I was trying to compute the following sum:
$$\sum_{n\le x}{\frac{r_2(n)}{n}}$$
where $r_2(n)=\vert\{(a,b)\in\mathbb{Z}^2:a^2+b^2=n\}\vert$. Using Abel's summation formula with $a_n=r_2(n)$, $\varphi(t)=\frac{1}{t}$, rememebering that $R_2(x)=\sum_{n\le x}{r_2(n)}=\pi x+O(\sqrt{x})$, we get:
$$\sum_{n\le x}{\frac{r_2(n)}{n}}=\frac{R_2(x)}{x}+\int_{1}^{x}\frac{R_2(t)}{t^2}dt=$$
$$=\pi+O(x^{-1/2})+\int_{1}^{x}\left(\frac{\pi}{t}+\frac{R_2(t)-\pi t}{t^2}\right)dt=$$
$$=\pi\log(x)+\pi+O(x^{-1/2})+\int_{1}^{+\infty}\frac{R_2(t)-\pi t}{t^2}dt-\int_{x}^{+\infty}{\frac{R_2(t)-\pi t}{t^2}dt}=$$
$$=\pi\log(x)+C+O(x^{-1/2})$$
where we also used the fact that the improper integral of $\frac{R_2(t)-\pi t}{t^2}$ converges since the integrand is $O(x^{-3/2})$, and the same estimate gives us an errore term for the integral starting from $x$.
Thus, this method provides us with a formula involving some unknown constant $C$, which in our calculation is given by $\pi$ and the improper integral. However, we can also use the explicit formula for $r_2(n)$ to try and get the exact value of $C$. It is well know that, if $\chi$ is the non-principal Dirichlet character modulo 4, then $r_2(n)=4\sum_{d\vert n}\chi(d)$, so we can say that:
$$\sum_{n\le x}{\frac{r_2(n)}{n}}=4\sum_{n\le x}\sum_{d\vert n}{\frac{\chi(d)}{n}}$$
If $\frac{n}{d}=m$, we can exchange the sums and reindex them in terms of $d$ and $m$ as it follows (Dirichlet's hyperbola method):
$$=4\sum_{d\le x}\sum_{m\le x/d}{\frac{\chi(d)}{dm}}=4\sum_{d\le x}{\frac{\chi(d)}{d}\sum_{m\le x/d}{\frac{1}{m}}}$$
To be more careful, I decided to break the sum in two parts, when $d\le x^{1/2}$ and when $x^{1/2}<d\le x$, such that in the first case we can estimate the second sum with the formula for $\sum_{k\le t}{\frac{1}{k}}=\log(t)+\gamma+O(\frac{1}{t})$ because $x/d\ge x^{1/2}$ so we can get a bound for the error. In the second case, we can estimate the sum in other ways to show that it is infinitesimal when $x\rightarrow +\infty$. Precisely:
$$\sum_{d\le x^{1/2}}{\frac{\chi(d)}{d}\sum_{m\le x/d}{\frac{1}{m}}}=\sum_{d\le x^{1/2}}{\frac{\chi(d)}{d}\left(\log(x)-\log(d)+\gamma+O(d/x)\right)}=$$
$$=\log(x)\sum_{d\le x^{1/2}}{\frac{\chi(d)}{d}}-\sum_{d\le x^{1/2}}{\frac{\chi(d)\log(d)}{d}}+\gamma\sum_{d\le x^{1/2}}{\frac{\chi(d)}{d}}+O\left(\frac{1}{x}\sum_{d\le x^{1/2}}{1}\right)$$
We now want to use the fact that $\sum_{d>t}{\frac{\chi(d)}{d}}=O(\frac{1}{t})$ and $\sum_{d>t}{\frac{\chi(d)\log(d)}{d}}=O\left(\frac{\log(t)}{t}\right)$, together with the convergence of $L(1,\chi)$ and $L'(1,\chi)$ to get:
$$\log(x)\left(L(1,\chi)+O(x^{-1/2})\right)+L'(1,\chi)+O\left(\frac{\log(x)}{x^{1/2}}\right)+\gamma\left(L(1,\chi)+O(x^{-1/2})\right)+O\left(\frac{1}{x^{1/2}}\right)=$$
$$=\frac{\pi}{4}\log(x)+\frac{\gamma\pi}{4}+L'(1,\chi)+O\left(\frac{\log(x)}{x^{1/2}}\right)$$
Using the well know fact that $L(1,\chi)=\frac{\pi}{4}$. For the other sum, notice that when $d$ is even $\chi(d)=0$, so we only care about odd terms. If we set $d=2k+1$, we get:
$$\sum_{x^{1/2}<d\le x}{\frac{\chi(d)}{d}\sum_{m\le x/d}{\frac{1}{m}}}=\sum_{\frac{x^{1/2}-1}{2}<k\le \frac{x-1}{2}}{\frac{\chi(2k+1)}{2k+1}\sum_{m\le x/(2k+1)}{\frac{1}{m}}}=\sum_{\frac{x^{1/2}-1}{2}<k\le \frac{x-1}{2}}{\frac{(-1)^k}{2k+1}\sum_{m\le x/(2k+1)}{\frac{1}{m}}}$$
If we let $a_k=\frac{1}{2k+1}\sum_{m\le x/(2k+1)}{\frac{1}{m}}$, then we have $a_k\ge 0$ for all $k$ and the sum of $(-1)^ka_k$, an alternating sum with $a_k$ strictly decreasing. Thus, we can bound it with the absolute value of the smallest term:
$$\sum_{\frac{x^{1/2}-1}{2}<k\le\frac{x-1}{2}}{(-1)^ka_k}=O\left(\frac{1}{x^{1/2}}\sum_{m\le x^{1/2}}{\frac{1}{m}}\right)=O\left(\frac{\log(x)}{x^{1/2}}\right)$$
Combining the result gotten so far, we can say that:
$$\sum_{n\le x}{\frac{r_2(n)}{n}}=\pi\log(x)+\pi\gamma+4L'(1,\chi)+O\left(\frac{\log(x)}{x^{1/2}}\right)$$
It now follows that $C=\pi\gamma+4L'(1,\chi)$, giving us also the following result:
$$\int_{1}^{+\infty}{\frac{R_2(t)-\pi t}{t^2}dt}=\pi(\gamma-1)+4L'(1,\chi)$$
Thus the only thing left to compute would be $L'(1,\chi)$. Does it converge to a nice, well known value?
