Zeta and Gamma function regularization with $\omega=1/0$

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).
• Why the downvotes? – Anixx Dec 8 '14 at 4:37
• Interesting question. I don't see the connection between Zeta regularization and your limit, could you provide a reference for that? – Antonio Vargas Dec 8 '14 at 4:39
• For the $\Gamma$ function, the values are the same as $~\displaystyle\lim_{\epsilon\to0}\frac{\Gamma(n+\epsilon)+\Gamma(n-\epsilon)}2$. – Lucian Dec 8 '14 at 5:23
• @Antonio Vargas there was a typo in the formula, sorry (actually I inserted the wrong formula). – Anixx Dec 8 '14 at 5:44
• Sorry, I wasn't clear. I wasn't commenting on the formula, I was wondering how it (or the new one) is connected to Zeta regularization. It's not obvious to me, so I was hoping you could provide a reference. Your other question doesn't say either. – Antonio Vargas Dec 8 '14 at 5:46

Well at least for the second part of the question, each function having pole at $x_0$ can be regularized the following way:
$$f(x_0)=\lim_{h\to 0}\frac{f(x_0+h)+f(x_0-h)}2+\omega \lim_{h\to 0}h\frac{f(x_0+h)-f(x_0-h)}2$$
$$f(x)=\lim_{h\to 0}\frac{f(x+h)+f(x-h)}2+\lim_{h\to 0}\frac{f(x+h)-f(x-h)}2$$