I have recently read about zeta function regularization, a way of ascribing values to functions having simple poles in a point and to divergent series. The values obtained are the same as those obtained using Ramanujan's summation.
In short this says that a function having pole in $x_0$ should be ascribed the following value:
$$f(x_0)=\lim_{h\to0}\frac{f(x_0+h)+f(x_0-h)}2=\lim_{x\to x_0} \left(\frac{f (x)^2 f''\left( x \right)}{2 f'\left(x\right)^2}-f (x)\right)$$ (see here for derivation).
This way we can obtain:
$$1/0=0$$ $$\zeta(1)=\gamma$$ $$\Gamma(0)=-\gamma$$ $$\Gamma(-1)=\gamma-1$$ $$\Gamma(-2)=\frac{3}{4}-\frac{\gamma }{2}$$ $$\Gamma(-3)=\frac{\gamma }{6}-\frac{11}{36}$$ $$\Gamma(-4)=\frac{25}{288}-\frac{\gamma }{24}$$ etc.
But as we can see, these values of Gamma function do not satisfy the functional equation. It seems like every value were missing a non-real term.
We can fix the issue by introducing a value $\omega$ with the following properties:
$$0\, \omega = 1$$ $$\omega+a \ne \omega$$ $$a \omega \ne \omega$$ $$\Re(\omega)=\Im(\omega)=0$$ $$|\omega|=\infty$$
This way the values of the Gamma function at negative arguments will satisfy the functional equation:
$$\Gamma(0)=-\gamma+\omega$$ $$\Gamma(-1)=\gamma-1-\omega$$ $$\Gamma(-2)=\frac{3}{4}-\frac{\gamma }{2}+\frac\omega 2$$ $$\Gamma(-3)=\frac{\gamma }{6}-\frac{11}{36}-\frac\omega 6$$
$$\zeta(1)=\gamma+\omega$$
etc. As you can see, these values perfectly satisfy the functional equation for Gamma.
Thus I wonder
- What algebraic properties of real and complex numbers get lost with introduction of such $\omega$?
- What is the general way of finding the coefficient at $\omega$ for an arbitrary function having a pole (I want an expression in derivatives, similar to the expression for the real part).
Well at least for the second part of the question, each function having pole at $x_0$ can be regularized the following way:
$$f(x_0)=\lim_{h\to 0}\frac{f(x_0+h)+f(x_0-h)}2+\omega \lim_{h\to 0}h\frac{f(x_0+h)-f(x_0-h)}2 $$
Or at any point,
$$f(x)=\lim_{h\to 0}\frac{f(x+h)+f(x-h)}2+\lim_{h\to 0}\frac{f(x+h)-f(x-h)}2 $$