Proving Series and Zeta Function Relationships

Question

In Section 2.15. of Titchmarsh's book The Theory of the Riemann Zeta-Function, after reaching the equality $$\int_0^{\infty} \psi^2(\frac{x}{\pi}) x^{w-1} dx = \int_0^{\infty} (\sum_{n=1}^{\infty} e^{-n^2 x})^2 x^{w-1} dx = \Gamma(w) \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \dfrac{1}{(m^2+n^2)^w},$$ it follows with two equalities without any proof that I failed to prove them after a long effort:

$1:$ $$\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \dfrac{1}{(m^2+n^2)^w} = \dfrac{1}{4} \sum_{n=1}^{\infty} \dfrac{r(n)}{n^w} – \zeta(2w),$$ where $r(n)$ is the number of ways of expressing $n$ as the sum of two squares.

$2:$ $$\dfrac{1}{4} \sum_{n=1}^{\infty} \dfrac{r(n)}{n^w} – \zeta(2w) = \zeta(w) \eta(w) – \zeta(2w),$$ where $$\eta(w) = 1^{-w} – 3^{-w} + 5^{-w} – 7^{-w} + \dots .$$

In $1$, I seperated terms $n=m$ but those terms can be included in $r(n)$ as well! Also, in Silverman's book A Friendly Introduction to Number Theory in all formulas related to $r(n)$, in the definition of it negative integers were included as well but I guess from factor $\frac{1}{4}$ only positive integers are included…

In $2$, any rearrangement does not yield the second equality either by beginning with $\dfrac{1}{4} \sum_{n=1}^{\infty} \dfrac{r(n)}{n^w} – \zeta(2w)$ or with $\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \dfrac{1}{(m^2+n^2)^w}$.

How these two equalities are derived?

Answer 1

1. You collect the terms where $m^2+n^2=d$ and there are, by definition, $\frac{1}{4}r(d)$ of them if for example $n$ would start at $0$ (instead of $1$). $\zeta$ arises because we don't count the lattice points on the let's say the real axis (since there $n$ (or $m$ depending how you choose your coordinates) would be $0$). The $\frac{1}{4}$ is there since we are only summing over lattice points in the let's say upper right quarter plane. (The lattice in question is $\mathbb{Z}^2$. So $r(d)$ counts the number of lattice points $a\in\mathbb{Z}^2$ with $\vert a\vert^2=d$.)
2. There is a video by 3blue1brown on youtube on exactly this. The Gaussian integers are an UFD with $4$ units. An odd rational prime $p$ factors into $2$ different (modulo units) Gaussian primes of norm $p$ if $p$ is $1$ mod $4$ and else it doesn't factor. $2$ factors into $2$ Gaussian primes which are the same (modulo units). From this we can see (see for example the video) that $$\sum_{n=1}^{\infty} \dfrac{r(n)}{n^s}=$$ $$4(1-2^{-s})^{-1}\prod_{(p\equiv_41)}(1-p^{-s})^{-2}\prod_{(p\equiv_43)}(1-p^{-2s})^{-1}=$$$$4\zeta(s)\prod_{(p\equiv_41)}(1-p^{-s})^{-1}\prod_{(p\equiv_43)}(1+p^{-s})^{-1}=4\zeta(s)\eta(s).$$ Also see the zeta function of the Gaussian integers/of number fields if you are interested.