Perron formula and contour shifting

For a positive real parameter $y$ and fixed $c>0$, integrate $y^s/s$ along the vertical segment from $c-iT$ to $c+iT$, oriented upward. Here $y^s$ uses the real logarithm. As the height tends to infinity symmetrically, the normalized integral tends to one when $y>1$, zero when $y<1$, and one half when $y=1$. Closing a finite rectangle to the left or right explains the first two values; the symmetric endpoint value can also be calculated directly.

At $y=1$, direct integration gives $K_T(1)=\pi^{-1}\arctan(T/c)$. This is not exactly one half at a finite height. For a Dirichlet series absolutely convergent to the right of its abscissa, choose $c$ greater than both that abscissa and zero. At finite height, absolute convergence justifies exchanging the coefficient sum and integral. The limiting kernel then selects coefficients with indices below the endpoint and half of the endpoint coefficient when the endpoint is an integer.

A truncated kernel is not an exact indicator. Away from $y=1$, its error is bounded, up to an absolute constant, by $y^c\min(1,(T|\log y|)^{-1})$; at one its error is at most a constant times $c/T$. Summing these coefficientwise yields the displayed bound for a fixed positive line parameter. The sum is finite as an absolute quantity because its minimum factor is at most one and the Dirichlet series converges absolutely on the chosen line.

The troublesome indices are those close to the endpoint, where the logarithmic denominator is small. A noninteger endpoint bounded away from integers is easier to estimate than an endpoint approaching a large coefficient. At an integer, the half-weight convention isolates that coefficient and its separate error explicitly. To prove convergence at a fixed endpoint, dominated convergence applies to the coefficient error sum with its absolutely summable majorant. To prove a uniform estimate as the endpoint grows requires additional arithmetic bounds; fixed-endpoint convergence alone is insufficient.

Suppose the integrand continues meromorphically across a finite rectangle and has no poles on its boundary. Apply the residue theorem with the right side oriented upward. The right vertical integral equals the left vertical integral, also written upward, plus the enclosed residues and the appropriately oriented horizontal contributions. A pole of $F$ can produce a main term, and the extra factor $1/s$ can create a pole at zero even if the original function is regular there. Enumerate both kinds before estimating the remaining sides.

For $F=\zeta$, crossing one contributes $x$, and crossing zero contributes $\zeta(0)=-1/2$. For $F=-\zeta'/\zeta$, crossing zeta zeros instead contributes the negative zero terms of the explicit formula. Heights must avoid poles, and bounds for the logarithmic derivative near zeros require care. A complete argument keeps finite rectangles until it has controlled horizontal pieces, the new vertical line, and the original truncation error. Meromorphic continuation alone supplies no growth estimate strong enough to dismiss all of them.