Distribution of sum of squares of T-distributed random variables I am looking at the distribution of the sum of squares of T-distributed random variables, with tail exponent $\alpha$. Where X is the r.v., the Fourier transform for $X^2$, $\mathscr{F}(t)$ gives me a solution for the square before the convolution $\mathscr{F}(t)^n$. 
$$\mathscr{F}(t)=\int_0^{\infty } \exp \left(i\, t\, x^2\right)\left(\frac{\left(\frac{\alpha }{\alpha +x^2}\right)^{\frac{\alpha +1}{2}} }{\sqrt{\alpha }\ B\left(\frac{\alpha }{2},\frac{1}{2}\right)}\right) \, \mathrm{d}x$$
With $\alpha=3$, the solution is possible but unwieldy and impossible to inverse to do an inverse Fourier for $\mathscr{F}(t)^n$. 
So the question is: has work been done on the distribution of the sample variance or standard deviation of T-distributed random variables?  (It would be to the StudentT what the Chi-square is to the Gaussian).
Thank you.
(Possible Solution) I figured out that $X^2$ is Fisher $F(1,\alpha)$ distributed, hence will look at the sum of Fisher distributed variables.
(Possible Solution) From the Characteristic Functions  the average of $n-$summed $X^2$  has the same first two moments of a $F(n,\alpha)$ distribution when these exist. Hence with u the square root and doing a change of variable inside a probability distribution, the density of the standard deviation of n-sample T variables can be approximated with:
$$g(u)=\frac{2 \alpha ^{\alpha /2} n^{n/2} u^{n-1} \left(\alpha +n u^2\right)^{-\frac{\alpha }{2}-\frac{n}{2}}}{B\left(\frac{n}{2},\frac{\alpha }{2}\right)}$$
 A: A clarification of your question (there seem to me to be two related, but different, parts): you are looking for (1) distribution of a sum of $n$ independent squared $t_{\alpha }$ random variables, and
(2) the sampling distribution of the variance (or the related standard deviation) of a random sample drawn from a $t_{\alpha }$ distribution (presumably your reason for asking about (1)).
Distribution of Sum of Independent Squared $t_{\alpha }$ Variables
If $T_i\sim t_{\alpha }$ are (independent) random $t$ variables with $\alpha$ d.f., then it is false that $\sum _{i=1}^n T_i^2\sim F(n,\alpha )$ (which is what you seem to be claiming in your second "possible solution"). This is easily verified by considering the first moment of each (the latter's first moment is $n$ times the first's).
The claim in your first "possible solution" is correct: $T_i^2\sim F(1,\alpha)$.  Rather than resorting to characteristic functions, I think this result is more transparent when considering the characterisation of the $t$ distribution as the distribution of the ratio $\frac{Z}{\sqrt{U/\alpha}}$ where $Z$ is a standard normal variable and $U$ is a chi-squared variable with $\alpha$ degrees of freedom, independent of $Z$. The square of this ratio is then the ratio of two independent chi-squared variables scaled by their respective degrees of freedom i.e. $\frac{V/1}{U/\alpha}$ with $V=Z^2$, which is a standard characterisation of an $F(1,\alpha)$ distribution (with numerator d.f. equal to 1 and denominator d.f. equal to $\alpha$).
Considering the note I made on first moments in the first paragraph above, it might seem that a better claim may be that $\sum _{i=1}^n T_i^2\sim n F(n,\alpha )$ [I have slightly abused notation here by using the same expression for the distribution as well as a random variable having that distribution.].  Whilst the first moments match, the second central moments do not (for $\alpha>4$ the variance of the first expression is less than the variance of the latter expression) - so this claim is false too.  [That being said, it is interesting to observe that $\lim_{\alpha \to \infty } \, n F(n,\alpha)= \chi _n^2$, which is the result we expect when summing squared (standard) normal variates.]
Sampling Distribution of Variance When Sampling from a $t_{\alpha }$ Distribution
Considering what I have written above, the expression you obtain for "the density of the standard deviation of n-sample T variables" is incorrect.  However, even if the $F(n,\alpha)$ were the correct distribution, the standard deviation is not simply the square root of the sum of squares (as you seem to have used to arrive at your $g(u)$ density).  You would instead be looking for the (scaled) sampling distribution of $\sum _{i=1}^n \left(T_i-\bar{T}\right){}^2=\sum _{i=1}^n T_i^2-n \bar{T}^2$.  In the normal case, the LHS of this expression can be re-written as a sum of squared normal variables (the term inside the square can be re-written as a linear combination of normal variables which is again normally distributed) which leads to the familiar $\chi^2$ distribution. Unfortunately, a linear combination of $t$ variables (even with the same degrees of freedom) is not distributed as $t$, so a similar approach can not be exploited.
Perhaps you should re-consider what it is you wish to demonstrate?  It may be possible to achieve the objective using some simulations, for example.  However, you do indicate an example with $\alpha=3$, a situation where only the first moment of $F(1,\alpha)$ is finite, so simulation won't help with such moment calculations.
A: You may want to check out Hotelling's T-distribution (http://en.wikipedia.org/wiki/Hotelling's_T-squared_distribution).  There is relationships with $T^2$ being a $F$-distribution (http://en.wikipedia.org/wiki/F-distribution#Related_distributions_and_properties), but I'm not sure this is exactly what you're asking for.  
