The average order of the two-square representation count is pi

By The divisor formula for the two-square representation count, $r_2=4(\mathbf{1}\ast {\chi }_4)$. Applying Dirichlet's hyperbola method for summatory convolutions with $f=\mathbf{1}$, $g={\chi }_4$, and $U=V=\sqrt{x}$ gives ${\sum }_{n\le x}{r}_2(n)=4({\sum }_{a\le N}G(x\mathrm{/}a)+{\sum }_{b\le N}{\chi }_4(b)\lfloor \frac{x}{b}\rfloor -NG(N))\mathrm{.}$

The values of ${\chi }_4$ repeat as $1,0,-1,0$, so every complete block of length $4$ has sum $0$ and every initial partial block has sum $0$ or $1$. Hence $G(y)=O(1)$ uniformly in $y$, and step 1.1 gives ${\sum }_{a\le N}G(x\mathrm{/}a)=O(N),NG(N)=O(N)\mathrm{.}$ Also $\lfloor x\mathrm{/}b\rfloor =x\mathrm{/}b+O(1)$, so ${\sum }_{b\le N}{\chi }_4(b)\lfloor \frac{x}{b}\rfloor =x{\sum }_{b\le N}\frac{{\chi }_4(b)}{b}+O(N)\mathrm{.}$

Deleting the zero even terms identifies ${\sum }_{b\le N}{\chi }_4(b)\mathrm{/}b$ with a partial Gregory-Leibniz sum. Therefore The Gregory-Leibniz series: pi over four equals 1-1/3+1/5-1/7+... gives ${\sum }_{b\le N}\frac{{\chi }_4(b)}{b}=\frac{\pi }{4}+O(1\mathrm{/}N)\mathrm{.}$

Substituting step 3.1 into step 2.1 and then into step 1.1 yields ${\sum }_{n\le x}{r}_2(n)=4(x\cdot \frac{\pi }{4}+O(N))=\pi x+O(\sqrt{x})\mathrm{.}$ Since ${\sum }_{n\le x}\pi =\pi \lfloor x\rfloor =\pi x+O(1)$, Summatory functions and average orders now says that the constant function $\pi$ is an average order of $r_2$.