To use classical Ito formula
\begin{equation}
f(t,B\_t) - f(0,B\_0) = \int\limits\_0^t f'\_s(s,B\_s)ds + \frac 12\int\limits\_0^t f''\_{xx}(s,B\_s)ds + \int\limits\_0^t f'\_x(s,B\_s)dB\_s
\end{equation}
$f(t,x)$ needs to be $C^{1,2}([0,\infty)\times\mathbb R)$. 


Is there any possibility to use it if $f(t,x)$ is piecewise continuously differentiable in $t$ and two times continuously differentiable in $x$? I mean, there exist $t\_i$, $i=1...n$, $f(t,x)\in C^{1,2}((t\_i,t\_{i+1})\times\mathbb R)$, $f$ and it's derivatives $f^\prime\_t$, $f^\prime\_x$ and $f^{\prime\prime}\_{xx}$ have jump discontinuity at $t\_i$.



Ito formula for discontinuous function
If $f$ is continuous in $t$ that still works. If not, you need to add the term $\sum\_i \Delta\_t f(t\_i,B\_{t\_i})$ to the right hand side where $\Delta\_t f(t,B\_t) = f(t\_+,B\_t)-f(t\_-,B\_t)$ and the sum goes over those $i$ where $t\_i \le t$. I guess.


