Interval with the shortest expected length How to find the interval with the shortest expected length for such a central function:
$Q(X;\sigma^2) = \frac{(n-1)S^2}{\sigma^2}$?
I determined only the class $C$- confidence intervals for the $\sigma^2$ parameter at the confidence level
$1-\alpha$:
$C={[T_L,T_U]: T_L = \frac{\sum_{i=1}^{n} (X_i - \overline{X})^2}{\chi^2_{\frac{\alpha}{2}}}, T_U = \frac{\sum_{i=1}^{n} (X_i - \overline{X})^2}{\chi^2_{1-{\frac{\alpha}{2}}}} }$ what needs to be done next to find the shortest interval?
 A: If I understand your question, you are looking for the shortest confidence interval on a parameter.  Your $T_L$ and $T_U$ seem to be central, eg same fraction of pdf above and below them, so not necessarily the shortest interval.
I would start by building the acceptance for a given value of the parameter: fix $\sigma^2$, then keep the values of $x$ such that the integral of your pdf over these values of $x$ is equal to your confidence level.  In general, this choice of $x$ is not unique, but since you want the shortest interval, chose values of $x$ by how probable they are.  This give an upper and lower acceptance,  $x_1(\sigma^2)$ and $x_2(\sigma^2)$ (assuming the pdf has a single maximum so that the acceptance is not a discontinuous set).  By construction, this acceptance is the shortest in terms of the distance between $x_1(\sigma^2)$ and $x_2(\sigma^2)$.  Repeating for other values of $\sigma^2$ gives you the acceptance curves $x_1(\sigma^2)$ and $x_2(\sigma^2)$.  Inverting these functions should give you the shortest possible confidence interval for parameter $\sigma^2$.
