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):

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

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

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-).

3 Improved explanation; added 6 characters in body

Maybe this perspective

Looking at Fourier transforms can give a more provide an intuitive feel context for the Hadamard finite part (F.P.) regularization. I think of the finite part in these types of divergent integrals intuitively by first thinking of it as a way of finessing a Fourier transform pair to agree.

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

To descend the ladder, formally take the derivative of both sides . You can ascend on above or of the R.H.Sexplicit F.P. expressions below, which is equivalent to multiplying the integrands above by integrating $i2\pi f$. To climb, integrate from $0$ to $x$, noting 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. On integral.) So, the L.H.Sexplicit F.P. you integrals below commute with differentiation and integration w.r.t.$x$and can simply divide be naturally defined in terms of the integrands by$i2\pi f$two ops, and the implicit symbolic formulas above allow us to climb formally retain the ladder and regard this operation representation of the two ops as an antiderivativemultiplication 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. So, the Hadamard finite part allows us to formally retain the natural operations on the Fourier transform. The OP's example is closely related to A) with$x=0$and is easier to accept more palatable within this context. In detail, in the limits$\varepsilon \to 0^+$and$L \to \infty,$$$B)\frac{sgn(x)}{2}=F.P.\int_{-\infty C)\displaystyle\delta(x)=\int_{-L }^{L }exp(i2\pi fx)df B)\displaystyle\frac{sgn(x)}{2}=F.P.\int_{-\infty }^{\infty }\frac{exp(i2\pi fx)}{i2\pi f}df=\left [ \int_{\varepsilon}^{L }+\int_{-L }^{-\varepsilon} \right ]\frac{exp(i2\pi fx)-1}{i2\pi f}df$$f}df$

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

where $F.P.$ denotes the Hadamard finite part and $C.P.V.$, the Cauchy principle value, and, of . (Of course, the $\frac{1}{f}$ terms pose no real serious problems since $\frac{1}{f}$ is an odd function and we are integrating symmetrically about $0$.0$.) $$A)\frac{sgn(x)}{2}x=F.P.\int_{-\infty A)\displaystyle\frac{sgn(x)}{2}x=F.P.\int_{-\infty }^{\infty }\frac{exp(i2\pi fx)}{(i2\pi f)^2}df=\left [ \int_{\varepsilon}^{L }+\int_{-L }^{-\varepsilon} \right ]\frac{exp(i2\pi fx)-(1+i2\pi fx)}{(i2\pi f)^2}df$$f)^2}df$

$$=\left =\displaystyle\left [ \int_{\varepsilon}^{L }+\int_{-L }^{-\varepsilon} \right ]\frac{exp(i2\pi fx)}{(i2\pi f)^2}df-\frac{2}{(i2\pi)^2 \varepsilon}=\frac{|x|}{2}.$$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.$2 Put in a missing factor Maybe this perspective can give a more intuitive feel for the Hadamard finite part regularization. I think of the finite part in these types of divergent integrals intuitively by first thinking of it as a way of finessing a Fourier transform pair to agree. Monkey around with this ladder of expressions (understood as finite parts): $$A)\int_{-\infty }^{\infty }\frac{exp(i2\pi fx)}{(i2\pi f)^2}df=\frac{sgn(x)}{2}x$$ $$B) \int_{-\infty }^{\infty }\frac{exp(i2\pi fx)}{i2\pi f}df=\frac{sgn(x)}{2}$$ $$C)\int_{-\infty }^{\infty }exp(i2\pi fx)df=\delta(x)$$ To descend the ladder, formally take the derivative of A) to obtain B) and do it again for B) to obtain C)both sides. You can ascend on the R.H.S. by integrating from$0$to$x$, noting that$x$can be negative or positive and that the delta function contributes only a value of$1/2$when evaluated on the boundary of the integral. On the L.H.S. you can simply divide the integrands by$i2\pi f$to climb the ladder and regard this operation as an antiderivative. For finite limits for the integrals, you'll end up with the expressions on the right being convolved with a sinc function with some phase, that should agree with the L.H.S. if the Hadamard finite finesse is applied. So, the Hadamard finite part allows us to formally retain the natural operations on the Fourier transform. The OP's example is closely related to A) with$x=0.$x=0$ and is easier to accept within this context. In detail, in the limits $\varepsilon \to 0^+$ and $L \to \infty,$

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

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

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

$$=C.P.V\int_{-\infty f}df=C.P.V\int_{-\infty }^{\infty }\frac{exp(i2\pi fx)}{i2\pi f}df$$

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

Similarly,

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

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

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

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