I have stated reading Linear integral equation.\
It is mention that for a function $u:\mathbb{R}^{N}\rightarrow\mathbb{R}$ and $x\in\mathbb{R}$ consider the linear integrodifferential operator defined as follows: \begin{equation}\tag{1} Lu(x)=\int_{\mathbb{R}^{N}}[u(x+y)-u(x)-(\nabla(u(x))y)\chi_{B_{1}}]K(y)dy, \end{equation}
where $k:\mathbb{R}^{N}\rightarrow[0,\infty)$ is a measurable and symmetric kernel.\ It is mentioned that with the symmetry of $k,$ $L$ can be written as follows: \begin{equation}\tag{2} Lu(x)=p.v.\int_{\mathbb{R}^{N}}[u(x+y)-u(x)]k(y)dy \end{equation} \begin{equation}\tag{3} =\frac{1}{2}\int_{\mathbb{R}^{N}}[u(x+y)+u(x-y)-2u(x)]K(y)dy. \end{equation} Question: In passing from equation (1) to (2) how does the principle value (p.v. )comes into the picture.
For clarity assume that $u\in C_c^\infty(\mathbb{R}^N)$.
Such operators arise for example in the theory of Lévy processes. This is the Lévy--Khintchine representation of the generator of a process with jump intensity given by (the Lévy measure) $K(y) dy$.
A prominent example, which I will focus on, is the fractional Laplacian, with $K(y) = |y|^{-N-\alpha}$, where $\alpha \in (0,2)$. In particular, operators with $\alpha \geq 1$ are considered.
Note that the expressions with $u$ in (1) and (3) are dominated by $\min \{1,|y|^2\}$ (Taylor expansion) and so they integrate $K$ (by the definition, the Lévy measure integrates $\min \{1,|y|^2\}$).
In (2) however, the integrand behaves only like $|y|$ for small $|y|$ and it may happen that the integral over $\mathbb{R}^N$ does not converge (for example, when $\alpha \geq 1$).
To pass from (1) to (2), you first represent the integral as the limit of integrals over $B(0,\epsilon)^c$ with $\epsilon \to 0$. You split as follows: $$\int_{B(0,\epsilon)^c} (u(x+y) - u(x) - y\nabla u(x)\chi_{B_1}(y)) K(y) dy = \int_{B(0,\epsilon)^c} (u(x+y) - u(x)) K(y) dy - \int_{B(0,\epsilon)^c}y\nabla u(x) K(y) \chi_{B_1}(y)dy.$$ The last integral is equal to zero by the symmetry of $K$. By passing to the limit you get that (1) = (2), but on the right hand side you have to write $p.v.$ because in general the integral over $\mathbb{R}^N$ does not exist.