Cramer Rao lower bound in Cauchy distribution

Suppose $X\sim C(\theta,1)$, a Cauchy distribution with location $\theta$ and scale unity.

For $\theta\in\mathbb R$, the pdf of $X$ is

$$f_{\theta}(x)=\frac{1}{\pi(1+(x-\theta)^2)}\qquad,\,x\in\mathbb R$$

Clearly, $$\frac{\partial}{\partial\theta}\ln f_{\theta}(x)=-\frac{2(x-\theta)}{1+(x-\theta)^2}$$

Therefore,

$$E_{\theta}\left[\frac{\partial}{\partial\theta}\ln f_{\theta}(X)\right]^2=4E_{\theta}\left[\frac{X-\theta}{1+(X-\theta)^2}\right]^2\tag{1}$$

Now for every $\theta$,

\begin{align} E_{\theta}\left[\frac{X-\theta}{1+(X-\theta)^2}\right]^2 &=\frac{1}{\pi}\int_{\mathbb R}\left[\frac{x-\theta}{1+(x-\theta)^2}\right]^2\frac{1}{1+(x-\theta)^2}\,\mathrm{d}x \\&=\frac{1}{\pi}\int_{\mathbb R}\frac{(x-\theta)^2}{(1+(x-\theta)^2)^3}\,\mathrm{d}x\\&=\frac{2}{\pi}\int_0^\infty\frac{t^2}{(1+t^2)^3}\,\mathrm{d}t \\&=\frac{1}{\pi}\int_0^\infty\frac{\sqrt u}{(1+u)^3}\,\mathrm{d}u \\&=\frac{1}{\pi}B\left(\frac{3}{2},\frac{3}{2}\right) \\&=\frac{1}{8} \end{align}

So from $(1)$, we have the Fisher information

$$I(\theta)=E_{\theta}\left[\frac{\partial}{\partial\theta}\ln f_{\theta}(X)\right]^2=\frac{1}{2}\quad,\forall\,\theta$$

And Cramer-Rao lower bound for $\theta$ is $$\text{CRLB}(\theta)=\frac1{I(\theta)}=2 \quad,\forall\,\theta$$

In case $X_1,X_2,\ldots,X_n$ are i.i.d with pdf $f_{\theta}$, Fisher information in $\mathbf X=(X_1,\ldots,X_n)$ is

$$ I_{\mathbf X}(\theta)=n I(\theta)=\frac n2 \quad,\forall\,\theta$$

The Cramer-Rao bound for $\theta$ is then $\frac2n$ for every $\theta$.