6
votes
0answers
144 views

Connection between integral expression and the factorial of infinity

Does the fact that $$\int_{-\infty}^{\infty}\exp\left(-\frac{1}{2}x^2\right)\mathrm{d}x=\sqrt{2\pi}$$ Have something to do with the fact that the regularized factorial of infinity is also ...
1
vote
1answer
104 views

Playing fast and loose with divergent series [closed]

I have been playing around recently with the regularization of infinite divergent sums and products, e.g. $$1+1+1+1+1+\ldots=\zeta(0)=-\frac{1}{2}$$ $$1+2+3+4+5+\ldots=\zeta(-1)=-\frac{1}{12}$$ ...
5
votes
4answers
186 views

Find the regularized sum $\frac{1}{\ln(2)}+\frac{1}{\ln(3)}+\frac{1}{\ln(4)}+…$

By considering the integral Zeta function $$F(s)=s+\frac{1}{2^s\ln(2)}+\frac{1}{3^s\ln(3)}+\frac{1}{4^s\ln(4)}+...$$ Evaluate $$\frac{1}{\ln(2)}+\frac{1}{\ln(3)}+\frac{1}{\ln(4)}+...$$ EDIT: ...
2
votes
1answer
164 views

Meaning of equality in zeta regularization

It is known that $$\sum\limits_{n = 1}^\infty{n = 1 + 2 + 3 + \cdots} = \infty$$ but it is also known that $$\sum\limits_{n = 1}^\infty{n = 1 + 2 + 3 + \cdots} = -\frac{1}{{12}}$$ which can obtained ...
9
votes
2answers
801 views

Values of the Riemann Zeta function and the Ramanujan Summation - How strong is the connection?

The Ramanujan Summation of some infinite sums is consistent with a couple of sets of values of the Riemann zeta function. We have, for instance, $$\zeta(-2n)=\sum_{n=1}^{\infty} n^{2k} = 0 ...