On Riemann zeta function and Dirac delta function/distribution

Question

Let $$I_{N} = \int_{n - \frac{1}{2}}^{n + \frac{1}{2}} \cos^{2N}(2\pi x)dx = \frac{2N-1}{2N} \int_{n - \frac{1}{2}}^{n + \frac{1}{2}} \cos^{2N-2}(2\pi x)dx $$ therefore (I think) $$ I_N = \frac{(2N-1)!!}{(2N)!!}$$ hence $$ \frac{(2N)!!}{(2N-1)!!} I_N = 1$$ Now let for $n \in \mathbb{N}$ $$J_{3,N}(n) = \frac{(2N)!!}{(2N-1)!!} \int_{n-\frac{1}{2}}^{n + \frac{1}{2}} \frac{1}{x^3} \cos^{2N}(2\pi x) dx $$ I think (but I was unable to prove so far) that $$\lim_{N\to \infty}J_{3,N}(n) = \frac{1}{n^3}$$ for all $n \in \mathbb{N}$

If this is the case then $$ \lim_{N \to \infty} \int_{\frac{1}{2}}^{\infty} \frac{1}{x^3} \cos^{2N}(2\pi x) dx = \sum_{n=1}^{\infty} \frac{1}{n^3} = \zeta(3)$$ Is this true? How can I provide a proof for this? Of course this can be generalised to other arguments of $\zeta$

Answer 1

Let $$C_n = \int_{-1}^1 (1-\frac{x^2}{2})^n dx \sim \int_{-\pi/2}^{\pi/2} |\cos(x)|^n dx$$ If $f$ is continuous $$\lim_{n \to \infty} \int_{-1}^{1} \frac{(1-\frac{x^2}{2})^{n} }{C_n} f(x)=\lim_{n \to \infty} \int_{-\pi/2}^{\pi/2} \frac{|\cos(x)|^{n} }{C_n} f(x) = f(0)$$ (So that $\frac{|\cos(x)|^n }{C_n}1_{|x| < \pi/2} \to \delta(x)$ in the sense of distributions)

And hence if $|f|$ is integrable, decreasing and $\sum_{n=1}^\infty |f(n\pi)| < \infty$, for $\epsilon > 0$ $$\lim_{n \to \infty}\int_{\epsilon}^\infty \frac{|\cos(x)|^n}{C_n} f(x)dx= \sum_{n=1}^\infty f(\pi n )$$ ie. for $\Re(s) > 1$ $$\lim_{n \to \infty}\int_{\epsilon}^\infty \frac{|\cos(x)|^n}{C_n} x^{-s}dx= \pi^{-s} \zeta(s)$$