How to consider an integral of the form $\int\limits_{aLet $X: [a,b] \to \mathbb R^{d}$ be some smooth path.
I am having trouble understanding what is actually being expressed in the integral of the form
$$\int\limits_{a<u_{1}<...<u_{n}<b}dX_{u_{1}}\otimes...\otimes dX_{u_{n}}\in (\mathbb R^{d})^{\otimes n}.$$
Is it just riemann integration in every "component" of the tensor? Any intuition behind the actual expression is much appreciated.
EDIT: For greater context, I am looking at the definition of a signature of a path from Differential Equations driven by Rough Paths by Terry Lyons on page 30. Here the signature is denoted as $\textbf{X}$, where
$$\textbf{X}=(1,\textbf{X}^{1},...,\textbf{X}^{n},....)$$
such that $$\textbf{X}^{n}=\int\limits_{a<u_{1}<...<u_{n}<b}dX_{u_{1}}\otimes...\otimes dX_{u_{n}}\in (\mathbb R^{d})^{\otimes n}$$
 A: Anthony Carapetis is right, I want to expand a bit on his comment, starting from the definition of signature and then moving to the notation used and how it makes sense.
I'll start looking at the problem the way my calculus teacher would look at it: purely analytical. I also think this is the way people started considering quantities that led to the integral you wrote, instead of trying to integrate funky tensor products hoping to find something.
What you want to define are iterated integrals of your path and you do it in an iterative fashion. One of the motivations to consider these types of integral comes from the Picard iteration for ODEs. For simplicity I'll assume $d=2$.
The first quantity we consider is a $2$-vector with coordinates $$S_{a, b}^i(X) := \int_a^bdX_s^i,$$
for $i=1, 2$. Notice that, for each $i$, the integral has meaning in the Stieltjes sense and for $X$ smooth enough it is equal to $\int_a^b\dot{X}^i(s)ds$.
Now we define vectors that live in higher dimensions, with components $$S_{a, b}^{ij}(X) := \int_a^b S_{a, s}^i(X)dX_s^j = \int_a^b\int_a^sdX^i_r dX^j_s = \int_{a<r<s<b}\dot{X}^i(r) \dot{X}^j(s)drds.$$
As the indices $i, j$ change you end up with a 4 dimensional object that can be written as
$$\begin{pmatrix} \int_{a<r<s<b}\dot{X}^1(r) \dot{X}^1(s)drds & \int_{a<r<s<b}\dot{X}^1(r) \dot{X}^2(s)drds\\\ \int_{a<r<s<b}\dot{X}^2(r) \dot{X}^1(s)drds & \int_{a<r<s<b}\dot{X}^2(r) \dot{X}^2(s)drds\end{pmatrix}.$$
Now, since the integral of a vector valued function is defined component-wise, one might be tempted to write the above matrix as $$\int_{a<r<s<b}\dot{X}(r)\otimes\dot{X}(s)drds.\tag{1}$$
This last expression is what you compute, but it is a bit different from what you wrote. If we use $$\int_{a<r<s<b}dX^i_r dX^j_s,\ \ \ \ i, j=1, 2$$ instead of $\int_{a<r<s<b}\dot{X}^i(r) \dot{X}^j(s)drds$ in the matrix above, then someone might be tempted to write (justified at least by notational convenience)
$$\int_{a<r<s<b}dX_r\otimes dX_s,\tag{2}$$
to indicate the same integral as $(1)$ (you can actually put some differential geometry in here, to make sense of tensor products of differentials and their integrals).
The same line of thought applies also to "higher order" integrals defined iteratively $S_{a, b}^I(x),$ for a multiindex $I$. For instance the next iterated integral is $9$-dimensional and has components $$S_{a, b}^{ijk}(X) := \int_a^b S_{a, s}^{ij}(X)dX_s^k = \int_a^b\int_a^s\int_a^r dX^i_t dX^j_rdX^k_s.$$ Moreover all of the above works as long as the path is sufficiently smooth. If the path is not smooth, you can still write $(2)$, to indicate the matrix with components $\int_{a<r<s<b}dX^i_r dX^j_s$ and then compute this relying on what this last integral actually means.
Here you can find some introductory material on signatures, which presents some examples and computations before introducing tensor products.
