Given a sample $ X_1,X_2 \dots X_{100}$ and the density function $ f(x;\theta) = \frac{1}{\pi \cdot \left(1+\left(x-\theta \right)^2\right)}$ , find Fisher information for $\hat{\theta}_{MLE.}$
My attempt:
$L(\theta;x_1,x_2\dots x_{100}) = \prod _{i=1}^{100}\left(\frac{1}{\pi \cdot \left(1+\left(x_i-\theta \right)^2\right)}\right)\:$
$l$ = $\log(L) = -100*\ln(\pi)-\sum^{100}_{i=1}(\ln(1+(x-\theta)^2)$.
$\frac{\partial }{\partial \theta}\left(\log(L)\right) = \sum_{i=1}^{100}(\frac{2(x_i-\theta)}{1+(x_i-\theta)^2}$
$\frac{\partial^2 }{\partial^2 \theta}\left(\log(L)\right) = \sum_{i=1}^{100}-\frac{2\left(-θ+x_i\right)}{1+\left(x_i-θ\right)^2}$
fisher's information : $I(\theta) = -E[\frac{\partial^2 }{\partial^2 \theta}\left(\log(L)\right)$
but I don't how to find the expected value, I tried to simplify it but that didn't help.