If we consider $\vec{E}$ the electric field in $R^n$ and we have : $$\int_{R}^{\infty}\vec{E}(\vec{r})d\vec{r}=0$$ where $R$ is in $R^n$ and $\vec{r}$ is in $R^n$ $$\vec{E}(\vec{r})=\frac{1}{4\pi \epsilon_0 \mid \vec{r}\mid^2}\vec{e_r}$$
Then do we have that $\vec{E}=0$ everywhere is we make the assumption that $E(\vec{r})=E(r)$ ? Thanks
On
The work required to bring a charge $Q$ from infinitely far away and place it at a point $\vec{R}$ is
$$W=Q(V(\vec{R})-V(\infty)$$
If we suppose the electric field is spherically symmetric (which, presumably,you are doing), then this is just
$$W=Q(V(R)-V(\infty))$$
The integral that you have is
$$\int_R^\infty E(r)dr=V(R)-V(\infty)$$
just based on the definition of the electric field (in the spherical symmetric case)
$$E(r)=-\frac{\text{d}}{\text{d}r}V(r)$$
Thus in your case, $W=0$ to bring a charge from $r=\infty$ to $r=R$. Because we had to make the assumption that the electric field was spherically symmetric, it follows that for $W$ to be zero, there must be no charges in the space because if it is spherically symmetric, we can write the field as having some net charge at the origin. If $W=0$, then this net charge at the origin must be zero, and thus $E(r)=0~\forall~r$.
Please note I say "spherical" but this works for all $n$.
The answer is "no". The integral can be $0$ also for non-zero field. For example consider two concentric charged spheres $R_1<R_2$ with $$\frac{Q_1}{R_1}=-\frac{Q_2}{R_2}.$$
Then for $R<R_1$ the integral in question is $0$ but the field outside the inner sphere is certainly not.