$V=\int _C\vec{F}d{x}$ where $C$ is a path that "goes to infinity". Does the path chosen matter? According to Wikipedia "The gravitational potential $V$ at a distance $x$ from a point mass of mass $M$ can be defined as the work $W$ that needs to be done by an external agent to bring a unit mass in from infinity to that point: $$V(\vec{x}) = \frac{1}{m}\int^x _{\infty} \vec{F}\cdot d\vec{x} = \frac{1}{m}\int^x _{\infty}G\frac{Mm}{x^2}dx=-\frac{GM}{x}$$
where $G$ is the gravitational constant, and $\vec{F}$ is the gravitational force."

Consider a point mass situated at $\vec{x}_0$ with a mass $M$. If we let $C$ be a path parametrized by $f:[0,\infty)\rightarrow \mathbb{R}^n$ where $f(0)=\vec{x}$ and $$\lim _{t\rightarrow \infty} |f(t)-\vec{x}|=\infty$$
does the equality $$\int _C\vec{F}\cdot d\vec{x}=V(\vec{x})$$
hold? That is, does the way in which the path goes to infinity matter?
 A: Your question actually has nothing to do with physics, if we strip it down to the essentials you're essentially asking "when does the (second) Fundamental theorem of calculus hold for improper integrals"?
tl;dr: As long as your path is reasonably enough behaved ($C^1$ and all the integrals converge in the improper case), it doesn't matter.

First of all, you write $\Bbb{R}^n$, but the formulas you write down are specific to $\Bbb{R}^3$ (in the general case the Newtonian potential is proportional to $\frac{1}{|\xi|^{n-2}}$). Note that on $\Omega=\Bbb{R}^3\setminus\{0\}$, we can always define the functions $V:\Omega\to\Bbb{R}$ and $\mathbf{F}:\Omega\to\Bbb{R}^3$ defined as
\begin{align}
V(\xi):=\frac{-GM}{|\xi|}\quad \text{and}\quad \mathbf{F}(\xi):=-\frac{GM\,\xi}{|\xi|^3}
\end{align}
(I ignored the $m$, so my $\mathbf{F}$ is the force per unit mass)
Then, a simple verification shows that $\mathbf{F}=-\text{grad}(V)$. Now when you want to consider line integrals of a vector field over a curve, usually one considers compact curves, which can be parametrized by a $C^1$ (or piecewise $C^1$) path $\gamma:[a,b]\to \Omega$, and then by definition,
\begin{align}
\int_{\gamma}\mathbf{F}\cdot d\mathbf{l}:=\int_a^b\mathbf{F}(\gamma(t))\cdot \gamma'(t)\,dt
\end{align}
The reason we typically assume $\gamma:[a,b]\to\Omega$ to be $C^1$ is so that the integrand $(\mathbf{F}\circ\gamma) \cdot \gamma'$ is continuous on $[a,b]$ and thus trivially Riemann-integrable on $[a,b]$, so the RHS is actually a proper object.
In the case when $\gamma$ has a domain which is an interval $I$ with lower endpoint $\alpha$, upper endpoint $\beta$ (either endpoint could be infinite, either coud be closed or open; so $(\alpha,\beta)$, $[\alpha,\beta)$ or $(-\infty,89]$ or literally interval), then a-priori, $(\mathbf{F}\circ \gamma)\cdot \gamma'$ is continuous but not necessarily Riemann-integrable on $I$. It will be Riemann-integrable over every compact subinterval $[a,b]\subset I$. Thus, we now have to make an extra assumption on the curve, namely that $\gamma:I\to \Omega$ is $C^1$ and that the following limit exists:
\begin{align}
\lim_{\substack{b\to \beta^-\\a\to \alpha^+}}\int_a^b\mathbf{F}(\gamma(t))\cdot \gamma'(t)\,dt
\end{align}
In this case, this limit is the definition of the symbol $\int_{\gamma}\mathbf{F}\cdot d\mathbf{l}$.
In other words, we require that $(\mathbf{F}\circ \gamma)\cdot \gamma'$ be improperly Riemann-integrable on $I=(\alpha,\beta)$. If we make this assumption, then certainly, by unwinding the definitions,
\begin{align}
\int_{\gamma}\mathbf{F}\cdot d \mathbf{l}&:=\lim_{\substack{b\to \beta^-\\a\to \alpha^+}}\int_a^b\mathbf{F}(\gamma(t))\cdot \gamma'(t)\,dt\\
&=\lim_{\substack{b\to \beta^-\\a\to \alpha^+}}\int_a^b-(V\circ \gamma)'(t)\,dt\tag{chain rule}\\
&=\lim_{\substack{b\to \beta^-\\a\to \alpha^+}}\left[V(\gamma(a))-V(\gamma(b))\right]\tag{FTC}
\end{align}
So far, we have simply dealt with the definition of the improper Riemann line integral. Finally, for your case of interest which is a path $\gamma:[0,\infty)\to\Omega$, such that $\gamma(0)=x_0$ and  $\lim\limits_{t\to \infty}|f(t)|=\infty$ (i.e $\gamma$ goes off to infinity). Let $\gamma_{op}$ denote the same curve, traversed in the opposite orientation. Then, we aim to show that $-\int_{\gamma_{op}}\mathbf{F}\cdot d\mathbf{l}=V(x_0)$ (i.e integrating from $\infty$ to $x_0$ of minus the gravitational force is the gravitational potential at $x_0$). This is simple (up to being careful of minus signs due to physics):
\begin{align}
-\int_{\gamma_{op}}\mathbf{F}\cdot d\mathbf{l}&:=\int_{\gamma}\mathbf{F}\cdot d\mathbf{l}\\
&=\text{as above}\\
&=\lim\limits_{\substack{b\to \infty \\ a\to 0^+}}\left[V(\gamma(a))-V(\gamma(b))\right]\\
&=\lim\limits_{\substack{b\to \infty \\ a\to 0^+}} \left[\frac{-GM}{|\gamma(a)|}+\frac{GM}{|\gamma(b)|}\right]\\
&=\frac{-GM}{|\gamma(0)|}+0\\
&=\frac{-GM}{|x_0|}\\
&=V(x_0)
\end{align}
Here, the limits were evaluated since $\gamma$ is continuous at $0$, and since $|\gamma(b)|\to \infty$ as $b\to \infty$, so the reciprocal converges to $0$.
A: I dont think it does, because gravity is a conservative force.
The only thing that matter are thr first and last positions
Whatever path you take, your work done by gravity remains the same, and your GPE also is the same at the point
