Given $c>0$ and define a function $f:(0,\infty)\mapsto \mathbb{R}$ by \begin{equation} f(\sigma)=\frac1{\sqrt {2\pi}\sigma}\int_{-\infty}^\infty\max\{-c,\min\{x,c\}\}^2e^{-\frac{x^2}{2\sigma^2}}. \end{equation} We can see $f$ as the second moment of $X_c=\max\{-c,\min\{X,c\}\}$ where $X\sim N(0,\sigma^2)$. I'd like to find the range of $f$. I have proved that $f$ is strictly increasing but I still don't know it limits when $\sigma\to 0,\infty$. Could anyone help me? Thanks in advance.
PS: we also be able to express $f$ by \begin{equation} f(\sigma)=2c^2(1-\phi(c))+\sigma^2(2\phi(c)-1) \end{equation} where $\phi(c)=P(X\leq c$).
Let us assume that $f$ is actually $$ f(\sigma)=\frac1{\sqrt {2\pi}\sigma}\int_{-\infty}^\infty\max\{-c,\min\{x,c\}\}^{\color{red}{\bf 2}}\mathrm e^{-x^2/(2\sigma^2)}\mathrm dx. $$ Define $u_c:(0,\infty)\to(0,\infty)$ by $u(t)=\min\{t,c^2\}$ and consider a standard normal random variable $Z$. Then, $$ f(\sigma)=E(u_c(\sigma^2Z^2)). $$ Since $u_c$ is nondecreasing, $f$ is nondecreasing (and actually increasing). Since $u_c$ is continuous and bounded with $u_c(0)=0$ and $u_c(\infty)=c^2$, one has $\lim\limits_{\sigma\to0}f(\sigma)=0$ and $\lim\limits_{\sigma\to\infty}f(\sigma)=c^2$.
For the limit when $\sigma\to\infty$, note that $\sigma^2Z^2\to\infty$ almost surely hence $u_c(\sigma^2Z^2)\to c^2$ almost surely and the domination $u_c(\sigma^2Z^2)\leqslant c^2$ allows to conclude.