Distribution of stochastic integral w.r. to brownian motion

Question

Let $B=(B_t)_{t \geq 0}$ be a standard brownian motion, $T > 0$ and $f : [0,T] \rightarrow \mathbb{R}$ a continuous function. I want to determine the distribution of the following integral:

$\int_{0}^{T} f(s) dB_s$.

My idea was to use

$\varphi(z) = \mathbb{E}(\exp(izY)) = \mathbb{E}(\exp(iz \lim_{n \rightarrow \infty} Y_n))=...$

with

$Y_n = \sum_{j=1}^{n} f(s_j) (B_{s_j} - B_{s_{j-1}})$

where {$s_0,...,s_n$} is a partition of the interval [0,T]. My question is whether I can write $Y_n$ in that way since we had the sum notation for (H.B) where H is an elementary process.

If I can't, how can I then determine the distribution of the given integral?

Thanks in advance!

Answer 1

Hint: $E(\int_0^T f(s)dB_s)=0$, and $var=\int_0^T f^2(s)ds$.