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 $0<t\leq 1$
$$\frac{|u(t)-u(-t)|}t\leq 2\sup_{|s|\leq 1}|u'(s)|+2\sup_{|s|\leq 1}|u''(s)|.$$
