Infinite sum involving number of solutions to $k=i^2+j^2$

I suppose we could start with the fact that: $$r_2(n) = 4(d_{1,4}(n) - d_{3,4}(n))$$

where, $d_{a,b}(n)$ is the number of divisors of $n$ that are congruent to $a\pmod{b}$.

Hence, $\displaystyle \sum\limits_{n=1}^{\infty} r_2(n)x^n = 4\sum\limits_{n=0}^{\infty} (-1)^{n}\frac{x^{2n+1}}{1-x^{2n+1}}$

Thus, $$\displaystyle \begin{align}\sum\limits_{n=1}^{\infty} (-1)^{n-1}\frac{r_2(n)}{n} &= 4\sum\limits_{n=0}^{\infty} (-1)^{n-1}\int_0^{-1} \frac{x^{2n}}{1-x^{2n+1}}\,dx \\&= 4\sum\limits_{n=0}^{\infty} (-1)^{n}\frac{\log (1-(-1)^{2n+1})}{2n+1}\\&= 4\log 2\sum\limits_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1} = \pi\log 2\end{align}$$

For any $a,b\in\mathbb{N}\setminus\{0\}$ we have: $$ \int_{0}^{+\infty}\frac{\sin(ax)}{a} e^{-bx}\,dx = \frac{1}{a^2+b^2}\tag{1}$$ so by multiplying both sides of $(1)$ by $(-1)^a (-1)^b$ and summing over $a\in[1,+\infty),b\in[1,+\infty)$ we get: $$\sum_{a,b\geq 1}\frac{(-1)^{a+b}}{a^2+b^2}=\int_{0}^{+\infty}\frac{\text{Im}\log(1+e^{ix})}{1+e^x}\,dx=\frac{\pi^2}{24}-\sum_{k=1}^{+\infty}\pi k\int_{(2k-1)\pi}^{(2k+1)\pi}\frac{dx}{1+e^x}\tag{2}$$ and: $$\begin{eqnarray*}\sum_{k=1}^{+\infty}\pi k\int_{(2k-1)\pi}^{(2k+1)\pi}\frac{dx}{1+e^x}&=&\pi\sum_{k=0}^{+\infty}\log\left(1+e^{-(2k+1)\pi}\right)\\&=&\pi\log\prod_{k\geq 0}\left(1+e^{-(2k+1)\pi}\right)\\&=&\pi\log\left(2^{1/4}e^{-\pi/24}\right)\tag{3}\end{eqnarray*}$$ where the last identity is kindly provided by a special value of the Dedekind eta function. $$ S = \pi\log 2 \tag{4}$$ readily follows.

Another possible approach is given by noticing that $(-1)^k r_2(k)$ is four times a multiplicative function, hence, by Dirichlet convolution: $$ S = 4\left(\sum_{n\geq 1}\frac{(-1)^n}{2n-1}\right)\cdot\left(\sum_{n\geq 1}\frac{(-1)^n}{n}\right)=4\cdot\frac{\pi}{4}\cdot\log 2.\tag{5}$$

We note the estimate of the summation function of $r_{2}(n)$:

$$S(n) = \sum\limits_{k=1}^{n} r_2(k) = \pi n + \mathcal{O}(\sqrt{n})$$

where, $S(n)$ is the number of integer points inside the disk of radius $\sqrt{n}$.

Since, $k = i^2+j^2 \iff 2k = (i+j)^2+(i-j)^2$, we have, $r_2(2k) = r_2(k)$.

Hence, $$\displaystyle \sum\limits_{k=1}^{2n} (-1)^{k-1}\frac{r_2(k)}{k} = \sum\limits_{k=1}^{2n}\frac{r_2(k)}{k} - 2\sum\limits_{k=1}^{n}\frac{r_2(2k)}{2k} = \sum\limits_{k=n+1}^{2n} \frac{r_2(k)}{k}$$

Now, by Abel Summation identity: $$\begin{align} \sum\limits_{k=n+1}^{2n} \frac{r_2(k)}{k} &= \sum\limits_{k=n}^{2n} \left(\frac{1}{k} - \frac{1}{k+1}\right)S(k) + \frac{S(2n)}{2n+1} - \frac{S(n)}{n} \\ &= \sum\limits_{k=n}^{2n} \frac{S(k)}{k(k+1)} + \frac{S(2n)}{2n+1} - \frac{S(n)}{n}\end{align}$$

Since, $\displaystyle \frac{S(n)}{n} = \pi + \mathcal{O}(n^{-1/2})$

$$\begin{align} \sum\limits_{k=n}^{2n} \frac{S(k)}{k(k+1)}&= \pi\sum\limits_{k=n}^{2n}\frac{1}{k+1} + \mathcal{O}\left(\sum\limits_{k=n}^{2n}\frac{1}{k^{1/2}(k+1)}\right)\\&=\pi\sum\limits_{k=n}^{2n}\frac{1}{k+1} + \mathcal{O}(n^{-1/2}) \end{align}$$

Hence, $$\sum\limits_{k=1}^{2n} (-1)^{k-1}\frac{r_2(k)}{k} = \pi\sum\limits_{k=n}^{2n}\frac{1}{k+1} + \mathcal{O}(n^{-1/2})$$

i.e., $$\sum\limits_{k=1}^{\infty} (-1)^{k-1}\frac{r_2(k)}{k} = \pi\log 2$$