TITLE: Anti-delta function?
QUESTION [23 upvotes]: Did anyone ever consider a "function" or "distribution" $F(x)$ with the following property:
its integral $\int_a^b F(x)\,dx=0$ for any finite interval $(a,b)$ but $\int_{-\infty}^\infty F(x)\,dx=1$? It definitely can be seen as a limit of smooth functions, and its Fourier transform will be a finction, equal to $1$ at $0$ and otherwise $0$.
I think such object for sure should be considered and named by someone.

REPLY [9 votes]: This is more a long comment following the ones of Mateusz Kwasnicki and Anixx and the nice answers given by Michael Hardy, Gro-Tsen, et al.
As stated by Mateusz, such a mathematical objiect cannot be a function nor a distribution: indeed in this last case, without a compactification of the real line (as suggested by Gro-Tsen'asnswer), there's no $\varphi\in C_0^\infty$ whose support is compact and includes the point at infinity, so the space of functionals on function spaces made of this functions is necessarily empty. However, this limitation is not inherited by hyperfunctions: there exist hyperfunctions, notably Fourier hyperfunctions, "supported" (it is better to say "carried", since the support of a distribution and the carrier of a hyperfunction are not exactly the same) at $x=+\infty$. The first explicit example of such kind of generalized functions was constructed by Morimoto and Yoshino [2], and was lated simplified and extended by Akira Kaneko in [1] (who also deals with the higher dimensional case). The method (as explained by Kaneko [1] pp. 99-100) consists in choosing a defining function for the sought for hyperfunction which grows very fast on the real axis, for example $e^{e^{z^2}}=\exp\big(e^{z^2}\big)$, and a path $\gamma$ to $+\infty$ along which it rapidly decreases: then the holomorphic function $F(z)$ defined as
$$\DeclareMathOperator{\dmu}{\operatorname{d}\!}
F(z) = \frac{1}{2\pi i}\int\limits_\gamma \frac{\exp\big(e^{\zeta^2}\big)}{\zeta-z} \dmu\zeta
$$
is such that its "boundary value" (i.e. the defined hyperfunction) $f(x) =F(x+i0)-F(x-i0)$ is $0$ for all finite $x\in\Bbb R$, as it can be proved by deforming the integration path and using the standard Cauchy integral theorem. But nevertheless it is
$$
F(x)= \exp\big(e^{x^2}\big) + \frac{1}{2\pi i}\int\limits_\gamma \frac{\exp\big(e^{\zeta^2}\big)}{\zeta -x} \dmu\zeta
$$
thus the "boundary value" of $F$ produces a non trivial Fourier hyperfunction carried at $+\infty$.
Bibliography
[1] Akira Kaneko, "Explicit construction of Fourier hyperfunctions supported at infinity" (English), Kawai, Takahiro (ed.) et al., Microlocal analysis and complex Fourier analysis, River Edge, NJ: World Scientific Publishers, ISBN 981-238-161-9/hbk, pp. 99-114 (2002), MR2068531, Zbl 1046.46031.
[2] Mitsuo Morimoto, Kunio Yoshino, "Some examples of analytic functionals with carrier at the infinity" (English),Proceedings of the Japan Academy, Series A 56, 357-361 (1980), MR0596004,  Zbl 0471.46024.