I have read the following statement about Integration by substitution on wikipedia (https://en.wikipedia.org/wiki/Integration_by_substitution):
Let $U$ be an open set in $\mathbb{R}^{n}$ and $\varphi:U\rightarrow\mathbb{R}^{n}$ an injective differentiable function with continuous partial derivatives, the $D\varphi$ Jacobian of which is nonzero for every $x$ in $U$. Then for any real-valued, compactly supported, continuous function $f$, with support contained in $\varphi(U)$,
$$\int_{\varphi\left(U\right)}f\left(\mathbf{v}\right)d\mathbf{v}=\int_{U}f\left(\varphi\left(\mathbf{u}\right)\right)\left|det\left(D\varphi\right)\left(\mathbf{u}\right)\right|d\mathbf{u}.$$
I was wondering if this statement holds when $f$ is not real valued, but $f:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$, and interpret $f\left(\mathbf{v}\right)d\mathbf{v}$ as a kind of dot product? (I guess for $f:\mathbb{R}^{n}\rightarrow\mathbb{R}^{m}$, $m\neq1$,$m\neq n$ this interpretation is incorrect because the dimensions wouldn't match, or I don't know if there is any other intepretation...)
If this can hold when $f:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$, then can you give me an example where this is used? Furthermore, is this relation holds for other abstract spaces, where dot product is defined?
Currently, I'm learning about line integral where $\mathbf{F}:U\subset\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$ is a vector field, and the line integral on $\Gamma$ curve is defined as $$\int_{\Gamma}\mathbf{F}\left(\mathbf{r}\right)\cdot d\mathbf{r}=\int_{a}^{b}\mathbf{F}\left(\mathbf{r}\left(t\right)\right)\cdot\mathbf{r}'\left(t\right)dt,$$
where $\cdot$ is the dot product, and $\mathbf{r}:\left[a,b\right]\rightarrow\Gamma$ is a parametrization of $\Gamma$. (See e.g.: https://en.wikipedia.org/wiki/Line_integral) I see the similarities between the theorem above and the last definition, but they are not exactly the same. On the latter, there is a concrete gradient vector of $\mathbf{r}$, but in the previous theorem there is a determinant, which is a scalar. So my main goal is to find the connection between these two equations if there is any...
The statement you quoted from Wikipedia holds only for real-valued functions. When dealing with vector-valued functions, we need to use a different approach.
One way to integrate a vector-valued function over a region is to integrate its components separately. That is, if f:Rn→Rn is a vector-valued function, we can write it as f=(f1,f2,...,fn), where each fi:Rn→R is a real-valued function. Then, we can integrate each fi over the region separately:
∫Rn f(v)dv = ∫Rn f1(v)dv + ∫Rn f2(v)dv + ... + ∫Rn fn(v)dv.
So, in general, we cannot interpret f(v)dv as a dot product because the dimensions do not match.
Regarding your question about the relationship between the integration by substitution formula and the line integral formula, there is indeed a connection. In fact, the integration by substitution formula is a generalization of the line integral formula.
To see this, let F:U⊂Rn→Rn be a vector field and let Γ be a smooth curve in U parametrized by r:[a,b]→Γ. Then, we can write the line integral of F over Γ as:
∫Γ F(r)⋅dr = ∫a^b F(r(t))⋅r'(t)dt
where ⋅ is the dot product. Now, let φ:U→Rn be a diffeomorphism (i.e., a bijective, differentiable function with a differentiable inverse) and let f:Rn→Rn be a vector field. Then, we can use the integration by substitution formula to write:
∫φ(U)f(v)dv = ∫U f(φ(u))|det(Dφ)(u)|du
where dv is the volume element in Rn and du is the volume element in U. Now, if we set f = F∘φ^-1, we get:
∫φ(U)F(r)⋅dr = ∫φ(U)F(φ(u))⋅|det(Dφ)(u)|du
where r = φ(u) is a point on Γ. Now, we can apply the change of variables u = φ^-1(r) and get:
∫φ(U)F(r)⋅dr = ∫U F(φ(u))⋅|det(Dφ)(u)|du = ∫Γ F(r)⋅n(r)ds
where n(r) is the unit normal vector to Γ at r and ds is the arc length element along Γ. This is precisely the line integral formula.
So, the integration by substitution formula is a generalization of the line integral formula, where we can integrate vector fields over arbitrary regions using diffeomorphisms.