Looking at Fourier transforms can provide an intuitive context for the Hadamard finite part (F.P.) regularization.

Monkey around with this ladder of expressions (understood as F.P.s):

$$A)\int_{-\infty }^{\infty }\frac{exp(i2\pi fx)}{(i2\pi f)^2}df=\frac{sgn(x)}{2}x = \int_{0}^{x}\frac{sgn(u)}{2}du$$

$$B) \int_{-\infty }^{\infty }\frac{exp(i2\pi fx)}{i2\pi f}df=\frac{sgn(x)}{2}$$
$$C)\int_{-\infty }^{\infty }exp(i2\pi fx)df=\delta(x) = \frac{d}{dx}\frac{sgn(x)}{2}$$

To descend the ladder, formally take the derivative of both sides above or of the explicit F.P. expressions below (second equalities), which is equivalent to multiplying the integrands above by $i2\pi f$. To climb, integrate from $0$ to $x$ both sides below, using the explicit expressions for the integrands for the F.P. given below in the second equalities, or simply divide the integrands on the L.H.S. above by $i2\pi f$. (Note that $x$ can be negative or positive and that the Dirac delta function contributes only a value of $1/2$ when evaluated on the boundary of the integral.) So, the explicit F.P. integrals below commute with differentiation and integration w.r.t. $x$ and can be *naturally defined* in terms of the two ops, and the implicit symbolic formulas above allow us to formally retain the representation of the two ops as multiplication and division operations in the Fourier transform integrands.

For finite limits for the integrals, you'll end up with the expressions on the right above being convolved with a sinc function with some phase, that should agree with the L.H.S. if the Hadamard finite finesse is applied.

The OP's example is closely related to A) with $x=0$ and is more palatable within this context. In detail, in the limits $\varepsilon \to 0^+$ and $L \to \infty,$

$C)\displaystyle\delta(x)=\int_{-L }^{L }exp(i2\pi fx)df$

$B)\displaystyle\frac{sgn(x)}{2}=F.P.\int_{-\infty }^{\infty }\frac{exp(i2\pi fx)}{i2\pi f}df=\left [ \int_{\varepsilon}^{L }+\int_{-L }^{-\varepsilon} \right ]\frac{exp(i2\pi fx)-1}{i2\pi f}df$

$=\displaystyle\left [ \int_{\varepsilon}^{L }+\int_{-L }^{-\varepsilon} \right ]\frac{exp(i2\pi fx)}{i2\pi f}df-\frac{ln(L/\varepsilon)}{i2\pi}-\frac{ln(\varepsilon/L)}{i2\pi}$

$=\displaystyle\left [ \int_{\varepsilon}^{L }+\int_{-L }^{-\varepsilon} \right ]\frac{exp(i2\pi fx)}{i2\pi f}df=C.P.V\int_{-\infty }^{\infty }\frac{exp(i2\pi fx)}{i2\pi f}df$

where $F.P.$ denotes the Hadamard finite part and $C.P.V.$, the Cauchy principle value. (Of course, the $\frac{1}{f}$ terms pose no serious problems since $\frac{1}{f}$ is an odd function and we are integrating symmetrically about $0$.)

Similarly,

$A)\displaystyle\frac{sgn(x)}{2}x=F.P.\int_{-\infty }^{\infty }\frac{exp(i2\pi fx)}{(i2\pi f)^2}df=\left [ \int_{\varepsilon}^{L }+\int_{-L }^{-\varepsilon} \right ]\frac{exp(i2\pi fx)-(1+i2\pi fx)}{(i2\pi f)^2}df$

$=\displaystyle\left [ \int_{\varepsilon}^{L }+\int_{-L }^{-\varepsilon} \right ]\frac{exp(i2\pi fx)}{(i2\pi f)^2}df-\frac{2}{(i2\pi)^2 \varepsilon}=\frac{|x|}{2}.$

It's even more convincing when you plot the integrals (including C) and observe how they evolve as $L$ increases for small $\varepsilon.$

Another context for the Hadamard finite limit is given in MSE-Q13956.

For a comparison of different methods of regularization for integrals of this type see http://arxiv.org/abs/hep-th/0202023 "Improved Epstein-Glaser renormalization in coordinate space I. Euclidean framework" by Gracia-Bondia (pg. 14-).