Suppose I have a ''nice'' function $f : \mathbb{R} \rightarrow \mathbb{C}$. Fix an integer $\alpha$ and $d$. Could someone please explain me how the following formula holds?
$$ \sum_{a \equiv \alpha (mod \ d)} f(a^2 + b^2) = \frac{1}{d} \sum_k e^{2 \pi i \alpha k/d} \int_{- \infty}^{\infty} f(t^2 + b^2) e^{2 \pi i t k / d} dt $$ I am really confused how to make this work because of the sum over a fixed congruence class. Thank you very much!
If you set $h(x):=f((xd+\alpha)^2+b^2)$ then your identity is the Poisson summation formula $$\sum_{n\in\mathbb Z} h(n)=\sum_{n\in\mathbb Z}\hat h(n)$$ where $\hat h(y)=\int_{-\infty}^\infty h(x)\exp(-2\pi i xy)\,dx$.
Indeed $\sum_{a \equiv \alpha (mod \ d)} f(a^2 + b^2)=\sum_{n\in\mathbb Z} h(n)$ and $$\hat h(n)=\int_{-\infty}^\infty f((xd+\alpha)^2+b^2)\exp(-2\pi i nx)\,dx$$ $$=\frac{1}{d} \exp(2 \pi i \alpha n/d) \int_{- \infty}^{\infty} f(t^2 + b^2) \exp(-2 \pi i t n / d) dt$$ (with $t=xd+\alpha$).