I found this question in a past exam for a course on Financial Economics.
Given the function $f(t,x)$, let $F(t,x)$ be a function such that $∂F/∂x = f$.
(a) By writing Itô’s formula in integral form, show that: $$\int^T_0 f(t,Bt)dB_t = F(T,B_T)−F(0,0)−\int^T_0( \frac{∂}{∂t} F(t,Bt) + \frac{1}{2}\frac{∂}{∂x} f(t,B_t))dt$$ where ${B_t}$ is a standard Brownian motion. (Here $\frac{∂}{∂x}$ denotes the derivative with respect to the second argument of the function.)
(b) Use (a) to simplify the following integral: $\int_0^T \sqrt{2t+B_t} dBt$
I managed to get 1(a) by applying Itô's Lemma to $dF(t,B_t$), and things rearranged without much trouble. I was reluctant to continue because something was bothering me here.
Is $\int_0^T \sqrt{2t+B_t} dBt$ even real? It will occur in some Brownian motion paths that $B_t < -2t$ for some $t \in [0,T]$. This made me wonder whether this integral can be evaluated. If it can be, what might the answer look like?
I did some more work and here's what I found. Let $f(t,x) = \sqrt{2t+x} = (2t+x)^{1/2}$. $$ F(t,x) = \int_0^x f(t,u)du = \frac{2}{3}[(2t+x)^{3/2}-(2t)^{3/2}]$$ $$ \frac{\partial}{\partial t}F(t,x) = 2[(2t+x)^{1/2} - (2t)^{1/2}] = 2[f(t,x) - (2t)^{1/2}] $$ $$\frac{\partial}{\partial x}f(t,x) = \frac{\partial^2}{\partial x^2}F(t,x) = \frac{1}{2}(2t+x)^{-1/2} = \frac{1}{2f(t,x)}$$
Plugging these into the formula from (a) gives $$\int_0^T(2t+B_t)^{1/2}dB_t $$ $$= \frac{2}{3}[(2T+B_T)^{3/2}-(2T)^{3/2}] - 0 -2\int_0^T [(2t+B_t)^{1/2} - (2t)^{1/2}]dt + \frac{1}{4}\int_0^T(2t+B_t)^{-1/2}dt = \frac{1}{4}\int_0^T(2t+B_t)^{-1/2}dt$$
This does seem consistent with my earlier problem (i.e. if the first integral has imaginary parts then the simplified version does too). Does this simplify any further?
Another question I have: Is $F(T,B_T)−F(0,0)−\int^T_0 \frac{∂}{∂t} F(t,Bt) dt$ always zero? Does the fundamental theorem of calculus apply here? It seems like it must. If so why do we bother having this term in the equation in (a)?
Ignoring for the moment the fact that $\sqrt{2t+B_t}$ will sometimes be negative, we have $F(t,x) = \frac23(2t+x)^\frac32,\qquad$ $\frac{\partial F}{\partial t}=2\sqrt{2t+x},\qquad$ $\frac{\partial f}{\partial x} = \frac{1}{2\sqrt{2t+x}}$,
so $\int_0^T\sqrt{2t+B_t}\;dB_t = \frac23(2T+B_T)^\frac32 -\int_0^T2\sqrt{2t+B_t}+\frac{1}{2\sqrt{2t+B_t}}\;dt$.
We now have converted a stochastic integral to a Riemann/Lebesgue integral. Given a sample path of $B_t$, this latter integral can be interpreted as "the area under a curve". We can't do better than this.
But all of this is purely formal, assuming that everything is defined. Probably, the exercise was meant in this spirit... However, it is well known that, with probability 1, a Brownian motion (starting at zero) will change sign infinitely often in the time interval $[0,h]$, no matter how small $h>0$. Now if $B_t$ is a $\mathbb P$--Brownian motion (for some probability measure $\mathbb P$), and if we perform the Girsanov transformation $\frac{d\mathbb Q}{d\mathbb P}=e^{-2B_T-2T}$, then $\tilde{B}_t:=2t+B_t$ is a $\mathbb Q$--Brownian motion (on $[0,T]$). So $\tilde{B}_t$ will change sign infinitely often on $[0,h]$ with $\mathbb Q$--probability 1. Since $\mathbb Q,\mathbb P$ are equivalent measures (i.e. have the same null sets, and thus sets of probability 1), we also have that $2t+B_t=\tilde{B}_t$ changes sign infinitely often with $\mathbb P$--probability 1.
BTW, $2\int_0^T (2t+B_t)^\frac12\;dt\not=\frac23(2T+B_T)^\frac32$, it is not possible to cancel the terms, as you have done in your solution. (If you regard the LHS and RHS as stochastic processes indexed by $T$, then the LHS yields a finite variation process, but the RHS has infinite variation.)