About infinitesimals and differentiating in high dimensions. I got a little confused here. Please take a look and help me to express correctly the notation below.
Let $f:\Re^n \rightarrow \Re$ be a real valued function over $n$-dimensional space of column vectors $\Re^n$. In addition, let $\mathbf{x}$ be an $n$-dimensional vector in $\Re^n$, which is written as $\mathbf{x} = (x_1,...,x_n)^T$, where $x_i\in\Re$, $i=1,...,n$. Moreover, let $I$ be the following integral:
$$
I(\mathbf{w}) = \int_{\Re^n} \! f(\mathbf{x}) \mathbf{x}^T\mathbf{w}
\,\mathrm{d}x_1\mathrm{d}x_2...\mathrm{d}x_n,
$$
or, if you prefer so, 
$$
I(\mathbf{w}) = \int_{\Re^n} \! f(\mathbf{x}) \mathbf{x}\cdot\mathbf{w}
\,\mathrm{d}x_1\mathrm{d}x_2...\mathrm{d}x_n,
$$
where $\mathbf{w} = (w_1,...,w_n)^T$ is another element of $\Re^n$.
First of all, how could I express the quantity $\mathrm{d}x_1\mathrm{d}x_2...\mathrm{d}x_n$ correctly? Is it equal to $\mathrm{d}^nx$ or $\mathrm{d}\mathbf{x}$? Does the integral $I(\mathbf{w})$ belong to $\Re$ or $\Re^n$? I would like that integral to be a vector, by the way. Also, what about the following integral:
$$
J = \int_{\Re^n} \! f(\mathbf{x}) \mathbf{x}
\,\mathrm{d}x_1\mathrm{d}x_2...\mathrm{d}x_n
$$
Next, by differentiating the above integral with respect to $\mathbf{w}$, I think that the result has to be a vector, no?
Thanks in advance!

Edit:
Thanks to @Anthony Carapetis the following holds:
$$
J = \int_{\Re^n} \! f(\mathbf{x}) \mathbf{x}
\,\mathrm{d}\mathbf{x} 
= \left(\int_{\Re^n} \! f(\mathbf{x})x_1 \,\mathrm{d}\mathbf{x},..., \int_{\Re^n}\! f(\mathbf{x})x_n \,\mathrm{d}\mathbf{x}\right)^T
$$

 A: First of all, both $d^nx$ and $d \bf x$ are common shorthands for the product measure $dx^1 \cdots dx^n$ - so long as you make it clear that $\bf x \in \mathbb R^n$ (which your integral notation does) it's fine.
As for the type of the integrals, a good rule of thumb is that $\int \cdots d\bf x$ is something like a sum $ \sum \cdots $; so you in fact have the incorrect expectations. Since $f(\mathbf x) \mathbf x \cdot \mathbf w$ is a scalar for each $\bf x$ and $\bf w$, its integral $I\bf(w)$ will be a scalar, and likewise the integral of the vectors $f(\mathbf x)\bf x$ will be a vector.
The derivative of a scalar function is indeed a vector - so if you had have been right about $J$ being a scalar then you would've been right about $\nabla J$. With your $J$ as defined (i.e. as a vector), the derivative would in fact be a matrix.
Regarding your edit: the integral $$\int_\mathbb R f(\mathbf x) x_1 dx_1$$ doesn't really make sense - if you're only integrating with respect to $x_1$ then you should be integrating a function of the scalar $x_1$, not a function of the vector $\mathbf x$. What is true is that 
$$J = \int_{\mathbb R^n} \! f(\mathbf{x}) \mathbf{x}
\,\mathrm{d}\mathbf{x} 
= \left(\int_{\mathbb R^n} \! f(\mathbf{x})x_1 \,\mathrm{d}\mathbf{x},..., \int_{\mathbb R^n}\! f(\mathbf{x})x_n \,\mathrm{d}\mathbf{x}\right)^T.$$
