $Y=X^2$ where $X$ is a non-central Gaussian distributed random variable. So $Y$ has a non-central chi-squared distribution with one degree of freedom. What is the entropy of $Y$? I write the integral formula of entropy, but I cannot simplify the integral at all.

Answer 1

The answer should be in this recent paper:

Stefan Moser (2020): "Expected Logarithm and Negative Integer Moments of a Noncentral χ2-Distributed Random Variable", Entropy, vol. 22, no. 9, https://doi.org/10.3390/e22091048

The result is provided for any odd or even integer number of degrees of freedom and is pretty complicated. It seems to be a novel result, not derived before.

Set the function $g_n(\xi)$ for an arbitrary $n$, defined for $\xi>0$ by

$$g_n(\xi) = \ln(\xi) - \mathrm{Ei}(-\xi) + \sum_{i=1}^{n-1}(-1)^i\left( e^{-\xi} (i-1)! - \frac{(n-1)!}{i(n-1-i)!} \right)\left(\frac{1}{\xi}\right)^i $$

Set the function $h_n(\xi)$ for an odd $n$, defined for $\xi>0$ by

$$h_n(\xi) = -\gamma - 2\ln(2)+2 \xi {}_2 F_2\left(1,1,\frac{3}{2},2,-\xi\right) +\sum_{i=1}^{(n-1)/2} (-1)^{i-1}\Gamma\left(i-\frac{1}{2}\right)\left(\sqrt{\xi}e^{-\xi}\mathrm{erfi(\sqrt{\xi})}+\sum_{j=1}^{i-1}\frac{(-1)^j\xi^j}{\Gamma\left(j+\frac{1}{2}\right)} \right)\left(\frac{1}{\xi}\right)^i$$

where $\mathrm{Ei}$ is the exponential integral function, $\gamma$ is the Euler constant, ${}_2 F_2$ is a generalised hypergeometric function and $\mathrm{erfi}$ is the imaginary error function.

Then the result for $X\sim\mathcal{X}'^2_n(\lambda)$ is, for $n$ odd, $$ \mathbb{E}\left[-\ln(X)\right]=\ln(2)+h_n\left(\frac{\lambda}{2}\right)$$ and, for $n$ even, $$ \mathbb{E}\left[-\ln(X)\right]=\ln(2)+g_{n/2}\left(\frac{\lambda}{2}\right)$$

Answer 2

I guess there is a minus missing in the answer. The correct form must be

$\mathbb{E}[-\ln(Y)] = -\ln(2)-h_n(\frac{\lambda}{2})$

am I right? (see Eq. 54 at Moser's paper)