Integration by part formula in Malliavin Calculus 
*

*The set $S$ of smooth random variables is the set of random variables $F : \Omega \rightarrow \mathbb R$ such that there exist a function $f$ in $ \mathcal C_p^{\infty}(\mathbb R^n)$
(for some $n \geq  1$) and
elements $h_1, \cdots , h_n$ of $L^2$ such that



$$F = f\left((\int_0^t h_1\, dW_s), \cdots , (\int_0^t
 h_n\,dW_s)\right) \qquad (*)$$



*The set $\mathcal P$ denotes the set of random variables of the form (*)
where $f$ is a polynomial.

*$S_b$ (resp. $S_0$) denotes the space of random variables of the
    form (*) with $f$ in  $ \mathcal C_p^{\infty}(\mathbb R^n)$ (resp. 
    $ \mathcal C_0^{\infty}(\mathbb R^n)$ ). 



We deﬁne the Malliavin
           derivative $DF$ of $F$ as the $L^2$-valued random variable  $$DF =\sum_{i=1}^n \frac{\partial f}{\partial x_i}f\left((\int_0^t 
 h_1\, dW_s), \cdots , (\int_0^t h_n\,dW_s)\right)h_i.$$



*Integration by part formula:
Let $F$ be a smooth random variable of the form (*) and let $h$ be
an element of $L^2$. The  integration by parts formula say that



$$\mathbb E[\langle DF, h\rangle_{L^2}] = \mathbb E[F(\int_0^t h\,
 dW_s)] \qquad (**).$$

By normalization the relation (**)  we can assume that $||h||_{L^2} = 1$ and  $F = f\left((\int_0^t e_1\, dW_s), \cdots , (\int_0^t
 e_n\,dW_s)\right) $ where $\{e_1, \cdots , e_n\}$ are ONB in $L^2$ , $e_1 = h$ and $f \in  \mathcal C_p^{\infty}(\mathbb R^n)$ We have that

\begin{align} \mathbb E\left[\langle DF, h\rangle_{L^2}\right] =&  \mathbb E
\left[ \frac{d}{dx_1}f\left((\int_0^t e_1\, dW_s), \cdots , (\int_0^t
 e_n\,dW_s)\right)\right] \tag{A}\\
 =& (2π)^{−n}\int_{\mathbb R^n}\frac{d}{dx_1}f(x_1, \cdots , x_n) \exp\left(-\frac{1}{2}\sum_{i=1}^nx_i^2\right)\,dx_1 · · · dx_n \tag{B} \\
 =& (2π)^{−n}\int_{\mathbb R^n} x_1f(x_1, \cdots , x_n) \exp\left(-\frac{1}{2}\sum_{i=1}^nx_i^2\right)\,dx_1 · · · dx_n \tag{C} \\
 =& \mathbb E[FW(e_1)] \tag{D}\\
 =& \mathbb E[FW(h)] . \end{align}

Question: 


*

*In A: Because $\langle e_1,h_i\rangle=0, \quad \forall i=2, \cdots, n$ then they neglect the terms $dx_2, \cdots dx_n.$. Am'I right?

*In B: They use the PDF of the standard law normal, but why?

*In C: I don't can't develop how they use the integration by part in $\mathbb R^n$. 

 A: Let me try to provide some insight...
A) Yes. To be more precise, you can, as you said, assume $||h||_{L^{2}}=1$. 
Now, there is an ONB $\{e_{n}\}_{n\geq 1}$, such that $h=e_{1}$ and 
$$F = f\left((\int_0^t e_1\, dW_s), \cdots , (\int_0^t
 e_n\,dW_s)\right) $$
This is because the mapping $W: h \rightarrow \int_{0}^{t}h(s)dW_{s}$ is linear.
In fact, let's denote $\int_{0}^{t}h(s)dW_{s} :=  W(h)$.
Now, since
$$DF =\sum_{i=1}^n \frac{\partial }{\partial x_i}f\left(W(e_{1}), \cdots , W(e_{n})\right)e_i.$$
we have that
$$\langle DF,e_{1}\rangle =\sum_{i=1}^n \frac{\partial }{\partial x_i}f\left(W(e_{1}), \cdots , W(e_{n})\right)\langle e_{1},e_{i}\rangle = \frac{\partial }{\partial x_1}f\left(W(e_{1}), \cdots , W(e_{n})\right)$$
B) Because the RV's 
$$\int_{0}^{t}e_{i}(s)dW_{s}$$ are normally distributed when $e_{i}$ is a deterministic function. Therefore, the vector $(W(e_{1}),\cdots, W(e_{n}))$ has an $n$-dimensional normal distribution.
C) This is just the usual integration-by-parts formula. We have, for
$$\phi(\vec{x}) = \exp\left(-\frac{1}{2}\sum_{i=1}^nx_i^2\right)$$ 
and
$$\vec{x} = (x_{1},\cdots, x_{n})$$
$$E[\langle DF, e_{1}\rangle] = (2π)^{−n}\int_{\mathbb R^n}\frac{\partial }{\partial x_1}f(\vec{x})\phi(\vec{x}) d\vec{x} = (2π)^{−n}\int_{\mathbb R^n}f(\vec{x})\phi(\vec{x})x_{1} d\vec{x}$$
because 
$$\lim_{\vec{x}\rightarrow \infty}\phi(\vec{x}) = \vec{0}$$
and the "boundary" element is zero.
You can think of the integration-by-parts as choosing $u$ and $v$ such that
$$u(x_{1})=\exp{\left(-\frac{1}{2}x^{2}_{1}\right)}$$
and
$$dv(x_{1}) = \frac{\partial}{\partial x_{i}}f(\vec{x})dx_{1}$$
Then, you have
$$du(x_{1}) = x_{1}\exp{\left(-\frac{1}{2}x^{2}_{1}\right)}$$
and
$$v(x_{1})=f(\vec{x})$$
Hope that helps.
