(Edit: see below... I misread the question...) One particular way to regularize that integral, relating to meaningful things, is by viewing it as Hilbert transform applied to one Schwartz function, producing (at worst) a tempered distribution, which is then applied to the second Schwartz function.

Part of the point is that the Hilbert transform, or already the principal value integral attached to $1/x$ in one variable, is not a literal integral (against a locally $L^1$ function), although it is a perfectly fine tempered distribution (the derivative of the locally integrable $\log|x|$).

Another implicit aspect, mentioned by @Abdelmalek Abdessalem, is to think about such an extension problem, by (uniquely?!?) characterizing the restricted operator by symmetry/homogeneity, etc., and explicitly ask whether there exists an extension that preserves those features, and whether the extension is unique. (This can be usefully written as a homological issue, as I first saw in a paper of Casselman about extended notions of automorphic forms.) For example, the p.v. integral against $1/x$, rather than $1/x$ itself, is the unique distribution (up to multiples) with that homogeneity and parity, by the classification of distributions on $\mathbb R$ supported at $\{0\}$ and what amounts to an application of the snake lemma. (As usual, what can be usefully written as homological algebra, e.g., a snake lemma application, can also be done directly, in effect reproving a special case of the short-to-long business.)

EDIT: ooops, (thanks to Jean Duchon for noticing my error!): I dropped the absolute value in the question. Thus, this is not the Hilbert transform. In fact, in that vein, the symmetry and mildly homological extension argument proves that there is no distribution extending integration against $1/|x|$ (note the absolute values now), preserving the homogeneity and parity. (The positive-homogeneity is the same as $1/x$, but the parity is opposite. The existence of $\delta$, with the same positive-homogeneity and parity is equivalent to the non-extendability, in effect. The same arguments apply to Hadamard's finie-partie functionals, too.)

(Edit: see below... I misread the question...) One particular way to regularize that integral, relating to meaningful things, is by viewing it as Hilbert transform applied to one Schwartz function, producing (at worst) a tempered distribution, which is then applied to the second Schwartz function.

Part of the point is that the Hilbert transform, or already the principal value integral attached to $1/x$ in one variable, is not a literal integral (against a locally $L^1$ function), although it is a perfectly fine tempered distribution (the derivative of the locally integrable $\log|x|$).

Another implicit aspect, mentioned by @Abdelmalek Abdesselam, is to think about such an extension problem, by (uniquely?!?) characterizing the restricted operator by symmetry/homogeneity, etc., and explicitly ask whether there exists an extension that preserves those features, and whether the extension is unique. (This can be usefully written as a homological issue, as I first saw in a paper of Casselman about extended notions of automorphic forms.) For example, the p.v. integral against $1/x$, rather than $1/x$ itself, is the unique distribution (up to multiples) with that homogeneity and parity, by the classification of distributions on $\mathbb R$ supported at $\{0\}$ and what amounts to an application of the snake lemma. (As usual, what can be usefully written as homological algebra, e.g., a snake lemma application, can also be done directly, in effect reproving a special case of the short-to-long business.)

EDIT: ooops, (thanks to Jean Duchon for noticing my error!): I dropped the absolute value in the question. Thus, this is not the Hilbert transform. In fact, in that vein, the symmetry and mildly homological extension argument proves that there is no distribution extending integration against $1/|x|$ (note the absolute values now), preserving the homogeneity and parity. (The positive-homogeneity is the same as $1/x$, but the parity is opposite. The existence of $\delta$, with the same positive-homogeneity and parity is equivalent to the non-extendability, in effect. The same arguments apply to Hadamard's finie-partie functionals, too.)

One particular way to regularize that integral, relating to meaningful things, is by viewing it as Hilbert transform applied to one Schwartz function, producing (at worst) a tempered distribution, which is then applied to the second Schwartz function.

Part of the point is that the Hilbert transform, or already the principal value integral attached to $1/x$ in one variable, is not a literal integral (against a locally $L^1$ function), although it is a perfectly fine tempered distribution (the derivative of the locally integrable $\log|x|$).

Another implicit aspect, mentioned by @Abdelmalek Abdessalem, is to think about such an extension problem, by (uniquely?!?) characterizing the restricted operator by symmetry/homogeneity, etc., and explicitly ask whether there exists an extension that preserves those features, and whether the extension is unique. (This can be usefully written as a homological issue, as I first saw in a paper of Casselman about extended notions of automorphic forms.) For example, the p.v. integral against $1/x$, rather than $1/x$ itself, is the unique distribution (up to multiples) with that homogeneity and parity, by the classification of distributions on $\mathbb R$ supported at $\{0\}$ and what amounts to an application of the snake lemma. (As usual, what can be usefully written as homological algebra, e.g., a snake lemma application, can also be done directly, in effect reproving a special case of the short-to-long business.)

(Edit: see below... I misread the question...) One particular way to regularize that integral, relating to meaningful things, is by viewing it as Hilbert transform applied to one Schwartz function, producing (at worst) a tempered distribution, which is then applied to the second Schwartz function.

Part of the point is that the Hilbert transform, or already the principal value integral attached to $1/x$ in one variable, is not a literal integral (against a locally $L^1$ function), although it is a perfectly fine tempered distribution (the derivative of the locally integrable $\log|x|$).

Another implicit aspect, mentioned by @Abdelmalek Abdessalem, is to think about such an extension problem, by (uniquely?!?) characterizing the restricted operator by symmetry/homogeneity, etc., and explicitly ask whether there exists an extension that preserves those features, and whether the extension is unique. (This can be usefully written as a homological issue, as I first saw in a paper of Casselman about extended notions of automorphic forms.) For example, the p.v. integral against $1/x$, rather than $1/x$ itself, is the unique distribution (up to multiples) with that homogeneity and parity, by the classification of distributions on $\mathbb R$ supported at $\{0\}$ and what amounts to an application of the snake lemma. (As usual, what can be usefully written as homological algebra, e.g., a snake lemma application, can also be done directly, in effect reproving a special case of the short-to-long business.)

EDIT: ooops, (thanks to Jean Duchon for noticing my error!): I dropped the absolute value in the question. Thus, this is not the Hilbert transform. In fact, in that vein, the symmetry and mildly homological extension argument proves that there is no distribution extending integration against $1/|x|$ (note the absolute values now), preserving the homogeneity and parity. (The positive-homogeneity is the same as $1/x$, but the parity is opposite. The existence of $\delta$, with the same positive-homogeneity and parity is equivalent to the non-extendability, in effect. The same arguments apply to Hadamard's finie-partie functionals, too.)