# Zeta regularization riddle

1. Aug 22, 2008

### mhill

examining the zeta regularization of divergent series i found

$$\int_{0}^{\infty} x^{m}dx = (m/2) \int_{0}^{\infty}x^{m-1}dx +\zeta (-m) - \sum_{r=1}^{\infty} \frac{B_ {2r} \Gamma (m+1) (m-2r+1)}{ (2r)! \Gamma (m-2r+2)} \int_{0}^{\infty} x^{m-2r}dx$$

of course this is only justified considering we take the sum of the series

$$1+2^{r} + 3^{r} + ................. = \zeta (-r)$$

B_2r are just the Bernoulli numbers.

Also another Brain Teaser...

Let be R(x) the Riesz function , see http://en.wikipedia.org/wiki/Riesz_criterion then we define $$R(x^{2} ) = f(x)$$

then prove that f(x) satisfy the Integral equation

$$\frac{1-e^{x^2} }{2} = \int_{0}^{\infty} \frac{dt}{t} [x/t] f(t)$$ , with $$\zeta(s)=s\int_{0}^{\infty} dx [x]x^{-s-1}$$