[proofplan] We first isolate the Perron kernel and prove that its symmetric vertical-line integral equals the indicator of the interval $(1,\infty)$. This is done by evaluating a truncated Bromwich integral using rectangular contours, closing to the left or to the right according as the parameter is larger or smaller than $1$, with the horizontal and far-vertical sides estimated explicitly. We then insert the absolutely convergent Dirichlet series into the finite-height integral, reduce the formula to a weighted sum of Perron kernels, and justify the limiting passage by splitting the $n$-sum into a finite part and a uniformly controlled tail. [/proofplan] [step:Evaluate the Perron kernel away from its discontinuity] For $y>0$ with $y \neq 1$, define the Perron kernel candidate \begin{align*} K_b: (0,\infty) \setminus \{1\} &\to \mathbb{C}\\ y &\mapsto \lim_{T \to \infty} \frac{1}{2\pi i} \int_{b-iT}^{b+iT} \frac{y^s}{s}\,ds, \end{align*} whenever this symmetric limit exists. We claim that \begin{align*} K_b(y) = \begin{cases} 1, & y>1,\\ 0, & 0<y<1. \end{cases} \end{align*} Let $\log y$ denote the real logarithm of $y$, and let $\mathcal{L}^1$ denote one-dimensional Lebesgue measure on $\mathbb{R}$. Define \begin{align*} G_y:\mathbb C\setminus\{0\}&\to\mathbb C\\ s&\mapsto \frac{y^s}{s}. \end{align*} Since $y^s=e^{s\log y}$ is an entire function of $s$, the map $G_y$ is meromorphic on $\mathbb{C}$ with a single simple pole at $s=0$, whose residue is $1$. Assume first that $y>1$. Let $R>b$ and $T>0$, and let $\Gamma_{R,T}$ be the positively oriented rectangle with vertical sides $\operatorname{Re}(s)=b$ and $\operatorname{Re}(s)=-R$, and horizontal sides $\operatorname{Im}(s)=\pm T$. The pole $s=0$ lies inside $\Gamma_{R,T}$ once $R>0$. By the [Residue Theorem](/page/Residue%20Theorem), \begin{align*} \frac{1}{2\pi i}\int_{\Gamma_{R,T}}\frac{y^s}{s}\,ds=1. \end{align*} With positive orientation, the side on $\operatorname{Re}(s)=b$ is traversed upward from $b-iT$ to $b+iT$, while the side on $\operatorname{Re}(s)=-R$ is traversed downward from $-R+iT$ to $-R-iT$. On the left vertical side, use the parametrization $s=-R+it$, where $t \in [-T,T]$. Since $|s| \ge R$ and $|y^s|=y^{-R}$, its absolute contribution is bounded by \begin{align*} \int_{-T}^{\,T}\frac{y^{-R}}{|-R+it|}\,d\mathcal{L}^1(t) \le \frac{2T}{R}y^{-R}. \end{align*} This bound tends to $0$ as $R\to\infty$ for each fixed $T$ because $y>1$. On the horizontal side with $\operatorname{Im}(s)=T$, use the parametrization $s=\sigma+iT$, where $\sigma \in [-R,b]$. Since $|\sigma+iT| \ge T$, its absolute contribution is bounded by \begin{align*} \int_{-R}^{b}\frac{y^\sigma}{|\sigma+iT|}\,d\mathcal{L}^1(\sigma) \le \frac{1}{T}\int_{-R}^{b}y^\sigma\,d\mathcal{L}^1(\sigma) \le \frac{1}{T}\int_{-\infty}^{b}y^\sigma\,d\mathcal{L}^1(\sigma) = \frac{y^b}{T\log y}. \end{align*} The same estimate holds for the horizontal side with $\operatorname{Im}(s)=-T$. Therefore the contour identity may be read as the upward integral on $\operatorname{Re}(s)=b$ plus the two horizontal side integrals plus the downward integral on $\operatorname{Re}(s)=-R$. For each fixed $T>0$, first let $R\to\infty$; the left vertical contribution tends to $0$, and the two horizontal contributions remain bounded in absolute value by $2y^b/(T\log y)$. Then let $T\to\infty$, so the horizontal bound tends to $0$. Hence the residue identity gives \begin{align*} \lim_{T\to\infty} \frac{1}{2\pi i} \int_{b-iT}^{b+iT}\frac{y^s}{s}\,ds = 1. \end{align*} Assume next that $0<y<1$. Let $R>b$ and close the contour to the right, using the positively oriented rectangle with vertical sides $\operatorname{Re}(s)=b$ and $\operatorname{Re}(s)=R$. This rectangle contains no pole of $s \mapsto y^s/s$, because $0$ lies to the left of the line $\operatorname{Re}(s)=b$ and $b>0$. By the [Residue Theorem](/page/Residue%20Theorem), the contour integral is zero. With positive orientation, the side on $\operatorname{Re}(s)=b$ is traversed downward from $b+iT$ to $b-iT$, so the upward truncated Perron integral appears with the opposite sign in the contour identity. On the right vertical side, using $s=R+it$ with $t \in [-T,T]$, the absolute contribution is bounded by \begin{align*} \int_{-T}^{\,T}\frac{y^R}{|R+it|}\,d\mathcal{L}^1(t) \le \frac{2T}{R}y^R, \end{align*} which tends to $0$ as $R\to\infty$ for each fixed $T$ because $0<y<1$. On the horizontal side with $\operatorname{Im}(s)=T$, using $s=\sigma+iT$ with $\sigma \in [b,R]$, we have \begin{align*} \int_b^R\frac{y^\sigma}{|\sigma+iT|}\,d\mathcal{L}^1(\sigma) \le \frac{1}{T}\int_b^R y^\sigma\,d\mathcal{L}^1(\sigma) \le \frac{1}{T}\int_b^\infty y^\sigma\,d\mathcal{L}^1(\sigma) = \frac{y^b}{T\log(1/y)}. \end{align*} The same estimate holds on the horizontal side with $\operatorname{Im}(s)=-T$. Taking first $R\to\infty$ eliminates the right vertical contribution and leaves the two horizontal contributions bounded by $2y^b/(T\log(1/y))$; taking then $T\to\infty$ eliminates the horizontal contributions. Accounting for the opposite orientation of the side $\operatorname{Re}(s)=b$ therefore gives \begin{align*} \lim_{T\to\infty} \frac{1}{2\pi i} \int_{b-iT}^{b+iT}\frac{y^s}{s}\,ds = 0. \end{align*} This proves the claimed formula for $K_b(y)$. [/step] [step:Insert the Dirichlet series into the truncated integral] For $T>0$, define the truncated Perron integral \begin{align*} I_x:(0,\infty) &\to \mathbb{C}\\ T &\mapsto \frac{1}{2\pi i} \int_{b-iT}^{b+iT} F(s)\frac{x^s}{s}\,ds. \end{align*} For brevity write $I_T(x):=I_x(T)$. Parametrize the vertical segment by the map \begin{align*} \gamma_T:[-T,T]&\to \mathbb{C},\\ t&\mapsto b+it. \end{align*} Since the Dirichlet series for $F$ converges absolutely on $\operatorname{Re}(s)=b$, \begin{align*} \sum_{n=1}^{\infty}|a_n|n^{-b}<\infty. \end{align*} For every $t\in[-T,T]$, \begin{align*} |a_n n^{-(b+it)}x^{b+it}(b+it)^{-1}| \le \frac{x^b}{b}|a_n|n^{-b}. \end{align*} The right-hand side is summable over $n$. Hence the [Weierstrass M-test](/page/Weierstrass%20M-test) gives [uniform convergence](/page/Uniform%20Convergence) of the integrand series on the finite interval $[-T,T]$, and uniform convergence permits termwise integration, giving \begin{align*} I_T(x) &= \sum_{n=1}^{\infty} a_n \frac{1}{2\pi i} \int_{b-iT}^{b+iT} \left(\frac{x}{n}\right)^s\frac{ds}{s}. \end{align*} [/step] [step:Pass from the kernel identity to the summatory function] For $T>0$ and $n \in \mathbb{N}$, define \begin{align*} K_{b,T}: (0,\infty) &\to \mathbb{C}\\ y &\mapsto \frac{1}{2\pi i} \int_{b-iT}^{b+iT}y^s\frac{ds}{s}. \end{align*} Because $x$ is not an integer, $x/n\neq1$ for every $n\in\mathbb{N}$. Applying the kernel computation with $y=x/n$ gives \begin{align*} \lim_{T\to\infty}K_{b,T}(x/n) = \begin{cases} 1, & n<x,\\ 0, & n>x. \end{cases} \end{align*} It remains to justify passing this limit through the infinite $n$-sum. Choose $N_0\in\mathbb{N}$ with $N_0>2x$. For $n\ge N_0$, define the positive number $\lambda_n:=\log(n/x)$, so $\lambda_n\ge \log 2>0$. The parametrization $s=b+it$, $t\in[-T,T]$, gives \begin{align*} K_{b,T}(x/n) &= \frac{1}{2\pi}\left(\frac{x}{n}\right)^b \int_{-T}^{\,T}\frac{e^{-it\lambda_n}}{b+it}\,d\mathcal{L}^1(t). \end{align*} Define the complex-valued amplitude \begin{align*} h_n:\mathbb{R} &\to \mathbb{C}\\ t &\mapsto \frac{1}{b+it}. \end{align*} Since $b>0$, the map $h_n$ is continuously differentiable and satisfies $h_n'(t)=-i(b+it)^{-2}$. Applying [integration by parts](/page/Integration%20By%20Parts) on the finite interval $[-T,T]$ to the product of $h_n(t)$ and $e^{-it\lambda_n}$ gives \begin{align*} \int_{-T}^{\,T}e^{-it\lambda_n}h_n(t)\,d\mathcal{L}^1(t) &= \left[\frac{e^{-it\lambda_n}}{-i\lambda_n}h_n(t)\right]_{-T}^{\,T} + \frac{1}{i\lambda_n}\int_{-T}^{\,T}e^{-it\lambda_n}h_n'(t)\,d\mathcal{L}^1(t). \end{align*} Taking absolute values and using $|h_n(\pm T)|\le 1/b$ and $|h_n'(t)|=(b^2+t^2)^{-1}$, we obtain \begin{align*} \left|\int_{-T}^{\,T}\frac{e^{-it\lambda_n}}{b+it}\,d\mathcal{L}^1(t)\right| &\le \frac{2}{b\lambda_n} + \frac{1}{\lambda_n}\int_{-T}^{\,T}\frac{1}{b^2+t^2}\,d\mathcal{L}^1(t)\\ &\le \frac{2}{b\lambda_n} + \frac{1}{\lambda_n}\int_{-\infty}^{\infty}\frac{1}{b^2+t^2}\,d\mathcal{L}^1(t)\\ &= \frac{\pi+2}{b\lambda_n} \le \frac{\pi+2}{b\log 2}. \end{align*} Therefore, for all $T>0$ and all $n\ge N_0$, \begin{align*} |a_n K_{b,T}(x/n)| \le \frac{(\pi+2)x^b}{2\pi b\log 2}|a_n|n^{-b}. \end{align*} The sequence on the right is summable over $n\ge N_0$ because the Dirichlet series for $F$ converges absolutely on $\operatorname{Re}(s)=b$. Let $\varepsilon>0$. Choose $N\ge N_0$ such that \begin{align*} \sum_{n>N}\frac{(\pi+2)x^b}{2\pi b\log 2}|a_n|n^{-b}<\varepsilon. \end{align*} The finite sum over $1\le n\le N$ converges term by term by the kernel computation, while the preceding estimate bounds the tail uniformly in $T$. Hence \begin{align*} \lim_{T\to\infty} I_T(x) &= \sum_{n=1}^{\infty} a_n\,\mathbb{1}_{\{n<x\}}. \end{align*} Since $x$ is non-integral, the conditions $n<x$ and $n\le x$ are equivalent for integers $n$, and therefore \begin{align*} \lim_{T\to\infty} I_T(x) &= \sum_{n\le x}a_n. \end{align*} Substituting the definition of $I_T(x)$ yields \begin{align*} \sum_{n\le x} a_n = \lim_{T\to\infty} \frac{1}{2\pi i} \int_{b-iT}^{b+iT} F(s)\frac{x^s}{s}\,ds, \end{align*} which is Perron's formula at every non-integral point $x>0$. [/step]