[proofplan]
We apply the internal [Truncated Perron Formula](/theorems/TEMP-24) to the absolutely convergent Dirichlet series for the logarithmic derivative of the zeta function in the half-plane $\operatorname{Re}(s)>1$. The coefficient sequence is the von Mangoldt function, so the Perron main term is exactly the Chebyshev function. The same formula gives the stated finite-height error term, and dominated convergence in the explicit error series gives the limiting vertical-integral representation.
[/proofplan]
[step:Record the truncated Perron input]
Let $\mathcal{L}^1$ denote one-dimensional Lebesgue measure. We will use the [Truncated Perron Formula](/theorems/TEMP-24) in the following form. If a Dirichlet series
\begin{align*}
F(s)=\sum_{n=1}^{\infty}a_n n^{-s}
\end{align*}
converges absolutely in a half-plane containing the line $\operatorname{Re}(s)=c>0$, then for every $T\geq1$ and every non-integral $x>0$,
\begin{align*}
\sum_{n\leq x}a_n
&=
\frac{1}{2\pi i}\int_{c-iT}^{c+iT}F(s)\frac{x^s}{s}\,ds \\
&\quad+
O\left(
\sum_{n=1}^{\infty}|a_n|
\left(\frac{x}{n}\right)^c
\min\left\{1,\frac{1}{T|\log(x/n)|}\right\}
\right),
\end{align*}
with an absolute implicit constant. The non-integrality of $x$ ensures that no summand lies at the Perron kernel jump.
Define the interval indicator
\begin{align*}
\mathbb{1}_{(1,\infty)}:(0,\infty)&\to\{0,1\}\\
y&\mapsto
\begin{cases}
1, & y>1,\\
0, & 0<y\leq1.
\end{cases}
\end{align*}
Equivalently, the theorem uses the truncated Perron kernel
\begin{align*}
K_T:(0,\infty)\setminus\{1\}&\to\mathbb C\\
y&\mapsto \frac{1}{2\pi i}\int_{c-iT}^{c+iT}\frac{y^s}{s}\,ds
\end{align*}
and the estimate
\begin{align*}
K_T(y)=\mathbb{1}_{(1,\infty)}(y)
+
O\left(y^c\min\left\{1,\frac{1}{T|\log y|}\right\}\right),
\end{align*}
where the implicit constant is absolute.
[guided]
This step isolates the only Perron-kernel input used in the proof. Let
\begin{align*}
F(s)=\sum_{n=1}^{\infty}a_n n^{-s}
\end{align*}
be a Dirichlet series that converges absolutely in a half-plane containing the line $\operatorname{Re}(s)=c>0$. The internal truncated Perron theorem says that for every $T\geq1$ and every non-integral $x>0$,
\begin{align*}
\sum_{n\leq x}a_n
&=
\frac{1}{2\pi i}\int_{c-iT}^{c+iT}F(s)\frac{x^s}{s}\,ds \\
&\quad+
O\left(
\sum_{n=1}^{\infty}|a_n|
\left(\frac{x}{n}\right)^c
\min\left\{1,\frac{1}{T|\log(x/n)|}\right\}
\right).
\end{align*}
The implicit constant is absolute.
The same theorem may be read through the truncated Perron kernel
\begin{align*}
K_T:(0,\infty)\setminus\{1\}&\to\mathbb C\\
y&\mapsto \frac{1}{2\pi i}\int_{c-iT}^{c+iT}\frac{y^s}{s}\,ds.
\end{align*}
For $y\neq1$, it gives
\begin{align*}
K_T(y)=\mathbb{1}_{(1,\infty)}(y)
+
O\left(y^c\min\left\{1,\frac{1}{T|\log y|}\right\}\right).
\end{align*}
In the present proof the coefficients will be
\begin{align*}
a_n=\Lambda(n),
\end{align*}
and the corresponding Dirichlet series will be the logarithmic derivative of the zeta function.
[/guided]
[/step]
[step:Expand $-\zeta'/\zeta$ into its absolutely convergent Dirichlet series]
For $\operatorname{Re}(s)>1$, the logarithmic derivative identity obtained by differentiating the Euler product for the zeta function gives the absolutely convergent Dirichlet series expansion
\begin{align*}
-\frac{\zeta'}{\zeta}(s)=\sum_{n=1}^{\infty}\frac{\Lambda(n)}{n^s},
\end{align*}
and this series converges absolutely on every vertical line $\operatorname{Re}(s)=c$ with $c>1$. Thus
\begin{align*}
\sum_{n=1}^{\infty}\Lambda(n)n^{-c}<\infty.
\end{align*}
Fix $T\geq 1$. Since
\begin{align*}
\int_{-T}^{\,T}\left|\frac{x^{c+it}}{c+it}\right|\,d\mathcal{L}^1(t)
\leq x^c\int_{-T}^{\,T}\frac{1}{|c+it|}\,d\mathcal{L}^1(t)<\infty,
\end{align*}
Tonelli's theorem gives absolute integrability of the sum of absolute values on the finite vertical segment, and [Fubini's theorem](/theorems/2961) therefore permits termwise integration:
\begin{align*}
\frac{1}{2\pi i}\int_{c-iT}^{c+iT}-\frac{\zeta'}{\zeta}(s)\frac{x^s}{s}\,ds
&=\sum_{n=1}^{\infty}\Lambda(n)\frac{1}{2\pi i}\int_{c-iT}^{c+iT}\frac{(x/n)^s}{s}\,ds \\
&=\sum_{n=1}^{\infty}\Lambda(n)K_T(x/n).
\end{align*}
[guided]
The point of choosing $c>1$ is that the logarithmic derivative of the zeta function is represented there by an absolutely convergent Dirichlet series:
\begin{align*}
-\frac{\zeta'}{\zeta}(s)=\sum_{n=1}^{\infty}\frac{\Lambda(n)}{n^s}.
\end{align*}
On the line $\operatorname{Re}(s)=c$, absolute convergence says
\begin{align*}
\sum_{n=1}^{\infty}\Lambda(n)n^{-c}<\infty.
\end{align*}
We now justify exchanging the sum and the finite vertical integral. For $s=c+it$,
\begin{align*}
\left|\frac{x^s}{s}\right|=\frac{x^c}{|c+it|}.
\end{align*}
Therefore
\begin{align*}
\sum_{n=1}^{\infty}\Lambda(n)n^{-c}
\int_{-T}^{\,T}\frac{x^c}{|c+it|}\,d\mathcal{L}^1(t)<\infty,
\end{align*}
because the integral over the finite interval $[-T,T]$ is finite and the coefficient series converges. This is exactly the absolute integrability condition needed for termwise integration: Tonelli's theorem applies to the non-negative majorant, and Fubini's theorem then justifies interchanging the series with the finite vertical integral. Hence
\begin{align*}
\frac{1}{2\pi i}\int_{c-iT}^{c+iT}-\frac{\zeta'}{\zeta}(s)\frac{x^s}{s}\,ds
&=\sum_{n=1}^{\infty}\Lambda(n)\frac{1}{2\pi i}\int_{c-iT}^{c+iT}\frac{(x/n)^s}{s}\,ds \\
&=\sum_{n=1}^{\infty}\Lambda(n)K_T(x/n).
\end{align*}
The last equality is precisely the definition of $K_T$ with $y=x/n$.
[/guided]
[/step]
[step:Apply the kernel estimate term by term]
Because $x$ is not an integer, $x/n\neq 1$ for every $n\in\mathbb{N}$. Applying the Perron kernel estimate with $y=x/n$ gives
\begin{align*}
K_T(x/n)
=\mathbb{1}_{(1,\infty)}(x/n)
+O\left(\left(\frac{x}{n}\right)^c
\min\left\{1,\frac{1}{T|\log(x/n)|}\right\}\right).
\end{align*}
Multiplying by $\Lambda(n)$ and summing over $n\geq 1$, we obtain
\begin{align*}
\sum_{n=1}^{\infty}\Lambda(n)K_T(x/n)
&=\sum_{n=1}^{\infty}\Lambda(n)\mathbb{1}_{(1,\infty)}(x/n) \\
&\quad +O\left(\sum_{n=1}^{\infty}\Lambda(n)\left(\frac{x}{n}\right)^c
\min\left\{1,\frac{1}{T|\log(x/n)|}\right\}\right).
\end{align*}
Since $x$ is not an integer,
\begin{align*}
\mathbb{1}_{(1,\infty)}(x/n)=\mathbb{1}_{\{n\leq x\}}(n),
\end{align*}
and therefore
\begin{align*}
\sum_{n=1}^{\infty}\Lambda(n)\mathbb{1}_{(1,\infty)}(x/n)
=\sum_{1\leq n\leq x}\Lambda(n)
=\psi(x).
\end{align*}
Combining this identity with the previous step gives the truncated formula.
[guided]
The reason for excluding integral $x$ is now visible. The Perron kernel has a jump at $y=1$, and the term $y=x/n$ equals $1$ exactly when $n=x$. Since $x$ is not an integer, no summand lies at the jump.
For each $n\in\mathbb{N}$, the kernel estimate gives
\begin{align*}
K_T(x/n)
=\mathbb{1}_{(1,\infty)}(x/n)
+O\left(\left(\frac{x}{n}\right)^c
\min\left\{1,\frac{1}{T|\log(x/n)|}\right\}\right).
\end{align*}
After multiplying by $\Lambda(n)$ and summing, the main term is
\begin{align*}
\sum_{n=1}^{\infty}\Lambda(n)\mathbb{1}_{(1,\infty)}(x/n).
\end{align*}
The inequality $x/n>1$ is equivalent to $n<x$. Because $x$ is not an integer, this is the same as $n\leq x$ for integer $n$. Thus
\begin{align*}
\sum_{n=1}^{\infty}\Lambda(n)\mathbb{1}_{(1,\infty)}(x/n)
=\sum_{1\leq n\leq x}\Lambda(n)
=\psi(x).
\end{align*}
The remaining terms are exactly controlled by
\begin{align*}
\sum_{n=1}^{\infty}\Lambda(n)\left(\frac{x}{n}\right)^c
\min\left\{1,\frac{1}{T|\log(x/n)|}\right\}.
\end{align*}
This proves the stated [truncated Perron formula](/theorems/4357).
[/guided]
[/step]
[step:Let the truncation height tend to infinity]
For every fixed $n\in\mathbb{N}$, since $x/n\neq 1$,
\begin{align*}
\min\left\{1,\frac{1}{T|\log(x/n)|}\right\}\to 0
\end{align*}
as $T\to\infty$. Also,
\begin{align*}
0\leq \Lambda(n)\left(\frac{x}{n}\right)^c
\min\left\{1,\frac{1}{T|\log(x/n)|}\right\}
\leq \Lambda(n)\left(\frac{x}{n}\right)^c,
\end{align*}
and
\begin{align*}
\sum_{n=1}^{\infty}\Lambda(n)\left(\frac{x}{n}\right)^c
=x^c\sum_{n=1}^{\infty}\Lambda(n)n^{-c}<\infty.
\end{align*}
By dominated convergence for series, the error term tends to $0$ as $T\to\infty$. Hence
\begin{align*}
\psi(x)=\frac{1}{2\pi i}\lim_{T\to\infty}\int_{c-iT}^{c+iT}-\frac{\zeta'}{\zeta}(s)\frac{x^s}{s}\,ds.
\end{align*}
This is the asserted Perron representation of $\psi$.
[guided]
The truncated formula has already identified the finite-height integral as $\psi(x)$ plus an explicit error. To pass to the infinite vertical integral, we must show that this error tends to $0$ as $T\to\infty$.
Fix $n\in\mathbb{N}$. Since $x$ is not an integer, $x/n\neq 1$, so $|\log(x/n)|>0$. Therefore
\begin{align*}
\min\left\{1,\frac{1}{T|\log(x/n)|}\right\}\to 0
\end{align*}
as $T\to\infty$. The summand in the error term is non-negative and satisfies the pointwise domination
\begin{align*}
0\leq \Lambda(n)\left(\frac{x}{n}\right)^c
\min\left\{1,\frac{1}{T|\log(x/n)|}\right\}
\leq \Lambda(n)\left(\frac{x}{n}\right)^c.
\end{align*}
The dominating series is summable because $c>1$ and the Dirichlet series for $-\zeta'/\zeta$ is absolutely convergent on $\operatorname{Re}(s)=c$:
\begin{align*}
\sum_{n=1}^{\infty}\Lambda(n)\left(\frac{x}{n}\right)^c
=x^c\sum_{n=1}^{\infty}\Lambda(n)n^{-c}<\infty.
\end{align*}
Thus the [dominated convergence theorem](/theorems/4) for series applies to the error series and gives
\begin{align*}
\sum_{n=1}^{\infty}\Lambda(n)\left(\frac{x}{n}\right)^c
\min\left\{1,\frac{1}{T|\log(x/n)|}\right\}\to 0.
\end{align*}
Taking the limit $T\to\infty$ in the truncated formula yields
\begin{align*}
\psi(x)=\frac{1}{2\pi i}\lim_{T\to\infty}\int_{c-iT}^{c+iT}-\frac{\zeta'}{\zeta}(s)\frac{x^s}{s}\,ds.
\end{align*}
This is exactly the Perron representation of $\psi$ stated in the theorem.
[/guided]
[/step]
