Rationale for Hadamard's finite part of a divergent integral (Note: I asked this question a few days ago on math.stackexchange but didn't get any responses. I've therefore decided to post it here instead.)
I have a problem justifying throwing away the divergent term in order to obtain Hadamard's finite part. I find this step to be highly unusual, and it is not obvious to me how the resulting expression can be valid. I'd appreciate, if possible, intuitive arguments -- I assume there are some. Its not the mathematics that I have problems with at this stage. Its the intuition.
For example (taken from the Wikipedia page) the finite part of the following integral
$$\int_a^b \frac{f(t)}{(t-x)^2}\, dt = \lim_{\varepsilon \to 0} \left[ \int_a^{x-\varepsilon}\frac{f(t)}{(t-x)^2}\,dt + \int_{x+\varepsilon}^b\frac{f(t)}{(t-x)^2}\,dt -\frac{2f(x)}{\varepsilon} \right]$$
involves throwing away the term $\frac{2f(x)}{\varepsilon}$. I find it hard to justify this step especially when the term is neither finite nor negligible.
 A: This is a regularization of an integral. Roughly speaking, you are looking for a generalized function which is equal to $(t-x)^{-2}$ on the open set where $t\neq x$. Such regularizations arise naturally when taking derivatives. For instance, the distributional derivative of the integrable function $\ln|t|$ is the Cauchy principal value regularization of $1/t$. The second derivative is precisely the distribution defined above (with $x=0$).
A: 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-).
Edit 2/11/21: Another example, in fractional calculus, of where the Hadamard finite part could obviously be invoked  where it is equivalent to another route of analytic continuation is in my reply to this MO-Q.
A: This can be viewed as a meromorphic continuation in s of the distribution which is integration against |x-t|^s in t. M. Riesz (1938)first observed that this is a meromorphic continuation of convergent integrals, Gelfand-Shilov (1958) formalized this in the context of Schwartz' distributions. Gelfand-Graev's volume I discusses many such examples. The exponent -2 is not at the boundary of the region re(s)>-1 where the integral converges (absolutely), so the integral does not have an obvious, natural evaluation as a limit.
The seemingly whimsical discarding of the 2f(x)/epsilon term is merely an obscured version of a correct computation of the meromorphic continuation. 
Meromorphic continuation is one way to ensure uniqueness of a 'regularization' procedure, which otherwise may be ambiguous, or accidentally fail to have verifiable continuity and other properties.
