Consider the following exponential family of distributions: $$ p(x ; \beta)=\frac{1}{Z(\beta)} \pi(x) \exp (-\beta h(x)), \;\beta \in\left[\beta_{n} ; \beta_{0}\right]. $$ ${Z(\beta)}$ -- the normalizing constant. $x \in \mathbb{R}$.
I want to prove the following identity: $$ \log Z\left(\beta_{n}\right)-\log Z\left(\beta_{0}\right)=\int_{\beta_{n}}^{\beta_{0}}\langle h(x)\rangle_{p(x ; \beta)} d \beta, $$
$\langle h(x)\rangle_{p(x ; \beta)}=\int_{-\infty}^{\infty}h(x)p(x ; \beta)dx$ here.
My attempt:
It's clear that $Z\left(\beta_{}\right)=\int \pi(x) \exp (-\beta h(x))dx$, it follows from properties of densities.
So, I need to prove $$ \log Z\left(\beta_{n}\right)-\log Z\left(\beta_{0}\right)=\int_{\beta_{n}}^{\beta_{0}} \int_{-\infty}^{\infty}h(x)p(x ; \beta)dx d \beta = \int_{\beta_{n}}^{\beta_{0}} \int_{-\infty}^{\infty}h(x) \frac{1}{Z(\beta)} \pi(x) \exp (-\beta h(x)) dx d \beta, $$ Also I noticed that $$\frac{\partial (\pi(x) \exp (-\beta h(x)))}{\partial \beta}=-h(x)\pi(x) \exp (-\beta h(x))$$
But I don't understand what to do next.
For $p(x;\beta)$ resp. that clumsy and misleading $\langle h(x)\rangle_{p(x;\beta)}$ (this is not a function of $x$) I am going to use the slightly different notation $$ p_\beta(x)\quad\text{ and }\quad\langle h,p_\beta\rangle\,. $$ Since $Z(\beta)$ is the normalizing constant it must satisfy $$ Z(\beta)=\int_{-\infty}^{+\infty}\pi(x)\exp(-\beta h(x))\,dx\,. $$ Therefore, $$ Z'(\beta)=\int_{-\infty}^{+\infty}-h(x)\,\pi(x)\exp(-\beta h(x))\,dx=-Z(\beta)\int_{-\infty}^{+\infty}h(x)\,p(x;\beta)\,dx=-Z(\beta)\langle h,p_\beta\rangle\,. $$ It follows that $$ (\log Z(\beta))'=\frac{Z'(\beta)}{Z(\beta)}=-\langle h,p_\beta\rangle $$ which is equivalent to $$ \log Z(\beta)-\log Z(\beta_0)=-\int_{\beta_0}^\beta\langle h,p_\beta\rangle\,d\beta\, $$ which is your formula.