= Solution
A precise endpoint version of the <Perron formula> is as follows. If $F(s)=\sum_{n\geq1}a_nn^{-s}$ converges absolutely on $\Re s=c>0$, then, for $x>0$,
$$
\boxed{\sum_{n<x}a_n+\frac12\mathbf1_{x\in\mathbb N}a_x
=\lim_{T\to\infty}\frac1{2\pi i}
\int_{c-iT}^{c+iT}F(s)\frac{x^s}{s}\,ds.}
$$
When $x$ is not an <integer>, the left side is simply the sum over $n\leq x$.
Prove first the scalar kernel formula, with symmetric vertical limits:
$$
K(y)=\frac1{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{y^s}{s}\,ds
=\begin{cases}1,&y>1,\\1/2,&y=1,\\0,&0<y<1.\end{cases}
$$
For $y>1$, close the contour to the left and apply the <residue theorem> at $s=0$; the exponential $e^{s\log y}$ makes the closing integral vanish. For $y<1$, close to the right, where there is no enclosed <pole> and the exponential decreases. The usual semicircle estimate, or integration by parts on the closing arcs, justifies these limits. At $y=1$ the truncated integral is $\pi^{-1}\arctan(T/c)$, tending to $1/2$.
One also needs a uniform bound to pass through the <Dirichlet series>. Pairing the two halves of the vertical integral gives, with $u=\log y$,
$$
K_T(y)=\frac{y^c}{\pi}\int_0^T
\frac{c\cos(tu)+t\sin(tu)}{c^2+t^2}\,dt.
$$
The cosine term is absolutely bounded. The sine term is uniformly bounded by integration by parts against the bounded sine integral $\int_0^v\sin(tu)\,dt/t$, since $t^2/(c^2+t^2)$ is increasing from zero to one. Thus $|K_T(y)|\leq C_cy^c$, independently of $T,y$.
For finite $T$, <absolute convergence> permits termwise integration, giving $\sum_na_nK_T(x/n)$. The uniform bound dominates this by $C_cx^c\sum_n|a_n|n^{-c}<\infty$. The <dominated convergence theorem> and the scalar kernel formula now prove the boxed <Perron formula>, including its half-weight endpoint convention.
For the link with <primes>, the <logarithmic derivative> of the <Euler product> gives
$$
-\frac{\zeta'(s)}{\zeta(s)}=\sum_{n\geq1}\frac{\Lambda(n)}{n^s}\qquad(\Re s>1).
$$
Applying <Perron formula> therefore expresses the second <Chebyshev function> $\psi_0(x)$, with half weight at a prime-power endpoint, as the vertical integral of $-\zeta'/\zeta(s)\,x^s/s$. Shift this contour left, choosing heights away from zeros and using the partial-fraction expansion to control the integrand. The <pole> at one contributes $x$, a nontrivial zero $\rho$ of multiplicity $m$ contributes $-m x^\rho/\rho$, the trivial zeros contribute $\sum_{k\geq1}x^{-2k}/(2k)$, and the <pole> from $1/s$ contributes $-\zeta'(0)/\zeta(0)=-\log(2\pi)$. Thus the resulting <Riemann–von Mangoldt explicit formula> is
$$
\psi_0(x)=x-\sum_\rho\frac{x^\rho}{\rho}
-\log(2\pi)-\frac12\log(1-x^{-2})\qquad(x>1),
$$
with zeros summed symmetrically in height. A truncated contour gives the corresponding truncated formula and remainder. \b[The <pole> at one gives the main term for <primes>; zeros farther left give smaller oscillating corrections.] Zero-free regions therefore translate into error bounds for $\psi(x)$ and, by <partial summation>, for the <prime-counting function>.
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