For all $n\ge 1$, let $$ a_n = \begin{cases} 1\quad&\text{if $n$ can be written as the sum of two squares;}\\ 0&\text{otherwise} \end{cases} $$ I am interested in $A(x):=\sum_{n\le x}a_n$.
Consider the $L$-function $F(s)=\sum_{n\ge 1}\frac{a_n}{n^s}$. We see that $F$ converges absolutely for $\operatorname{Re}(s)>1$. Moreover, let $\chi$ be the non-trivial character modulo $4$, then $$ F(s)^2 = L(\chi,s)\zeta(s)(1-2^{-2s})^{-1}\prod_{p\equiv 3\pmod 4}(1-p^{-2s})^{-1} =: L(\chi,s)\zeta(s)f(s)^2, $$ so $F(s)^2$ has a meromorphic continuation to $\{s\in\mathbb{C}:\operatorname{Re}(s)>1/2\}$, with a simple pole at $s=1$.
Now, Exercise $6.21(d)$ in Montgomery and Vaughan, which I need help with, is to show $$ A(x) = \frac1{2\pi i}\int_{\mathcal{C}}f(s)\sqrt{\zeta(s)L(\chi s)}\frac{x^s}{s}\,ds + O\left(x\exp\left(-c\sqrt{\log x}\right)\right), $$ where $\mathcal{C}$ is the contour from $1-c-i\delta$ in a straight line to $1-i\delta$, along the semi-circle $1+\delta e^{i\theta}$ for $-\pi/2\le \theta\le \pi/2$ to $1+i\delta$, and from there to $1-c+i\delta$, where $\delta = 1/\log x$ and $\delta$ is sufficiently small.
My first, and thus far only, idea was to consider the closed contour $\mathcal{C}'$, which is $\mathcal{C}$ together with the line segment from $1-c+i\delta$ to $1-c-i\delta$, and to compute $$ \frac1{2\pi i}\oint_{\mathcal{C}'}f(s)\sqrt{\zeta(s)L(\chi s)}\frac{x^s}{s}\,ds $$ using the residue formula, and to relate $A(x)$ to $$ \int_{1-c-i\delta}^{1-c+i\delta}f(s)\sqrt{\zeta(s)L(\chi s)}\frac{x^s}{s}\,ds $$ using Perron's formula.
However, I don't think I can apply the residue theorem to the contour integral around $\mathcal{C}'$, since $f(s)\sqrt{\zeta(s) L(\chi,s)}$ is nowhere near holomorphic around $s=1$ (the square of this function has a pole of order $1$ at $s=1$). And I don't think I can apply Perron's formula because $1-c<1=\sigma_c(F)$.
You still can use Perron's formula, but Hankel contour has to be used instead of closed rectangles because the the function in question is only one-valued in a cut plane. The resulting formula is
$$ A(x)\sim{Kx\over\sqrt{\log x}}\left(1+{b_1\over\log x}+{b_2\over\log^2x}+{b_3\over\log^3x}+\dots\right), $$
where the right hand side is an asymptotic series. I have written a detailed proof of this result in this blog article.