[proofplan] The proof reduces Perron's formula to the Mellin inversion kernel for the indicator of $(1,\infty)$, with half-weight at the endpoint. First we expand the absolutely convergent Dirichlet series on the vertical line $\operatorname{Re}(s)=c$ and justify the finite-height interchange of summation and integration. Then we evaluate the kernel \begin{align*} K_T(y)=\frac{1}{2\pi i}\int_{c-iT}^{c+iT}\frac{y^s}{s}\,ds \end{align*} as $T \to \infty$. Finally, a dominated convergence argument against the absolutely summable sequence $|a_n|n^{-c}$ passes the kernel limit through the Dirichlet series. [/proofplan]

[step:Define the Perron kernel and justify termwise integration at finite height]Let $\mathcal{L}^1$ denote one-dimensional Lebesgue measure on $\mathbb{R}$. For $T \geq 1$, define the truncated Perron kernel \begin{align*} K_T:(0,\infty) &\to \mathbb{C} \\ y &\mapsto \frac{1}{2\pi i}\int_{c-iT}^{c+iT}\frac{y^s}{s}\,ds. \end{align*} Parametrising the vertical segment by the map \begin{align*} \gamma_T:[-T,T] &\to \mathbb{C} \\ t &\mapsto c+it, \end{align*} we have $ds=i\,d\mathcal{L}^1(t)$ and therefore \begin{align*} K_T(y) = \frac{1}{2\pi}\int_{-T}^{\,T}\frac{y^{c+it}}{c+it}\,d\mathcal{L}^1(t). \end{align*} Since $c>\sigma_a$, the series $\sum_{n=1}^{\infty} |a_n|n^{-c}$ converges. For each fixed $T \geq 1$, \begin{align*} \sum_{n=1}^{\infty} \int_{-T}^{\,T} \left| a_n n^{-c-it}\frac{x^{c+it}}{c+it} \right|\,d\mathcal{L}^1(t) &= x^c \left(\sum_{n=1}^{\infty}|a_n|n^{-c}\right) \int_{-T}^{\,T}\frac{1}{|c+it|}\,d\mathcal{L}^1(t) \\ &< \infty. \end{align*} Thus the sum and the finite-height integral may be interchanged by absolute integrability on $\mathbb{N}\times[-T,T]$. Hence \begin{align*} I_T(x) &= \sum_{n=1}^{\infty} a_n \frac{1}{2\pi i} \int_{c-iT}^{c+iT} \frac{(x/n)^s}{s}\,ds \\ &= \sum_{n=1}^{\infty}a_n K_T(x/n). \end{align*}[/step]

[guided]Let $\mathcal{L}^1$ denote one-dimensional Lebesgue measure on $\mathbb{R}$. The purpose of introducing $K_T$ is to isolate the part of the integral that depends only on the ratio $x/n$. Define \begin{align*} K_T:(0,\infty) &\to \mathbb{C} \\ y &\mapsto \frac{1}{2\pi i}\int_{c-iT}^{c+iT}\frac{y^s}{s}\,ds. \end{align*} The vertical path is parametrised by \begin{align*} \gamma_T:[-T,T] &\to \mathbb{C} \\ t &\mapsto c+it. \end{align*} Under this parametrisation, $ds=i\,d\mathcal{L}^1(t)$, so \begin{align*} K_T(y) = \frac{1}{2\pi}\int_{-T}^{\,T}\frac{y^{c+it}}{c+it}\,d\mathcal{L}^1(t). \end{align*} We now justify the interchange of summation and integration. The hypothesis $c>\sigma_a$ says precisely that the Dirichlet series is absolutely convergent on the line $\operatorname{Re}(s)=c$, so \begin{align*} \sum_{n=1}^{\infty}|a_n|n^{-c}<\infty. \end{align*} For fixed $T\geq 1$, the absolute value of the integrand after expanding $F$ is \begin{align*} \left| a_n n^{-c-it}\frac{x^{c+it}}{c+it} \right| = |a_n|n^{-c}\frac{x^c}{|c+it|}. \end{align*} Therefore \begin{align*} \sum_{n=1}^{\infty} \int_{-T}^{\,T} \left| a_n n^{-c-it}\frac{x^{c+it}}{c+it} \right|\,d\mathcal{L}^1(t) &= x^c \left(\sum_{n=1}^{\infty}|a_n|n^{-c}\right) \int_{-T}^{\,T}\frac{1}{|c+it|}\,d\mathcal{L}^1(t) \\ &<\infty. \end{align*} This finite quantity is the exact condition needed to exchange the countable sum with the finite vertical integral. After the interchange, the $n$th term becomes \begin{align*} a_n \frac{1}{2\pi i} \int_{c-iT}^{c+iT} \frac{(x/n)^s}{s}\,ds = a_nK_T(x/n). \end{align*} Hence \begin{align*} I_T(x)=\sum_{n=1}^{\infty}a_nK_T(x/n). \end{align*}[/guided]

[step:Evaluate the limiting Perron kernel][claim:Kernel limit] For every $y>0$, \begin{align*} \lim_{T\to\infty}K_T(y) = \begin{cases} 1, & y>1,\\ \frac{1}{2}, & y=1,\\ 0, & 0<y<1. \end{cases} \end{align*} Moreover, there exists a constant $C_c>0$, depending only on $c$, such that \begin{align*} |K_T(y)|\leq C_c y^c \end{align*} for all $T\geq 1$ and all $y>0$. [/claim] [proof] Write $\lambda=\log y$. If $y=1$, then \begin{align*} K_T(1) &= \frac{1}{2\pi}\int_{-T}^{\,T}\frac{1}{c+it}\,d\mathcal{L}^1(t) \\ &= \frac{1}{2\pi}\int_{-T}^{\,T}\frac{c-it}{c^2+t^2}\,d\mathcal{L}^1(t). \end{align*} The odd part integrates to $0$, and hence \begin{align*} K_T(1) = \frac{1}{2\pi}\int_{-T}^{\,T}\frac{c}{c^2+t^2}\,d\mathcal{L}^1(t) = \frac{1}{\pi}\arctan(T/c) \to \frac{1}{2}. \end{align*} Assume next that $y\neq 1$. The function \begin{align*} G_y:\mathbb{C}\setminus\{0\} &\to \mathbb{C} \\ s &\mapsto \frac{y^s}{s} \end{align*} is holomorphic away from the simple pole at $s=0$, where the residue is $1$. If $y>1$, let $R>0$ and close the segment from $c-iT$ to $c+iT$ by the positively oriented rectangle with left side $\operatorname{Re}(s)=-R$. On that rectangle, $|y^s|=y^{\operatorname{Re}(s)}$, so the left vertical side tends to $0$ as $R\to\infty$. On the horizontal sides, writing $s=\sigma\pm iT$ with $-R\leq \sigma\leq c$, we have $|s|\geq T$ and therefore \begin{align*} \int_{-R}^{c}\left|\frac{y^{\sigma\pm iT}}{\sigma\pm iT}\right|\,d\mathcal{L}^1(\sigma) &\leq \frac{1}{T}\int_{-R}^{c}y^\sigma\,d\mathcal{L}^1(\sigma) \\ &\leq \frac{y^c}{T\log y}. \end{align*} Thus, after first letting $R\to\infty$ and then $T\to\infty$, the added sides contribute $0$. The enclosed pole at $s=0$ contributes residue $1$ by the [Residue Theorem](/theorems/352), so \begin{align*} \lim_{T\to\infty}K_T(y)=1. \end{align*} If $0<y<1$, let $R>c$ and close the segment to the right with right side $\operatorname{Re}(s)=R$. Since $y^R\to 0$ as $R\to\infty$, the right vertical side tends to $0$. On the horizontal sides, writing $s=\sigma\pm iT$ with $c\leq \sigma\leq R$, we have $|s|\geq T$ and hence \begin{align*} \int_{c}^{R}\left|\frac{y^{\sigma\pm iT}}{\sigma\pm iT}\right|\,d\mathcal{L}^1(\sigma) &\leq \frac{1}{T}\int_c^R y^\sigma\,d\mathcal{L}^1(\sigma) \\ &\leq \frac{y^c}{T|\log y|}. \end{align*} The positively oriented right-closing rectangle traverses the original vertical segment downward, from $c+iT$ to $c-iT$, so its vertical contribution is the negative of the segment defining $K_T(y)$. Since the rectangle encloses no pole, because $s=0$ lies to the left of the line $\operatorname{Re}(s)=c>0$, the sum of all four contour integrals is $0$. The added sides vanish after first letting $R\to\infty$ and then $T\to\infty$, and therefore the original upward vertical integral has limit $0$: \begin{align*} \lim_{T\to\infty}K_T(y)=0. \end{align*} It remains to prove the uniform domination. Define \begin{align*} H_T:\mathbb{R} &\to \mathbb{C} \\ \lambda &\mapsto \frac{1}{2\pi}\int_{-T}^{\,T}\frac{e^{it\lambda}}{c+it}\,d\mathcal{L}^1(t). \end{align*} We prove that there is a constant $C_c>0$, depending only on $c$, such that $|H_T(\lambda)|\leq C_c$ for every $T\geq1$ and every $\lambda\in\mathbb{R}$. Pairing the terms at $t$ and $-t$ gives \begin{align*} H_T(\lambda) = \frac{1}{\pi}\int_0^{\,T}\frac{c\cos(t\lambda)+t\sin(t\lambda)}{c^2+t^2}\,d\mathcal{L}^1(t). \end{align*} The cosine contribution satisfies \begin{align*} \left|\frac{1}{\pi}\int_0^{\,T}\frac{c\cos(t\lambda)}{c^2+t^2}\,d\mathcal{L}^1(t)\right| \leq \frac{1}{\pi}\int_0^\infty\frac{c}{c^2+t^2}\,d\mathcal{L}^1(t) = \frac{1}{2}. \end{align*} For the sine contribution, define \begin{align*} b_c:[0,\infty)&\to\mathbb{R} \\ t&\mapsto \frac{t}{c^2+t^2}. \end{align*} The map $b_c$ is absolutely continuous, increases on $[0,c]$, decreases on $[c,\infty)$, and satisfies \begin{align*} \|b_c\|_{L^\infty([0,\infty))}=\frac{1}{2c}, \qquad \operatorname{Var}_{[0,\infty)}(b_c)=\frac{1}{c}. \end{align*} If $|\lambda|\geq1$, [integration by parts](/theorems/2098) on each interval of monotonicity, applied to the absolutely continuous primitive $t\mapsto -\cos(t\lambda)/\lambda$ of $\sin(t\lambda)$, gives \begin{align*} \left|\int_0^{\,T} b_c(t)\sin(t\lambda)\,d\mathcal{L}^1(t)\right| \leq \frac{2\|b_c\|_{L^\infty([0,\infty))}+\operatorname{Var}_{[0,\infty)}(b_c)}{|\lambda|} \leq \frac{2}{c}. \end{align*} If $0<|\lambda|<1$, set $u=|\lambda|t$ and $\beta=c|\lambda|$. The sine contribution has absolute value \begin{align*} \left|\int_0^{|\lambda|T}\frac{u\sin u}{\beta^2+u^2}\,d\mathcal{L}^1(u)\right|. \end{align*} On $[0,1]$ this is at most $1$. On $[1,\infty)$ the function $u\mapsto u/(\beta^2+u^2)$ has total variation at most $2$ uniformly for $0<\beta<c$, and the primitive of $\sin u$ is bounded by $2$ on every interval. The Dirichlet integration-by-parts estimate therefore gives a bound at most $4$ on the part over $[1,|\lambda|T]$. Thus the sine contribution is bounded by $5$ when $0<|\lambda|<1$, and it is $0$ when $\lambda=0$. Hence \begin{align*} |H_T(\lambda)|\leq \frac{1}{2}+\frac{1}{\pi}\max\left\{\frac{2}{c},5\right\}=:C_c \end{align*} for all $T\geq1$ and all $\lambda\in\mathbb{R}$. Since $K_T(y)=y^cH_T(\log y)$, this proves \begin{align*} |K_T(y)|\leq C_c y^c \end{align*} for all $T\geq1$ and all $y>0$. [/proof] Applying the claim with $y=x/n$ gives, for each $n\in\mathbb{N}$, \begin{align*} \lim_{T\to\infty}K_T(x/n) = \begin{cases} 1, & n<x,\\ \frac{1}{2}, & n=x,\\ 0, & n>x. \end{cases} \end{align*}[/step]

[guided]The Perron kernel is designed to behave like a sharp cutoff. The claim says exactly that: \begin{align*} K_T(y)\to \mathbb{1}_{(1,\infty)}(y)+\frac{1}{2}\mathbb{1}_{\{1\}}(y). \end{align*} For $y=1$, there is no oscillation, and the computation is direct: \begin{align*} K_T(1) &= \frac{1}{2\pi}\int_{-T}^{\,T}\frac{1}{c+it}\,d\mathcal{L}^1(t) \\ &= \frac{1}{2\pi}\int_{-T}^{\,T}\frac{c-it}{c^2+t^2}\,d\mathcal{L}^1(t). \end{align*} The imaginary part is odd as a function of $t$, so it integrates to $0$ over the symmetric interval $[-T,T]$. Thus \begin{align*} K_T(1) = \frac{1}{2\pi}\int_{-T}^{\,T}\frac{c}{c^2+t^2}\,d\mathcal{L}^1(t) = \frac{1}{\pi}\arctan(T/c) \to \frac{1}{2}. \end{align*} For $y\neq 1$, the sign of $\log y$ determines which side of the vertical line makes the exponential decay. Define \begin{align*} G_y:\mathbb{C}\setminus\{0\} &\to \mathbb{C} \\ s &\mapsto \frac{y^s}{s}. \end{align*} This function has a single simple pole at $s=0$, and its residue there is $1$ because $y^s=e^{s\log y}=1+O(s)$ as $s\to0$. If $y>1$, then $y^{\operatorname{Re}(s)}$ decays as $\operatorname{Re}(s)\to-\infty$. Let $R>0$ and close the contour to the left by the rectangle whose left side is $\operatorname{Re}(s)=-R$. The pole $s=0$ lies inside the resulting contour because $c>0$. The left side vanishes as $R\to\infty$ because $y^{-R}\to0$. On a horizontal side $s=\sigma\pm iT$, with $-R\leq\sigma\leq c$, the bound $|s|\geq T$ gives \begin{align*} \int_{-R}^{c}\left|\frac{y^{\sigma\pm iT}}{\sigma\pm iT}\right|\,d\mathcal{L}^1(\sigma) \leq \frac{1}{T}\int_{-R}^{c}y^\sigma\,d\mathcal{L}^1(\sigma) \leq \frac{y^c}{T\log y}. \end{align*} Thus the horizontal sides vanish as $T\to\infty$. By the [Residue Theorem](/theorems/352), the limiting vertical integral equals the residue contribution, namely $1$. If $0<y<1$, the exponential decays to the right. Let $R>c$ and close the contour to the right by the rectangle whose right side is $\operatorname{Re}(s)=R$. The pole $s=0$ is not enclosed, since it lies to the left of the original line $\operatorname{Re}(s)=c$. The right side vanishes as $R\to\infty$ because $y^R\to0$. On a horizontal side $s=\sigma\pm iT$, with $c\leq\sigma\leq R$, the bound $|s|\geq T$ gives \begin{align*} \int_c^R\left|\frac{y^{\sigma\pm iT}}{\sigma\pm iT}\right|\,d\mathcal{L}^1(\sigma) \leq \frac{1}{T}\int_c^R y^\sigma\,d\mathcal{L}^1(\sigma) \leq \frac{y^c}{T|\log y|}. \end{align*} For the right-closing rectangle, the positively oriented contour traverses the original vertical segment downward, from $c+iT$ to $c-iT$. Thus the contour integral contains the negative of the vertical integral defining $K_T(y)$. The pole $s=0$ is not enclosed, the added sides vanish as above, and therefore the limiting original vertical integral is $0$. We also need a domination estimate before passing the limit through the infinite sum. Define \begin{align*} H_T:\mathbb{R} &\to \mathbb{C} \\ \lambda &\mapsto \frac{1}{2\pi}\int_{-T}^{\,T}\frac{e^{it\lambda}}{c+it}\,d\mathcal{L}^1(t). \end{align*} The estimate to prove is a uniform bound for $H_T(\lambda)$, independent of both $T$ and $\lambda$. Pair the integrand values at $t$ and $-t$. Since \begin{align*} \frac{e^{it\lambda}}{c+it}+\frac{e^{-it\lambda}}{c-it} = 2\frac{c\cos(t\lambda)+t\sin(t\lambda)}{c^2+t^2}, \end{align*} we have \begin{align*} H_T(\lambda) = \frac{1}{\pi}\int_0^{\,T}\frac{c\cos(t\lambda)+t\sin(t\lambda)}{c^2+t^2}\,d\mathcal{L}^1(t). \end{align*} The cosine part is uniformly bounded because \begin{align*} \left|\frac{1}{\pi}\int_0^{\,T}\frac{c\cos(t\lambda)}{c^2+t^2}\,d\mathcal{L}^1(t)\right| \leq \frac{1}{\pi}\int_0^\infty\frac{c}{c^2+t^2}\,d\mathcal{L}^1(t) = \frac{1}{2}. \end{align*} For the sine part, define the absolutely continuous amplitude \begin{align*} b_c:[0,\infty)&\to\mathbb{R} \\ t&\mapsto \frac{t}{c^2+t^2}. \end{align*} It has $\|b_c\|_{L^\infty([0,\infty))}=1/(2c)$ and total variation $1/c$. When $|\lambda|\geq1$, [integration by parts](/theorems/210) against the primitive $t\mapsto-\cos(t\lambda)/\lambda$ of $\sin(t\lambda)$ gives \begin{align*} \left|\int_0^{\,T} b_c(t)\sin(t\lambda)\,d\mathcal{L}^1(t)\right| \leq \frac{2\|b_c\|_{L^\infty([0,\infty))}+\operatorname{Var}_{[0,\infty)}(b_c)}{|\lambda|} \leq \frac{2}{c}. \end{align*} When $0<|\lambda|<1$, the substitution $u=|\lambda|t$ gives \begin{align*} \left|\int_0^{\,T} b_c(t)\sin(t\lambda)\,d\mathcal{L}^1(t)\right| = \left|\int_0^{|\lambda|T}\frac{u\sin u}{c^2|\lambda|^2+u^2}\,d\mathcal{L}^1(u)\right|. \end{align*} The part over $[0,1]$ is at most $1$, using $|\sin u|\leq u$. On $[1,\infty)$ the amplitude $u\mapsto u/(c^2|\lambda|^2+u^2)$ has uniformly bounded total variation, and the primitive of $\sin u$ is uniformly bounded; the same integration-by-parts estimate bounds this tail by a universal constant. Thus there is a constant $C_c>0$, depending only on $c$, such that $|H_T(\lambda)|\leq C_c$ for all $T\geq1$ and all $\lambda\in\mathbb{R}$. Since \begin{align*} K_T(y) = y^cH_T(\log y), \end{align*} we obtain \begin{align*} |K_T(y)|\leq C_c y^c. \end{align*} Substituting $y=x/n$ now gives \begin{align*} \lim_{T\to\infty}K_T(x/n) = \begin{cases} 1, & n<x,\\ \frac{1}{2}, & n=x,\\ 0, & n>x. \end{cases} \end{align*}[/guided]

[step:Pass the kernel limit through the absolutely convergent Dirichlet series] From the previous steps, \begin{align*} I_T(x)=\sum_{n=1}^{\infty}a_nK_T(x/n). \end{align*} The kernel bound gives \begin{align*} |a_nK_T(x/n)| \leq C_c x^c |a_n|n^{-c}. \end{align*} The dominating series \begin{align*} \sum_{n=1}^{\infty}C_c x^c |a_n|n^{-c} \end{align*} converges because $c>\sigma_a$. Therefore the [Dominated Convergence Theorem](/theorems/4) applied to the [measure space](/page/Measure%20Space) $(\mathbb{N},2^{\mathbb{N}},\#)$, where $\#$ denotes counting measure, yields \begin{align*} \lim_{T\to\infty}I_T(x) &= \sum_{n=1}^{\infty}a_n\lim_{T\to\infty}K_T(x/n) \\ &= \sum_{n<x}a_n+\frac{1}{2}\mathbb{1}_{\mathbb{N}}(x)a_x. \end{align*} If $x\notin\mathbb{N}$, there is no integer $n$ with $n=x$, and the condition $n<x$ is equivalent to $n\leq x$ for $n\in\mathbb{N}$. Hence \begin{align*} \lim_{T\to\infty}I_T(x)=\sum_{n\leq x}a_n. \end{align*} If $x\in\mathbb{N}$, the term $n=x$ contributes exactly $\frac{1}{2}a_x$, so \begin{align*} \lim_{T\to\infty}I_T(x)=\sum_{n<x}a_n+\frac{1}{2}a_x. \end{align*} This is the asserted Perron formula. [/step]