Let $(B_t)$ a Brownian motion. I know that $$\mathbb E[f(X)]=\int_{\mathbb R}f(x)\mu_X(dx),$$
So, logically if $$f(B_s)=\exp\left(\int_0^t B_s^2ds\right)$$ then $$\mathbb E[f(B_s)]=\int_{\mathbb R}e^{\int_0^tx^2dt}\mu_{B_s}(dx)=\int_{\mathbb R}e^{x^2t}\mu_{B_s}(dx).$$
But this doesn't really look correct. Indeed, if $f(x)=e^{\int_0^t x^2ds}$, then $$f(B_s)=e^{B_s^2\int_0^tds}=e^{B_s^2t}$$ and $$f(B_s)=e^{\int_0^t B_s^2ds}\neq e^{B_s^2t},$$ so something goes wrong here. Any idea ?
Note that I just want to prove that $$\mathbb E\left[\exp\left(\int_0^t B_s^2ds\right)\right]<\infty .$$