proof of Feynman–Kac formula the article given by wikipedia 
http://en.wikipedia.org/wiki/Feynman%E2%80%93Kac_formula#Proof
states at some point of the proof that:
(line 7) ''the third term is  o(dtdu)  and can be dropped''
Can anyone see why the cross-variation term can be dropped, i.e is equal to zero ?
 A: This is a non-rigorous way to think about this.  On the other hand, the first time I learned about Brownian motion was using non-standard analysis, so this can be made rigorous with some effort.  I wrote this using the notation $B_t$, which on the web page is written as $W_t$ ($B$ for Brown, $W$ for Weiner).
So I think of $dt$ as being an infinitesimal quantity.  And $dB_t = \pm \sqrt{dt}$.  In this way
$$ B_t = \int_0^t dB_t = \sum_{n=0}^{t/dt} \pm \sqrt{dt} $$
is a reasonable sized value, i.e., neither infinitesimal nor infinite.  This is because
$$ E \left|\sum_{n=1}^N \pm 1\right| \approx \left(E \left|\sum_{n=1}^N \pm 1\right|^2\right)^{1/2} = \sqrt{N} $$
and so
$$ E\left|\sum_{n=0}^{t/dt} \pm \sqrt{dt}\right| \approx \sqrt{t/dt} \times \sqrt{dt} \approx \sqrt t .$$
(Here all the $\pm$ are independent and take either value with probability $1/2$.)
Then
$$ \int_0^t dt du \approx \sum_{n=0}^{t/dt} dt(U dt \pm V \sqrt{dt}) $$
for some $U$ and $V$ (which might not be constant, but depend upon $t$ and $B_t$, and this is of order $dt^2\times(t/dt) + dt\sqrt{dt}\times \sqrt{t/dt}$, and this is clearly infinitesimal, and hence in non-standard analysis converts to zero.
