principal value as distribution, written as integral over singularity

Let $C_0^\infty(\mathbb{R})$ be the set of smooth functions with compact support on the real line $\mathbb{R}.$ Then, the map

$$\operatorname{p.\!v.}\left(\frac{1}{x}\right)\,: C_0^\infty(\mathbb{R}) \to \mathbb{C}$$

defined via the Cauchy principal value as

$$ \operatorname{p.\!v.}\left(\frac{1}{x}\right)(u)=\lim_{\varepsilon\to 0+} \int_{\mathbb{R}\setminus [-\varepsilon;\varepsilon]} \frac{u(x)}{x} \, \mathrm{d}x \quad\text{ for }u\in C_0^\infty(\mathbb{R})$$

Now why is $$ \lim_{\varepsilon\to 0+} \int_{\mathbb{R}\setminus [-\varepsilon;\varepsilon]} \frac{u(x)}{x} \, \mathrm{d}x = \int_0^{+\infty} \frac{u(x)-u(-x)}{x}\, \mathrm{d}x $$ why is the integral defined on the left.

We can write $$I(\varepsilon):=\int_{\Bbb R\setminus [-\varepsilon,\varepsilon]}\frac{u(x)}xdx=\int_{-\infty}^{-\varepsilon}\frac{u(x)}xdx+\int_{\varepsilon}^{\infty}\frac{u(x)}xdx.$$ In the first integral of the RHs, we do the substitution $t=-x$, then $$I(\varepsilon)=-\int_{\varepsilon}^{+\infty}\frac{u(t)}tdt+\int_{\varepsilon}^{\infty}\frac{u(x)}xdx=\int_{\varepsilon}^{+\infty}\frac{u(t)-u(-t)}tdt.$$ Now we can conclude, since, by fundamental theorem of analysis, the integral $\int_0^{+\infty}\frac{u(t)-u(-t)}tdt$ is convergent. Indeed, $$u(t)-u(-t)=\int_{-t}^tu'(s)ds=\left[su'(s)\right]_{-t}^t-\int_{-t}^tsu''(s)ds\\= t(u'(t)+u'(-t))-\int_{-t}^tsu''(s)ds$$ hence, for $0<t\leq 1$ $$\frac{|u(t)-u(-t)|}t\leq 2\sup_{|s|\leq 1}|u'(s)|+2\sup_{|s|\leq 1}|u''(s)|.$$

Because $1/x$ is an odd function. So, decomposing $u(x)$ in its odd and even parts, that is

$$u(x)=\frac{u(x)+u(-x)}{2}+\frac{u(x)-u(-x)}{2}$$

we have

$$\lim_{\varepsilon \to 0} \int_{\lvert x \rvert > \varepsilon} \frac{u(x)}{x}\, dx= \lim_{\varepsilon \to 0} \left(\int_{\lvert x \rvert > \varepsilon} \frac{u(x)+u(-x)}{2x}\, dx + \int_{\lvert x \rvert > \varepsilon} \frac{u(x)-u(-x)}{2x}\, dx\right)$$

and the first integral on the right hand side vanishes because its argument is odd. On the contrary, the second integral has an even argument, so we can rewrite it as follows:

$$\lim_{\varepsilon \to 0}\int_{\lvert x \rvert > \varepsilon} \frac{u(x)-u(-x)}{2x}\, dx = \lim_{\varepsilon \to 0} \int_\varepsilon^\infty \frac{u(x)-u(-x)}{x}\, dx.$$