CMC - Problem 10

A completion of the starting step with the Poisson summation formula:

Fix $b$ and consider $f(a) = \frac{1}{a^4 + 4b^4}.$ First, we compute the (continuous) Fourier transform $\hat{f}(c) = \int_{-\infty}^{\infty} e^{-2i\pi ct}\cdot \frac{1}{t^4 + 4b^4}\,dt$ with a contour integral.

Denote the integrand as $g(t)$, i.e. $g(t) = e^{-2i\pi ct}\cdot \frac{1}{t^4 + 4b^4}.$

Suppose $c \geq 0$. Then if $\Im(t) \leq 0$, we have $\Re(-2i\pi ct) \leq 0$ and $|e^{-2i\pi ct}| \leq 1$. Consider the contour integral of $g$ around a large semicircle centered at the origin in the half-plane $\Im(t) \leq 0$ of the complex plane. Since $\frac{1}{t^4 + 4b^4}$ decays quickly enough if $t$ has large magnitude, if the semicircle's radius goes to infinity, the integral over the arc of the semicircle goes to 0, so that the contour integral converges to the integral over the real line, $\int_{\infty}^{-\infty} g(t)\,dt.$ (Note the orientation, since we need to go around the semicircle counterclockwise.)

At the same time, the integral can be evaluated with the residue formula. Let $\omega = e^{i\pi/4}$, a primitive eighth root of unity. $g$ has four simple poles at $t = \sqrt{2}\omega^k b$ for $k = 1, 3, 5, 7$; the relevant poles inside our contour are $\sqrt{2}\omega^5b$ and $\sqrt{2}\omega^7b$. The residues at those points can be evaluated in an L'Hôpital-esque manner to be:

$\begin{aligned} \operatorname{res}_{\sqrt{2}\omega^5b} g &= e^{-2i\pi c(\sqrt{2}\omega^5b)}\cdot \frac{1}{4(\sqrt{2}\omega^5b)^3} \\ &= e^{2\pi bc(-1+i)}\cdot \frac{1+i}{16b^3} \\ \operatorname{res}_{\sqrt{2}\omega^7b} g &= e^{-2i\pi c(\sqrt{2}\omega^7b)}\cdot \frac{1}{4(\sqrt{2}\omega^7b)^3} \\ &= e^{2\pi bc(-1-i)}\cdot \frac{-1+i}{16b^3} \\ \end{aligned}$

Additionally supposing $c$ is an integer, we then have $e^{2\pi bci} = 1$ and can further simplify to $\begin{aligned} \operatorname{res}_{\sqrt{2}\omega^5b} g &= e^{-2\pi bc}\cdot \frac{1+i}{16b^3} \\ \operatorname{res}_{\sqrt{2}\omega^7b} g &= e^{-2\pi bc}\cdot \frac{-1+i}{16b^3} \end{aligned}$

So, $\begin{aligned} \int_{\infty}^{-\infty} g(t)\,dt &= 2\pi i\left( e^{-2\pi bc}\cdot \frac{1+i}{16b^3} + e^{-2\pi bc}\cdot \frac{-1+i}{16b^3}\right) \\ &= -\frac{\pi}{4b^3} \cdot e^{-2\pi bc} \\ \hat{f}(c) &= \frac{\pi}{4b^3} \cdot e^{-2\pi bc} \\ \end{aligned}$

Note that since $f$ is even, $\hat{f}(c) = \hat{f}(-c)$. Now the Poisson summation formula can be evaluated as two geometric series.

$\sum_{a=-\infty}^\infty f(a) = \sum_{c=-\infty}^\infty \hat{f}(c) = \frac{\pi}{4b^3} \left(\frac{1}{1 - e^{-2\pi b}} + \frac{e^{-2\pi b}}{1 - e^{-2\pi b}}\right) = \frac{\pi \coth(b\pi)}{4b^3}$

Thus, again since $f$ is even, $\begin{aligned} \sum_{a=1}^\infty f(a) &= \frac{1}{2}\left(\sum_{a=-\infty}^\infty f(a) - f(0)\right) \\ &= \frac{\pi \coth(b\pi)}{8b^3} - \frac{1}{8b^4} \\ &= \frac{1}{8b^4}\left(b\pi \coth(b\pi) - 1\right), \end{aligned}$ as desired.

I have no idea how the parent post managed to continue from here, or how Sophie-Germain can help, however.

Excellent progress, 7 points. Can you finish the proof using the hint?

Has anyone proven the first step? If not, I'd like to take a shot.

how do you write the solution

Do I really need the knowledge of Fourier to really understand or maybe Riemann Zeta Function?

Fixed.

Just post your solution to this thread.

I'm still waiting for an explanation of how this helps, and I bet I'm not the only one ;-)

I'll award 2 points for this answer, considering the magnitude of this problem. But the solution's where it's at.

$\displaystyle \sum_{a=1}^m \displaystyle\sum_{b=1}^n f(a, b)$ means $\displaystyle \sum_{a=1}^m \left(\displaystyle\sum_{b=1}^n f(a, b)\right),$ or $(f(1, 1) + f(1, 2) + \ldots + f(1, n)) \\+ (f(2, 1) + f(2, 2) + \ldots + f(2, n)) \\+ \ldots \\ + (f(m, 1) + f(m, 2) + \ldots + f(m, n)).$

In other words, we evaluate the inner sums first, taking outside variables as constant.

Note that (in this example at least) we could switch the order of the summation symbols, to get $\displaystyle \sum_{b=1}^n \displaystyle\sum_{a=1}^m f(a, b)$ or $(f(1, 1) + f(2, 1) + \ldots + f(m, 1)) \\+ (f(1, 2) + f(2, 2) + \ldots + f(m, 2)) \\+ \ldots \\ + (f(1, n) + f(2, n) + \ldots + f(m, n)).$

See how these two are actually the same?