# Ito Integration by parts

Recall that if $$u(t)$$ and $$v(t)$$ are deterministic functions which are once differentiable then the classic integration by parts formula states that
$\int_0^t u(s) (\frac{dv}{ds})(s)\,ds = u(t)v(t) – u(0)v(0) – \int_0^t v(s) (\frac{du}{ds})(s)\,ds$

As is suggested by the formal relations

$(\frac{dv}{ds})(s)\,ds=dv(s) \qquad\text{and}\qquad (\frac{du}{ds})(s)\, ds=du(s)$

this can be rearranged  to state

$u(t)v(t)- u(0)v(0)= \int_0^t u(s) dv(s) + \int_0^t v(s) du(s)$

which holds for more general Riemann–Stieltjes integrals. Now consider two Ito processes $$X_t$$ and $$Y_t$$ given by

$dX_t=b_s ds + \sigma_s dW_t \qquad\text{and}\qquad dY_t=f_s ds + g_s dW_t$

where $$W_t$$ is a standard Brownian Motion. Derive the “Integration by Parts formula” for Ito calculus by applying Ito’s formula to $$X_tY_t$$. Compare this the the classical formula given above.