Computing $E\left(\exp\left(B_t+\int_0^tB_sdB_s - \frac12\int_0^tB^2_sds\right)\right)$ if $B$ is a standard Brownian Motion Let $B_t$ be a standard Brownian motion under probability $P$. I'm considering computing this:
$$E^P\left[ e ^ { B_t + \int_{0}^{t}B_s\mathrm dB_s - \frac{1}{2}\int_{0}^{t}B^2_s\mathrm ds  } \right], t\in[0,T]$$
I'm suggested that we should use Girsanov's Theorem, but I am wondering if it works:
Let $Z_t=e ^ {\int_{0}^{t}B_s\mathrm dB_s - \frac{1}{2}\int_{0}^{t}B^2_s\mathrm ds  }$, then $E[Z_T]=1$. Let $\mathscr{F}_T=\sigma(B_t,0\leq t \leq T)$, then we can introduce a new measure $\tilde{P}$ by defining as follow:
$$
\tilde P(A)=\int_{A}Z_T\mathrm dP, \forall A\in \mathscr{F}_T
$$
Obviously $\tilde P$ is also a probability over $\mathscr{F}_T$. Thus $E^{\tilde P}[X]=E^P[XZ_t]$ for $X\in \mathscr{F}_t$.
Then by Girsanov's Theorem, $\tilde{B}_t=B_t-\int_{0}^{t}B_s\mathrm ds$ is a standard Brownian motion under $\tilde P$. Thus we have:
$$\begin{align*}
E^P\left[ e ^ { B_t + \int_{0}^{t}B_s\mathrm dB_s - \frac{1}{2}\int_{0}^{t}B^2_s\mathrm ds  } \right] &= E^P[e^{B_t}Z_t] 
\\&=E^{\tilde P}[e^{B_t}]\\
&=E^{\tilde P}[e^{\tilde B_t+\int_{0}^{t}B_s\mathrm ds}]
\end{align*}$$
Then I don't know what to do next. Is it possible to use Girsanov's Theorem here? Or is there a probabilistic way or pde way to solve this?
Thank you!
 A: Yes, this is computable now that you have edited the problem by putting a minus sign in front of the second integral term (rather than a plus sign).
You are on the right track by using Girsanov's theorem. I will use the same change-of-measure and the same notation that you have used in the question.
Since $\tilde B_t = B_t - \int_0^t B_s ds$, it follows from the product rule that that $$\tilde B_t e^{-t} = \frac{d}{dt} \bigg[ e^{-t} \int_0^t B_sds \bigg] \;\;\;\; \Longrightarrow\;\;\;\; \int_0^t \tilde B_se^{-s}ds = e^{-t} \int_0^t B_sds $$ where I integrated both sides to get the implication. Now multiplying both sides by $e^t$ and then differentiating yields \begin{align*} B_t &= \frac{d}{dt} \bigg[ e^t \int_0^t \tilde B_s e^{-s}ds \bigg] \\ &= \tilde B_t +\int_0^t \tilde B_s e^{t-s}ds \\ &= \int_0^t e^{t-s}d \tilde B_s\end{align*}
where I used a stochastic integration-by-parts in the last equality. 
Thus we have written $B_t$ in terms of the process $(\tilde B_s)$. But since $(\tilde B_s)$ is a Brownian motion under $\tilde P$, the above computation reveals that $B_t$ is distributed as a mean-zero Gaussian under $\tilde P$. Moreover its variance can be easily computed by using the Itô isometry: $$\sigma^2=\Bbb E^{\tilde P} [ B_t^2 ]  = \Bbb E^{\tilde P} \bigg[ \bigg( \int_0^t e^{t-s}d\tilde B_s \bigg)^2 \bigg] = \int_0^t e^{2(t-s)}ds = \frac{1}{2}(e^{2t}-1)$$
Now, if $Z$ is a mean-zero Gaussian, then it is well-known that $E[e^Z] = e^{\frac{1}{2}\sigma^2}$, therefore we find that $$\Bbb E^P\left[ e ^ { B_t + \int_{0}^{t}B_s\mathrm dB_s - \frac{1}{2}\int_{0}^{t}B^2_s\mathrm ds  } \right]=\Bbb E^{\tilde P}[e^{B_t}] = e^{\frac{1}{4}(e^{2t}-1)}$$ I may have made an error in some of the computations, maybe someone can post a different method to verify.
