If $W_t$ is Brownian motion, is it true that for any Ito integrable function $f$, $$\int_0^t f(u) dW_u = W_tf(t) - \int_0^t W_uf'(u)du$$?
I've seen this represntation used a couple of times without proof, and I was wondering if it was true in general or if there are some specific properties $f$ must satisfy.
If true, how can it be proved? I've tried applying Ito's formula to the right hand side, but no luck.
This is a result of the Stochastic Leibniz Rule, i.e., $$ {\rm d}\left(X_tY_t\right)=Y_t{\rm d}X_t+X_t{\rm d}Y_t+{\rm d}\left<X,Y\right>_t, $$ where $\left<X,Y\right>$ denotes the co-variation of $X$ and $Y$.
Back to your question, simply put $X_t=f(t)$ and $Y_t=W_t$, and $$ {\rm d}\left(f(t)W_t\right)=W_t{\rm d}f(t)+f(t){\rm d}W_t+{\rm d}\left<f,W\right>_t. $$ Note that $f$ is deterministic, for which ${\rm d}f(t)=f'(t){\rm d}t$ and $\left<f,W\right>=0$. The above result reduces to $$ {\rm d}\left(f(t)W_t\right)=f'(t)W_t{\rm d}t+f(t){\rm d}W_t. $$ The integration of this last equation immediately leads to the formula you expect.