The most simple infinite number $\omega$ is the germ at infinity of the function $f(x)=x$. Denoted in Levi-Civita field as $\varepsilon^{-1}$ (so has inverse $\varepsilon$ in that field, as well as in Hardy field). Can be represented as divergent integrals $\int_0^\infty dx$ and $\int_0^\infty\frac1{x^2}dx$ (via Laplace transform). Half the numerocity of integers, equal to numerocity of even or odd numbers. Divergent sum representation: $1/2+\sum_{k=1}^\infty 1$. Has finite part (regularized value) $0$. Also, the germ of the function $\frac1x$ at zero (from above). If we modify Delta distribution to behave like a function, $\omega=\pi\tilde{\delta}(0)$ (via Fourier transform).
$\lambda=-\int_0^1 \frac1x dx$. This is negatively infinite constant. Via its definition can be used to extend logarithmic function to zero (thus $\lambda=\ln 0$). Has finite part (regularized value) $-\gamma$. Since Harmonic series is the Riemann sum of this integral, $\lambda$ is the negative of the sum of Harmonic series: $\lambda=-\sum_{k=1}^\infty \frac1k$. From this equality (or via Laplace transform), $\int_1^\infty\frac1xdx=-\lambda-\gamma$. Thus, $\int_0^\infty \frac1x dx=-2\lambda-\gamma$. Other integral representations include: $\int_0^\infty\frac{e^{-x}-1}{x}dx,-\int_0^\infty \frac{1}{x^2+x}dx$. Can be expressed via $\omega$: $\lambda=-\ln\omega-\gamma$. Some divergent integrals can be expressed in terms of $\lambda$: $\int_0^\infty \frac{\ln t}{t} \, dt=\gamma\lambda+\gamma^2/2,$ $\int_0^{\infty } | \tan x| \, dx=-\lambda\omega$. If we generalize the notions of periods and Chow's $EL$-numbers to our extended set, then $\lambda$ would be both, because it can be represented as $\int_0^1 \frac{-1}t dt$ (integral of an algebraic function over algebraic domain) and $\ln 0$ respectively. Logarithms of zero divisors can be expressed via $\lambda$, in split-complex numbers: $\ln \left(\frac{a j}{2}+\frac{a}{2}\right)=\frac{j}{2} (\ln a-\lambda)+\frac{1}{2} (\ln a+\lambda)$, in dual numbers $\ln \varepsilon=\lambda+\varepsilon \omega$, $\varepsilon^\varepsilon=1+\varepsilon(1+\lambda)$.
$-\lambda-\gamma$. Equal to $\ln \omega$ and germ at infinity of logarithmic function $f(x)=\ln x$. Has finite part $0$. Unlike $\lambda$, is neither generalized period, nor EL-number. Has representations $\int_1^\infty\frac1xdx,$ $\int_0^\infty \frac{e^{-x}}{x}dx,$ $\int_0^\infty \frac{x-\log (x)-1}{(x-1)^2}dx$.
$\int_{-\infty}^\infty e^x dx$. Equal to $e^\omega$. Has finite part $0$. Germ of the function $f(x)=e^x$ at infinity.
$\sum_{k=1}^\infty k=\frac{\omega^2}2-\frac1{12}$ Has finite part $-1/12$.
The most simple infinite number $\omega$ is the germ at infinity of the function $f(x)=x$. Denoted in Levi-Civita field as $\varepsilon^{-1}$ (so has inverse $\varepsilon$ in that field, as well as in Hardy field). Can be represented as divergent integrals $\int_0^\infty dx$ and $\int_0^\infty\frac1{x^2}dx$ (via Laplace transform). Half the numerocity of integers, equal to numerocity of even or odd numbers. Divergent sum representation: $1/2+\sum_{k=1}^\infty 1$. Has finite part (regularized value) $0$. Also, the germ of the function $\frac1x$ at zero (from above). If we modify Delta distribution to behave like a function, $\omega=\pi\tilde{\delta}(0)$ (via Fourier transform).
$\lambda=-\int_0^1 \frac1x dx$. This is negatively infinite constant. Via its definition can be used to extend logarithmic function to zero (thus $\lambda=\ln 0$). Has finite part (regularized value) $-\gamma$. Since Harmonic series is the Riemann sum of this integral, $\lambda$ is the negative of the sum of Harmonic series: $\lambda=-\sum_{k=1}^\infty \frac1k$. From this equality (or via Laplace transform), $\int_1^\infty\frac1xdx=-\lambda-\gamma$. Thus, $\int_0^\infty \frac1x dx=-2\lambda-\gamma$. Other integral representations include: $\int_0^\infty\frac{e^{-x}-1}{x}dx,-\int_0^\infty \frac{1}{x^2+x}dx$. Can be expressed via $\omega$: $\lambda=-\ln\omega-\gamma$. Some divergent integrals can be expressed in terms of $\lambda$: $\int_0^\infty \frac{\ln t}{t} \, dt=\gamma\lambda+\gamma^2/2,$ $\int_0^{\infty } | \tan x| \, dx=-\lambda\omega$. If we generalize the notions of periods and Chow's $EL$-numbers to our extended set, then $\lambda$ would be both, because it can be represented as $\int_0^1 \frac{-1}t dt$ (integral of an algebraic function over algebraic domain) and $\ln 0$ respectively. Logarithms of zero divisors can be expressed via $\lambda$, in split-complex numbers: $\ln \left(\frac{a j}{2}+\frac{a}{2}\right)=\frac{j}{2} (\ln a-\lambda)+\frac{1}{2} (\ln a+\lambda)$, in dual numbers $\ln \varepsilon=\lambda+\varepsilon \omega$, $\varepsilon^\varepsilon=1+\varepsilon(1+\lambda)$.
$-\lambda-\gamma$. Equal to $\ln \omega$ and germ at infinity of logarithmic function $f(x)=\ln x$. Has finite part $0$. Unlike $\lambda$, is neither generalized period, nor EL-number. Has representations $\int_1^\infty\frac1xdx,$ $\int_0^\infty \frac{e^{-x}}{x}dx,$ $\int_0^\infty \frac{x-\log (x)-1}{(x-1)^2}dx$.
$\int_{-\infty}^\infty e^x dx$. Equal to $e^\omega$. Has finite part $0$. Germ of the function $f(x)=e^x$ at infinity.
The most simple infinite number $\omega$ is the germ at infinity of the function $f(x)=x$. Denoted in Levi-Civita field as $\varepsilon^{-1}$ (so has inverse $\varepsilon$ in that field, as well as in Hardy field). Can be represented as divergent integrals $\int_0^\infty dx$ and $\int_0^\infty\frac1{x^2}dx$ (via Laplace transform). Half the numerocity of integers, equal to numerocity of even or odd numbers. Divergent sum representation: $1/2+\sum_{k=1}^\infty 1$. Has finite part (regularized value) $0$. Also, the germ of the function $\frac1x$ at zero (from above). If we modify Delta distribution to behave like a function, $\omega=\pi\tilde{\delta}(0)$ (via Fourier transform).
$\lambda=-\int_0^1 \frac1x dx$. This is negatively infinite constant. Via its definition can be used to extend logarithmic function to zero (thus $\lambda=\ln 0$). Has finite part (regularized value) $-\gamma$. Since Harmonic series is the Riemann sum of this integral, $\lambda$ is the negative of the sum of Harmonic series: $\lambda=-\sum_{k=1}^\infty \frac1k$. From this equality (or via Laplace transform), $\int_1^\infty\frac1xdx=-\lambda-\gamma$. Thus, $\int_0^\infty \frac1x dx=-2\lambda-\gamma$. Other integral representations include: $\int_0^\infty\frac{e^{-x}-1}{x}dx,-\int_0^\infty \frac{1}{x^2+x}dx$. Can be expressed via $\omega$: $\lambda=-\ln\omega-\gamma$. Some divergent integrals can be expressed in terms of $\lambda$: $\int_0^\infty \frac{\ln t}{t} \, dt=\gamma\lambda+\gamma^2/2,$ $\int_0^{\infty } | \tan x| \, dx=-\lambda\omega$. If we generalize the notions of periods and Chow's $EL$-numbers to our extended set, then $\lambda$ would be both, because it can be represented as $\int_0^1 \frac{-1}t dt$ (integral of an algebraic function over algebraic domain) and $\ln 0$ respectively. Logarithms of zero divisors can be expressed via $\lambda$, in split-complex numbers: $\ln \left(\frac{a j}{2}+\frac{a}{2}\right)=\frac{j}{2} (\ln a-\lambda)+\frac{1}{2} (\ln a+\lambda)$, in dual numbers $\ln \varepsilon=\lambda+\varepsilon \omega$, $\varepsilon^\varepsilon=1+\varepsilon(1+\lambda)$.
$-\lambda-\gamma$. Equal to $\ln \omega$ and germ at infinity of logarithmic function $f(x)=\ln x$. Has finite part $0$. Unlike $\lambda$, is neither generalized period, nor EL-number. Has representations $\int_1^\infty\frac1xdx,$ $\int_0^\infty \frac{e^{-x}}{x}dx,$ $\int_0^\infty \frac{x-\log (x)-1}{(x-1)^2}dx$.
$\int_{-\infty}^\infty e^x dx$. Equal to $e^\omega$. Has finite part $0$. Germ of the function $f(x)=e^x$ at infinity.
$\sum_{k=1}^\infty k=\frac{\omega^2}2-\frac1{12}$ Has finite part $-1/12$.
The most important is Bernoulli's umbra $B^-$ and $B^+=B^-+1$. $\operatorname{eval}B^-=-1/2$$\operatorname{eval}(B^-)=-1/2$, $\operatorname{eval}(B^+)=1/2$ (here "eval" is evaluation, akin to taking real part). Logarithm of $B^-$ is undefined, but $\operatorname{eval}\ln(B^+)=-\gamma$$\operatorname{eval}(\ln B^+)=-\gamma$. Links logarithms with trigonometric functions: $\operatorname{eval}\frac1{\pi }\ln \left(\frac{B^+-\frac{x}{\pi }}{B^-+\frac{x}{\pi }}\right)=\cot x$.
$e^{B^-}$ and $e^{B^+}$ respectively have evaluations $\frac 1{e-1}$ and $\frac1{1-1/e}$. $(-1)^{B^-+1/2}$ has evaluation $\frac{\pi}2$.
Euler's umbra $E$. $\operatorname{eval}E=0;$ $\operatorname{eval}\ln E=-\pi/2$
The most simple infinite number $\omega$ is the germ at infinity of the function $f(x)=x$. Denoted in Levi-Civita field as $\varepsilon^{-1}$ (so has inverse $\varepsilon$ in that field, as well as in Hardy field). Can be represented as divergent integrals $\int_0^\infty dx$ and $\int_0^\infty\frac1{x^2}dx$ (via Laplace transform). Half the numerocity of integers, equal to numerocity of even or odd numbers. Divergent sum representation: $1/2+\sum_{k=1}^\infty 1$. Has finite part (regularized value) $0$. Also, the germ of the function $\frac1x$ at zero (at positive directionfrom above). If we modify Delta distribution to behave like a function, $\omega=\pi\tilde{\delta}(0)$ (via Fourier transform).
$\lambda=-\int_0^1 \frac1x dx$. This is negatively infinite constant. Via its definition can be used to extend logarithmic function to zero (thus $\lambda=\ln 0$). Has finite part (regularized value) $-\gamma$. Since Harmonic series is the Riemann sum of this integral, $\lambda$ is the negative of the sum of Harmonic series: $\lambda=-\sum_{k=1}^\infty \frac1k$. From this equality (or via Laplace transform), $\int_1^\infty\frac1xdx=-\lambda-\gamma$. Thus, $\int_0^\infty \frac1x dx=-2\lambda-\gamma$. Other integral representations include: $\int_0^\infty\frac{e^{-x}-1}{x}dx,-\int_0^\infty \frac{1}{x^2+x}dx$. Can be expressed via $\omega$: $\lambda=-\ln\omega-\gamma$. Some divergent integrals can be expressed in terms of $\lambda$: $\int_0^\infty \frac{\ln t}{t} \, dt=\gamma\lambda+\gamma^2/2,$ $\int_0^{\infty } | \tan x| \, dx=-\lambda\omega$. If we generalize the notions of periods and Chow's $EL$-numbers to our extended set, then $\lambda$ would be both, because it can be represented as $\int_0^1 \frac{-1}t dt$ (integral of an algebraic function over algebraic domain) and $\ln 0$ respectively. Logarithms of zero divisors can be expressed via $\lambda$, in split-complex numbers: $\ln \left(\frac{a j}{2}+\frac{a}{2}\right)=\frac{j}{2} (\ln a-\lambda)+\frac{1}{2} (\ln a+\lambda)$, in dual numbers $\ln \varepsilon=\lambda+\varepsilon \omega$, $\varepsilon^\varepsilon=1+\varepsilon(1+\lambda)$.
$-\lambda-\gamma$. Equal to $\ln \omega$ and germ at infinity of logarithmic function $f(x)=\ln x$. Has finite part $0$. Unlike $\lambda$, is neither generalized period, nor EL-number. Has representations $\int_1^\infty\frac1xdx,$ $\int_0^\infty \frac{e^{-x}}{x}dx,$ $\int_0^\infty \frac{x-\log (x)-1}{(x-1)^2}dx$.
$\int_{-\infty}^\infty e^x dx$. Equal to $e^\omega$. Has finite part $0$. Germ of the function $f(x)=e^x$ at infinity.
The most important is Bernoulli's umbra $B^-$ and $B^+=B^-+1$. $\operatorname{eval}B^-=-1/2$, $\operatorname{eval}(B^+)=1/2$ (here "eval" is evaluation, akin to taking real part). Logarithm of $B^-$ is undefined, but $\operatorname{eval}\ln(B^+)=-\gamma$. Links logarithms with trigonometric functions: $\operatorname{eval}\frac1{\pi }\ln \left(\frac{B^+-\frac{x}{\pi }}{B^-+\frac{x}{\pi }}\right)=\cot x$.
$e^{B^-}$ and $e^{B^+}$ respectively have evaluations $\frac 1{e-1}$ and $\frac1{1-1/e}$. $(-1)^{B^-+1/2}$ has evaluation $\frac{\pi}2$.
Euler's umbra $E$. $\operatorname{eval}E=0;$ $\operatorname{eval}\ln E=-\pi/2$
The most simple infinite number $\omega$ is the germ at infinity of the function $f(x)=x$. Denoted in Levi-Civita field as $\varepsilon^{-1}$ (so has inverse $\varepsilon$ in that field, as well as in Hardy field). Can be represented as divergent integrals $\int_0^\infty dx$ and $\int_0^\infty\frac1{x^2}dx$ (via Laplace transform). Half the numerocity of integers, equal to numerocity of even or odd numbers. Divergent sum representation: $1/2+\sum_{k=1}^\infty 1$. Has finite part (regularized value) $0$. Also, the germ of the function $\frac1x$ at zero (at positive direction). If we modify Delta distribution to behave like a function, $\omega=\pi\tilde{\delta}(0)$ (via Fourier transform).
$\lambda=-\int_0^1 \frac1x dx$. This is negatively infinite constant. Via its definition can be used to extend logarithmic function to zero (thus $\lambda=\ln 0$). Has finite part (regularized value) $-\gamma$. Since Harmonic series is the Riemann sum of this integral, $\lambda$ is the negative of the sum of Harmonic series: $\lambda=-\sum_{k=1}^\infty \frac1k$. From this equality (or via Laplace transform), $\int_1^\infty\frac1xdx=-\lambda-\gamma$. Thus, $\int_0^\infty \frac1x dx=-2\lambda-\gamma$. Other integral representations include: $\int_0^\infty\frac{e^{-x}-1}{x}dx,-\int_0^\infty \frac{1}{x^2+x}dx$. Can be expressed via $\omega$: $\lambda=-\ln\omega-\gamma$. Some divergent integrals can be expressed in terms of $\lambda$: $\int_0^\infty \frac{\ln t}{t} \, dt=\gamma\lambda+\gamma^2/2,$ $\int_0^{\infty } | \tan x| \, dx=-\lambda\omega$. If we generalize the notions of periods and Chow's $EL$-numbers to our extended set, then $\lambda$ would be both, because it can be represented as $\int_0^1 \frac{-1}t dt$ (integral of an algebraic function over algebraic domain) and $\ln 0$ respectively. Logarithms of zero divisors can be expressed via $\lambda$, in split-complex numbers: $\ln \left(\frac{a j}{2}+\frac{a}{2}\right)=\frac{j}{2} (\ln a-\lambda)+\frac{1}{2} (\ln a+\lambda)$, in dual numbers $\ln \varepsilon=\lambda+\varepsilon \omega$, $\varepsilon^\varepsilon=1+\varepsilon(1+\lambda)$.
$-\lambda-\gamma$. Equal to $\ln \omega$ and germ at infinity of logarithmic function $f(x)=\ln x$. Has finite part $0$. Unlike $\lambda$, is neither generalized period, nor EL-number. Has representations $\int_1^\infty\frac1xdx,$ $\int_0^\infty \frac{e^{-x}}{x}dx,$ $\int_0^\infty \frac{x-\log (x)-1}{(x-1)^2}dx$.
$\int_{-\infty}^\infty e^x dx$. Equal to $e^\omega$. Has finite part $0$. Germ of the function $f(x)=e^x$ at infinity.
The most important is Bernoulli's umbra $B^-$ and $B^+=B^-+1$. $\operatorname{eval}(B^-)=-1/2$, $\operatorname{eval}(B^+)=1/2$ (here "eval" is evaluation, akin to taking real part). Logarithm of $B^-$ is undefined, but $\operatorname{eval}(\ln B^+)=-\gamma$. Links logarithms with trigonometric functions: $\operatorname{eval}\frac1{\pi }\ln \left(\frac{B^+-\frac{x}{\pi }}{B^-+\frac{x}{\pi }}\right)=\cot x$.
$e^{B^-}$ and $e^{B^+}$ respectively have evaluations $\frac 1{e-1}$ and $\frac1{1-1/e}$. $(-1)^{B^-+1/2}$ has evaluation $\frac{\pi}2$.
Euler's umbra $E$. $\operatorname{eval}E=0;$ $\operatorname{eval}\ln E=-\pi/2$
The most simple infinite number $\omega$ is the germ at infinity of the function $f(x)=x$. Denoted in Levi-Civita field as $\varepsilon^{-1}$ (so has inverse $\varepsilon$ in that field, as well as in Hardy field). Can be represented as divergent integrals $\int_0^\infty dx$ and $\int_0^\infty\frac1{x^2}dx$ (via Laplace transform). Half the numerocity of integers, equal to numerocity of even or odd numbers. Divergent sum representation: $1/2+\sum_{k=1}^\infty 1$. Has finite part (regularized value) $0$. Also, the germ of the function $\frac1x$ at zero (from above). If we modify Delta distribution to behave like a function, $\omega=\pi\tilde{\delta}(0)$ (via Fourier transform).
$\lambda=-\int_0^1 \frac1x dx$. This is negatively infinite constant. Via its definition can be used to extend logarithmic function to zero (thus $\lambda=\ln 0$). Has finite part (regularized value) $-\gamma$. Since Harmonic series is the Riemann sum of this integral, $\lambda$ is the negative of the sum of Harmonic series: $\lambda=-\sum_{k=1}^\infty \frac1k$. From this equality (or via Laplace transform), $\int_1^\infty\frac1xdx=-\lambda-\gamma$. Thus, $\int_0^\infty \frac1x dx=-2\lambda-\gamma$. Other integral representations include: $\int_0^\infty\frac{e^{-x}-1}{x}dx,-\int_0^\infty \frac{1}{x^2+x}dx$. Can be expressed via $\omega$: $\lambda=-\ln\omega-\gamma$. Some divergent integrals can be expressed in terms of $\lambda$: $\int_0^\infty \frac{\ln t}{t} \, dt=\gamma\lambda+\gamma^2/2,$ $\int_0^{\infty } | \tan x| \, dx=-\lambda\omega$. If we generalize the notions of periods and Chow's $EL$-numbers to our extended set, then $\lambda$ would be both, because it can be represented as $\int_0^1 \frac{-1}t dt$ (integral of an algebraic function over algebraic domain) and $\ln 0$ respectively. Logarithms of zero divisors can be expressed via $\lambda$, in split-complex numbers: $\ln \left(\frac{a j}{2}+\frac{a}{2}\right)=\frac{j}{2} (\ln a-\lambda)+\frac{1}{2} (\ln a+\lambda)$, in dual numbers $\ln \varepsilon=\lambda+\varepsilon \omega$, $\varepsilon^\varepsilon=1+\varepsilon(1+\lambda)$.
$-\lambda-\gamma$. Equal to $\ln \omega$ and germ at infinity of logarithmic function $f(x)=\ln x$. Has finite part $0$. Unlike $\lambda$, is neither generalized period, nor EL-number. Has representations $\int_1^\infty\frac1xdx,$ $\int_0^\infty \frac{e^{-x}}{x}dx,$ $\int_0^\infty \frac{x-\log (x)-1}{(x-1)^2}dx$.
$\int_{-\infty}^\infty e^x dx$. Equal to $e^\omega$. Has finite part $0$. Germ of the function $f(x)=e^x$ at infinity.
The most simple infinite number $\omega$ is the germ at infinity of the function $f(x)=x$. Denoted in Levi-Civita field as $\varepsilon^{-1}$ (so has inverse $\varepsilon$ in that field, as well as in Hardy field). Can be represented as divergent integrals $\int_0^\infty dx$ and $\int_0^\infty\frac1{x^2}dx$ (via Laplace transform). Half the numerocity of integers, equal to numerocity of even or odd numbers. Divergent sum representation: $1/2+\sum_{k=1}^\infty 1$. Has finite part (regularized value) $0$. Also, the germ of the function $\frac1x$ at zero (at positive direction). If we modify Delta distribution to behave like a function, $\omega=\pi\tilde{\delta}(0)$ (via Fourier transform).
$\lambda=-\int_0^1 \frac1x dx$. This is negatively infinite constant. Via its definition can be used to extend logarithmic function to zero (thus $\lambda=\ln 0$). Has finite part (regularized value) $-\gamma$. Since Harmonic series is the Riemann sum of this integral, $\lambda$ is the negative of the sum of Harmonic series: $\lambda=-\sum_{k=1}^\infty \frac1k$. From this equality (or via Laplace transform), $\int_1^\infty\frac1xdx=-\lambda-\gamma$. Thus, $\int_0^\infty \frac1x dx=-2\lambda-\gamma$. Other integral representations include: $\int_0^\infty\frac{e^{-x}-1}{x}dx,-\int_0^\infty \frac{1}{x^2+x}dx$. Can be expressed via $\omega$: $\lambda=-\ln\omega-\gamma$. Some divergent integrals can be expressed in terms of $\lambda$: $\int_0^\infty \frac{\ln t}{t} \, dt=\gamma\lambda+\gamma^2/2$$\int_0^\infty \frac{\ln t}{t} \, dt=\gamma\lambda+\gamma^2/2,$ $\int_0^{\infty } | \tan x| \, dx=-\lambda\omega$. If we generalize the notions of periods and Chow's $EL$-numbers to our extended set, then $\lambda$ would be both, because it can be represented as $\int_0^1 \frac{-1}t dt$ (integral of an algebraic function over algebraic domain) and $\ln 0$ respectively. Logarithms of zero divisors can be expressed via $\lambda$, in split-complex numbers: $\ln \left(\frac{a j}{2}+\frac{a}{2}\right)=\frac{j}{2} (\ln a-\lambda)+\frac{1}{2} (\ln a+\lambda)$, in dual numbers $\ln \varepsilon=\lambda+\varepsilon \omega$, $\varepsilon^\varepsilon=1+\varepsilon(1+\lambda)$.
$-\lambda-\gamma$. Equal to $\ln \omega$ and germ at infinity of logarithmic function $f(x)=\ln x$. Has finite part $0$. Unlike $\lambda$, is neither generalized period, nor EL-number. Has representations $\int_1^\infty\frac1xdx,$ $\int_0^\infty \frac{e^{-x}}{x}dx,$ $\int_0^\infty \frac{x-\log (x)-1}{(x-1)^2}dx$.
$\int_{-\infty}^\infty e^x dx$. Equal to $e^\omega$. Has finite part $0$. Germ of the function $f(x)=e^x$ at infinity.
The most simple infinite number $\omega$ is the germ at infinity of the function $f(x)=x$. Denoted in Levi-Civita field as $\varepsilon^{-1}$ (so has inverse $\varepsilon$ in that field, as well as in Hardy field). Can be represented as divergent integrals $\int_0^\infty dx$ and $\int_0^\infty\frac1{x^2}dx$ (via Laplace transform). Half the numerocity of integers, equal to numerocity of even or odd numbers. Divergent sum representation: $1/2+\sum_{k=1}^\infty 1$. Has finite part (regularized value) $0$. Also, the germ of the function $\frac1x$ at zero (at positive direction). If we modify Delta distribution to behave like a function, $\omega=\pi\tilde{\delta}(0)$ (via Fourier transform).
$\lambda=-\int_0^1 \frac1x dx$. This is negatively infinite constant. Via its definition can be used to extend logarithmic function to zero (thus $\lambda=\ln 0$). Has finite part (regularized value) $-\gamma$. Since Harmonic series is the Riemann sum of this integral, $\lambda$ is the negative of the sum of Harmonic series: $\lambda=-\sum_{k=1}^\infty \frac1k$. From this equality (or via Laplace transform), $\int_1^\infty\frac1xdx=-\lambda-\gamma$. Thus, $\int_0^\infty \frac1x dx=-2\lambda-\gamma$. Other integral representations include: $\int_0^\infty\frac{e^{-x}-1}{x}dx,-\int_0^\infty \frac{1}{x^2+x}dx$. Can be expressed via $\omega$: $\lambda=-\ln\omega-\gamma$. Some divergent integrals can be expressed in terms of $\lambda$: $\int_0^\infty \frac{\ln t}{t} \, dt=\gamma\lambda+\gamma^2/2$. If we generalize the notions of periods and Chow's $EL$-numbers to our extended set, then $\lambda$ would be both, because it can be represented as $\int_0^1 \frac{-1}t dt$ (integral of an algebraic function over algebraic domain) and $\ln 0$ respectively. Logarithms of zero divisors can be expressed via $\lambda$, in split-complex numbers: $\ln \left(\frac{a j}{2}+\frac{a}{2}\right)=\frac{j}{2} (\ln a-\lambda)+\frac{1}{2} (\ln a+\lambda)$, in dual numbers $\ln \varepsilon=\lambda+\varepsilon \omega$, $\varepsilon^\varepsilon=1+\varepsilon(1+\lambda)$.
$-\lambda-\gamma$. Equal to $\ln \omega$ and germ at infinity of logarithmic function $f(x)=\ln x$. Has finite part $0$. Unlike $\lambda$, is neither generalized period, nor EL-number. Has representations $\int_1^\infty\frac1xdx,$ $\int_0^\infty \frac{e^{-x}}{x}dx,$ $\int_0^\infty \frac{x-\log (x)-1}{(x-1)^2}dx$.
$\int_{-\infty}^\infty e^x dx$. Equal to $e^\omega$. Has finite part $0$. Germ of the function $f(x)=e^x$ at infinity.
The most simple infinite number $\omega$ is the germ at infinity of the function $f(x)=x$. Denoted in Levi-Civita field as $\varepsilon^{-1}$ (so has inverse $\varepsilon$ in that field, as well as in Hardy field). Can be represented as divergent integrals $\int_0^\infty dx$ and $\int_0^\infty\frac1{x^2}dx$ (via Laplace transform). Half the numerocity of integers, equal to numerocity of even or odd numbers. Divergent sum representation: $1/2+\sum_{k=1}^\infty 1$. Has finite part (regularized value) $0$. Also, the germ of the function $\frac1x$ at zero (at positive direction). If we modify Delta distribution to behave like a function, $\omega=\pi\tilde{\delta}(0)$ (via Fourier transform).
$\lambda=-\int_0^1 \frac1x dx$. This is negatively infinite constant. Via its definition can be used to extend logarithmic function to zero (thus $\lambda=\ln 0$). Has finite part (regularized value) $-\gamma$. Since Harmonic series is the Riemann sum of this integral, $\lambda$ is the negative of the sum of Harmonic series: $\lambda=-\sum_{k=1}^\infty \frac1k$. From this equality (or via Laplace transform), $\int_1^\infty\frac1xdx=-\lambda-\gamma$. Thus, $\int_0^\infty \frac1x dx=-2\lambda-\gamma$. Other integral representations include: $\int_0^\infty\frac{e^{-x}-1}{x}dx,-\int_0^\infty \frac{1}{x^2+x}dx$. Can be expressed via $\omega$: $\lambda=-\ln\omega-\gamma$. Some divergent integrals can be expressed in terms of $\lambda$: $\int_0^\infty \frac{\ln t}{t} \, dt=\gamma\lambda+\gamma^2/2,$ $\int_0^{\infty } | \tan x| \, dx=-\lambda\omega$. If we generalize the notions of periods and Chow's $EL$-numbers to our extended set, then $\lambda$ would be both, because it can be represented as $\int_0^1 \frac{-1}t dt$ (integral of an algebraic function over algebraic domain) and $\ln 0$ respectively. Logarithms of zero divisors can be expressed via $\lambda$, in split-complex numbers: $\ln \left(\frac{a j}{2}+\frac{a}{2}\right)=\frac{j}{2} (\ln a-\lambda)+\frac{1}{2} (\ln a+\lambda)$, in dual numbers $\ln \varepsilon=\lambda+\varepsilon \omega$, $\varepsilon^\varepsilon=1+\varepsilon(1+\lambda)$.
$-\lambda-\gamma$. Equal to $\ln \omega$ and germ at infinity of logarithmic function $f(x)=\ln x$. Has finite part $0$. Unlike $\lambda$, is neither generalized period, nor EL-number. Has representations $\int_1^\infty\frac1xdx,$ $\int_0^\infty \frac{e^{-x}}{x}dx,$ $\int_0^\infty \frac{x-\log (x)-1}{(x-1)^2}dx$.
$\int_{-\infty}^\infty e^x dx$. Equal to $e^\omega$. Has finite part $0$. Germ of the function $f(x)=e^x$ at infinity.