User:Mike Battaglia/ Mike's Zeta Scratch Pad

Suppose that we define a sawtooth wave as follows:

[math]\displaystyle \text{saw}(t) = \sin(t) + \tfrac{1}{2} \sin(2t) + ... + \tfrac{1}{n} \sin(nt) + ... .[/math]

then the Fourier transform is as follows:

[math]\displaystyle \mathcal{F} \left\{ \text{saw} \right\} (ω) = \left[\frac{1}{2i} \sum_m \frac{\delta(ω+m)}{-m}\right] + \left[\frac{1}{2i} \sum_n \frac{\delta(ω-n)}{n}\right][/math]

Now, let's take the Mellin transform of the Fourier transform. Since the Mellin integral is one-sided, we throw all of the negative frequencies away, completely nixing the first summation, and yielding

[math]\displaystyle \mathcal{M} \left\{ \mathcal{F} \left\{ \text{saw} \right\} \right\} (s) = \int_0^\infty \left[\frac{1}{2i} \sum_n \frac{\delta(ω-n)}{n}\right] \omega^{s-1} d\omega[/math]

Since integration is linear, we can expand the summation out as follows:

[math]\displaystyle \mathcal{M} \left\{ \mathcal{F} \left\{ \text{saw} \right\} \right\} (s) = \frac{1}{2i} \cdot \sum_n \left[ \frac{1}{n} \int_0^\infty \delta(ω-n) \omega^{s-1} d\omega \right][/math]

Now, each individual integral is just the Mellin transform of a Dirac delta distribution, yielding

[math]\displaystyle \mathcal{M} \left\{ \mathcal{F} \left\{ \text{saw} \right\} \right\} (s) = \frac{1}{2i} \cdot \sum_n \left[ \frac{1}{n} n^{s-1} \right] = \frac{1}{2i} \cdot \sum_n n^{s-2}[/math]

Finally, noting the relation of the series on the right to the zeta function, we get

[math]\displaystyle \mathcal{M} \left\{ \mathcal{F} \left\{ \text{saw} \right\} \right\} (s) = \frac{1}{2i} \cdot \zeta(2-s)[/math]

Word.