I know the circle method can be used to count solutions to equations in a set. For example, if $P$ is the set of prime numbers and $eq_n$ is the equation $x^2+y^2=n$ then the solutions of $eq_n$ with $x,y\in P$ and $n\in \mathbb{N}$ can be counting using the fact that $$\int_0^1 e(\alpha h)\,d\alpha$$ equals $1$ when $h=0$ and equals $0$ otherwise. Taking $h=(p_1^2+p_2^2)-n$ for primes $p_1,p_2$ we get that the number of solutions is $$\sum_{p_1\;\text{prime}}\;\sum_{p_2\;\text{prime}}\int_0^1 e(\alpha(p_1^2+p_2^2-n))\,d\alpha=$$ $$=\cdots=$$ $$=\int_0^1 F(n)e(-\alpha n)\, d\alpha$$ for some function $F$.
Now lets say I want to count the number of solutions to $x^2+y^2<n$. In the same way that computing $$\int_0^1e(\alpha h)\,d\alpha$$ tells us whether $h=0$ or not, is there a similar integral or formula that tells us if $h\ge 0$ or not? A function $G$ of $h$ which equals 1 only if $h\ge 0$ ?
Would a function $f$ defined as $$f(h)=0\quad\text{if }h\ge 0$$ $$f(h)=x\quad\text{if }h<0$$ work? Maybe a smoother function ($e^{-1/x} type)$ is better?
Thanks!
I found a pretty neat answer: I realized that $$h\ge 0\implies\exists d\in\{0,1,2,\dots\}\quad\text{such that}\quad h=d$$ (crazy)
And, because $$\int_0^1 e(\alpha(h-d))\,d\alpha$$ is $=1$ iff $h=d$, and $h$ can only be equal to one and only one $d$ we have that $$\sum_{d=0}^\infty\int_0^1 e(\alpha(h-d))\,d\alpha$$ is equal to $1$ iff the integral is $=1$ for some $d=0,1,\dots$ iff $h=d$ for some $d=0,1\dots$ iff $h\ge 0$.
I am yet to prove rigorously that the sum and integral can be interchanged so that $$\int_0^1e(\alpha h)\sum_{d=0}^\infty e(-\alpha d)\,d\alpha=\int_0^1e(\alpha h)\frac{e(\alpha)}{e(\alpha)-1}\,d\alpha$$ is equal to one iff $h\ge 0$ and 0 otherwise. (Maybe the simplification of the sum is wrong but the details are not important)
EDIT: As Paul observed in his comment, the series $$\sum_{d=0}^\infty e(-\alpha d)$$ doesn't converge. However, I realized that, for my particular problem, I don't need to sum from $d=0$ to $\infty$ but to a finite $D$, so the sum $$\sum_{d=0}^D e(-\alpha d)$$ can be used instead without problems (and without justifying thesum-integral interchange)