Distribution of indefinite integral of Brownian motion multiplied by an exponential decay term w.r.t time

Question

Consider the indefinite integral \begin{equation} I=\int_{0}^{\infty}f(t)e^{-t} W_t\mathop{dt}, \end{equation} such that $W_t$ is a standard Brownian motion. To ensure convergence, we'll suppose that a continuously differentiable $f:\mathbb{R}_{>0}\rightarrow\mathbb{R}$ is bounded. What is the density of $I$? I simulated a trivial case with $f(t)=1$ and found that $I$ conforms to a normal density with mean $0$ and variance $\approx 0.7016^2$. For what, if any nontrivial, choices of $f$ is $I$ normal?

Answer 1

Hint $$\begin{align} I &= \int_0^{+\infty} \left( f(t)e^{-t}\int_0^t dW_s \right) dt\\ &=\iint_{0\le s \le t \le +\infty} f(t)e^{-t} dt dW_s\\ &=\int_0^{+\infty} \underbrace{\left( \int_s^{+\infty} f(t)e^{-t} dt \right)}_{\text{converge as } f \text{ is bounded by }(\sup |f|)\cdot e^{-s}} dW_s\\ &\stackrel{\mathcal D}{=}\mathcal{N} \left(0,\underbrace{\int_0^{+\infty}\left( \int_s^{+\infty} f(t)e^{-t} dt \right)^2ds}_{\text{converge } } \right) \end{align}$$