Ito Representation for Integral of Brownian Motion

Question

By Ito's representation theorem because the integral of Brownian Motion is mean zero we should have some random function $\phi$ such that

$$\int_0^T W_t dt = \int_0^T\phi(\omega,t)dW_t$$

I'm a little uncomfortable about the left hand side being finite variation and the right a local martingale even for fixed $T$, I just can't get any closer to the form above than the obvious Ito formula result

$$\int_0^T W_t dt = TW_T-\int_0^TtdW_t$$

Thanks

Answer 1

The identity

$$\int_0^T W_t \, dt = T W_T - \int_0^T t \, dW_t$$

is a good start. Now note that

$$T W_T = \int_0^T T \, dW_t$$

which implies that

$$\int_0^T W_t \, dt = \int_0^T (T-t) \, dW_t.$$

This is exactly the representation you are looking for.