Expected Fisher's information matrix for Student's t-distribution? I am having trouble finding a resource online that derives the expected Fisher's Information Matrix for the uni-variate Student's t-distribution.  Does anyone know of such a resource?
In absence of any existing resource that derives the expected Fisher's information matrix for the t-distribution, I am trying to derive it myself but I'm stuck.  Here is my work so far:
$y_i \sim t(\mu, \sigma^2, v)$ where $v$ is the degrees of freedom (df) parameter (assumed fixed).  Then:
\begin{align*}
f(y_i) &= \frac{\Gamma(\frac{v+1}{2})}{\Gamma(\frac{v}{2})\sqrt{\pi v \sigma^2}}\big(1+\frac{1}{v\sigma^2}(y_i-\mu)^2\big)^{\frac{-(v+1)}{2}} 
\end{align*}
Thus we have the following log-likelihood function:
\begin{align*}
log f(y_i)=log\Gamma(\frac{v+1}{2})-log\Gamma(\frac{v}{2})-\frac{1}{2}log(\pi v \sigma^2)+ \frac{-(v+1)}{2}log\big[1+\frac{1}{v\sigma^2}(y_i-\mu)^2\big]
\end{align*}
Here the first derivative equations:
\begin{align*}
&\frac{\partial}{\partial \mu}logf(y_i)=\frac{v+1}{2}\frac{\frac{2}{v\sigma^2}(y_i-\mu)}{1+\frac{1}{v\sigma^2}(y_i-\mu)^2} \\ & \frac{\partial}{\partial \sigma^2}logf(y_i)= \frac{-1}{2\sigma^2}-\frac{(v+1)}{2} \frac{\frac{-1}{v\sigma^4}(y_i-\mu)^2}{1+\frac{1}{v\sigma^2}(y_i-\mu)^2}
\end{align*}
And here are the 2nd derivative equations:
\begin{align*}
&\frac{\partial}{\partial \mu^2}logf(y_i)=\frac{v+1}{2}\frac{\frac{-2}{v\sigma^2}+\frac{2}{dv^2\sigma^4}(y_i-\mu)^2}{\big(1+\frac{1}{v\sigma^2}(y_i-\mu)^2\big)^2} \\ & \frac{\partial}{\partial \mu \partial \sigma^2}logf(y_i) =\frac{v+1}{2} \Big\{[\frac{2}{v\sigma^2}-\frac{4}{v^2\sigma^6}(y_i-\mu)^2][1+\frac{1}{v\sigma^2}(y_i-\mu)^2]^2-[\frac{-2}{v\sigma^2}+\frac{2}{v^2\sigma^4}(y_i-\mu)^2]*2[1+\frac{1}{v\sigma^2}(y_i-\mu)^2][\frac{-1}{v\sigma^4}(y_i-\mu)^2]\Big\}/\Big\{ [1+\frac{1}{v\sigma^2}(y_i-\mu)^2]^4\Big\}.....\text{really messy!} \\ &\frac{\partial}{\partial (\sigma^2)^2}logf(y_i)=\frac{1}{2\sigma^4}-\frac{(v+1)}{2}\frac{\frac{1}{v\sigma^6}(y_i-\mu)^2}{[1+\frac{1}{v\sigma^2}(y_i-\mu)^2]^2}
\end{align*}
Finally, the expected fisher's information matrix is calculated as follows:
\begin{align*}
\boldsymbol{I}= -\mathbb{E}\Big(\begin{bmatrix}
\frac{\partial^2}{\partial \mu^2}logf(y_i) & \frac{\partial}{\partial \mu \partial \sigma^2}logf(y_i) \\ 
\frac{\partial}{\partial \mu \partial \sigma^2}logf(y_i) & \frac{\partial^2}{\partial (\sigma^2)^2}logf(y_i)
\end{bmatrix}\Big)
\end{align*}
However, I have no idea how to calculate these expectations.  Is anyone aware of a resource that has done this?  Honestly, the only quantity I'm interested in is: $-\mathbb{E}\big[\frac{\partial^2}{\partial (\sigma^2)^2}logf(y_i)\big]$, would someone at least be able to help me calculate this?
 A: It was brought to my attention that Lange et al 1989 derived the expected Fisher's Information for the multivariate t-distribution in Appendix B.  Therefore, I got the answer I wanted, you can regard this question as answered!
In particular, using the result of Lange et al, I derived the following Fisher's Information Matrix for the univariate t-distribution (with fixed degrees of freedom parameter $v$):
\begin{align*}
\boldsymbol{I}=\begin{bmatrix}
\frac{v+1}{(v+3)\sigma^2} & 0 \\
0 & \frac{v}{2(v+3)\sigma^4}
\end{bmatrix}
\end{align*}
A: It is not difficult (but a bit tedious) by using the formula
$$
\mathcal{I}(\mu, \sigma^2) = \mathbb{E}\left[\begin{pmatrix}{\left(\frac{\partial}{\partial\mu} \log f(Y)\right)}^2 & \left(\frac{\partial}{\partial\mu} \log f(Y)\right)\left(\frac{\partial}{\partial\sigma} \log f(Y)\right) \\ \left(\frac{\partial}{\partial\mu} \log f(Y)\right)\left(\frac{\partial}{\partial\sigma} \log f(Y)\right) & {\left(\frac{\partial}{\partial\sigma^2} \log f(Y)\right)}^2 \end{pmatrix} \right].
$$
First, observe that by the change of variables $y \mapsto y-\mu$ in any involved integral, one can take $\mu=0$ in the calculations.
The calculations rely on the following integral:
$$
I(\lambda, a, b) := \int_0^\infty y^{2a-1}\left(1+\frac{1}{\lambda}y^2\right)^{-\frac{2a+b}{2}} \textrm{d}y
= \frac{\lambda^a}{2} B\left(a, \frac{b}{2}\right).
$$
This equality is obtained by the change of variables $y \mapsto y^2$ and with the help of the density of the Beta prime distribution.
Observe that the integrand is an even function when $2a-1$ is an even integer, hence
$$
J(\lambda, a, b) := \int_{-\infty}^{+\infty} y^{2a-1}\left(1+\frac{1}{\lambda}y^2\right)^{-\frac{p+1+b}{2}} \textrm{d}y = 2 I(\lambda, a, b) = \lambda^a B\left(a, \frac{b}{2}\right).
$$
I will detail only the first calculation. Set
$$
K(\nu, \sigma)=\frac{1}{B\left(\frac12,\frac{\nu}{2}\right)}\frac{1}{\sqrt{\nu \sigma^2}},
$$
the normalization constant of the density.
One has
$$
\begin{align}
\mathbb{E}\left[{\left(\frac{\partial}{\partial\mu} \log f(Y)\right)}^2 \right] = 
K(\nu, \sigma){\left(\frac{\nu+1}{\nu\sigma^2}\right)}^2 
J\left(\nu\sigma^2, \frac{3}{2}, \nu+2\right).
\end{align}
$$
Since $\frac{B\left(\frac12,\frac{\nu}{2}\right)}{B\left(\frac{3}{2},\frac{\nu+2}{2}\right)} = \frac{B\left(\frac12,\frac{\nu}{2}\right)}{B\left(\frac{3}{2},\frac{\nu}{2}\right)}\frac{B\left(\frac{3}{2},\frac{\nu}{2}\right)}{B\left(\frac{3}{2},\frac{\nu+2}{2}\right)}  = \frac{(\nu+1)}{1}\frac{(\nu+3)}{\nu}$, we find
$$
\mathbb{E}\left[{\left(\frac{\partial}{\partial\mu} \log f(Y)\right)}^2 \right] = \frac{\nu}{\nu+3}(\nu+1){(\nu\sigma^2)}^{-1/2-2+3/2}
= \frac{\nu+1}{(\nu+3)\sigma^2}.
$$
The second calculation is easy:
$$
\mathbb{E}\left[ \left(\frac{\partial}{\partial\mu} \log f(Y)\right)\left(\frac{\partial}{\partial\sigma} \log f(Y)\right)\right] = 0
$$
because it only involves integrals of odd functions.
Finally the calculation of 
$$
\mathbb{E}\left[{\left(\frac{\partial}{\partial\sigma^2} \log f(Y)\right)}^2 \right]
$$
is the more tedious and I skip it. Its calculation involves integrals $J(\nu\sigma^2, a, b)$ with $2a-1$ even integer, whose value is given above.
I've done the calculations and I've found
$$
\frac{{(\nu+1)}^2}{4{(\nu\sigma^4)}^2} K(\nu, \sigma^2) J\left(\nu\sigma^2, \frac{5}{2},\nu\right) - \frac{\nu+1}{2\nu\sigma^6} K(\nu, \sigma^2) J\left(\nu\sigma^2, \frac{3}{2},\nu\right) + \frac{1}{4\sigma^4}
$$
and this simplifies to 
$$
\frac{\nu}{2(\nu+3)\sigma^4}.
$$
