My idea is that the Expectation of Cauchy Distribution is infinite, and I find the following lemma
If $X$ were from the exponential family, it would have finite expectation.
The question is solved if I can prove the lemma, but I have no idea how to prove it.
I could get that if $f_X(x;θ)$ can be written as $$f_X(x;\eta)=h(x)\exp\left(\sum_{i=1}^s\eta_iT_i(x)−A(\eta)\right)$$ then $E[T_i(X)]=\frac{\partial A(\eta)}{\partial η_i}$ is finite
But I don't know how to apply this to show that the expectation $E[X]$ is finite.
Distributions of the Exponential Family can be written in the form: $$ f_X(x; \theta) = h(x)g(\theta)\exp\biggl(\eta(\theta)\cdot T(x)\biggr)$$
The Cauchy distribution has the following expression: $$\begin{align} f_X(x; \lambda) &= \frac{\lambda}{\pi(\lambda^2+x^2)} \\ &= \frac{\lambda}{\pi}\exp\biggl(-\log (\lambda^2+x^2) \biggr) \tag{*}\label{*}\\ &= \frac{\lambda}{\pi}\exp\biggl(-2\log (\lambda) \log \bigl(1+\frac{x^2}{\lambda^2}\bigr)\biggr) \tag{*} \end{align}$$
Can you fill in the final step?