Perron's formula

Perron's formula recovers coefficient partial sums from a Dirichlet series by the Mellin inversion formula. Suppose $F(s)={\sum }_{n\ge 1}{a}_n{n}^{-s}$ is absolutely convergent at a fixed real $c>0$. For $x>0$,The prime on the sum means that the term with $n=x$ receives weight $1/2$ if $x$ is an integer; for noninteger $x$ it is the ordinary sum over $n<x$. The integral is taken as a symmetric vertical limit, not asserted absolutely convergent on the infinite line.

The scalar truncated Perron kernel estimate tends to $1$ when $x/n>1$, to $0$ when $x/n<1$, and to $1/2$ when $x/n=1$. Absolute convergence on the line allows the coefficient sum to be interchanged with the finite integral and justifies passing to this limit. In analytic number theory, one then shifts the contour through poles and past controlled singularities: their residues supply main terms and error terms in the summatory function. The truncated Perron formula makes the finite-height error explicit.

The Perron's formula kernel $y^{s+1}/(s(s+1))$ integrates to $(y-1{)}_{+}$, giving the smoothed sum as an integral of $x^{s+1}/(\zeta (s)s(s+1))$. For each fixed height $T$, a zero-free rectangle permits shifting the central contour to a real part below one; its vertical contribution is $o(x^2)$, and its horizontal contributions are $O_T(x^2/\log x)$ when the original real part is $1+1/\log x$. A bound $1/\zeta (\sigma +it)=O({\log }^7(2+|t|))$ for $\sigma \ge 1$ bounds the remaining tails by $C{x}^2{\int }_T^{\infty }{\log }^7(2+t)t^{-2}dt$. Letting $x$ grow first and $T$ afterward proves the displayed cancellation, without uniform estimates to the left of one.

If the Dirichlet series $F(s)={\sum }_n{a}_n{n}^{-s}$ is absolutely convergent at $\sigma >0$, thenThe scalar kernel is $(2\pi {)}^{-1}{\int }_{\mathbb{R}}{e}^{v(\sigma +it)}(\sigma +it{)}^{-2}dt=v_{+}$. This follows by Fourier inversion applied to $u{e}^{-\sigma u}{1}_{u\ge 0}$, whose Fourier transform is $(\sigma +it{)}^{-2}$. Absolute convergence of the series and of $\int |\sigma +it{|}^{-2}dt$ justifies interchange. The factor $ds=idt$ is essential.

Suppose $|a_n|\le (\log (2n){)}^k$ and $A(x)={\sum }_{n\le x}{a}_n\log (x/n)=O(x(\log x{)}^{\beta })$, with $0\le \beta \le k$. For $0<h\le 1$, direct subtraction givesThe last term has absolute value $O((x{h}^2+h)(\log x{)}^k)$. Choose $h=(\log x{)}^{(\beta -k)/2}$ to obtain the displayed bound, with an additional $O((\log x{)}^k)$ term harmless as $x\to \infty$ for fixed $k$. This elementary finite-difference argument converts a logarithmically smoothed Perron formula bound to cancellation in its original partial sum.

Suppose the Dirichlet series $F(s)={\sum }_{n\ge 1}{a}_n{n}^{-s}$ converges absolutely at a fixed real $c>0$. For $x>0$ and $T\ge 2$, putThenAs in Perron's formula, the prime denotes half weight at an integer endpoint. If $n=x$, interpret the minimum as $1$; the sharper endpoint kernel error is $O_c(1/T)$.

To prove the bound, interchange the absolutely convergent coefficient sum with the finite integral and apply the truncated Perron kernel estimate term by term. The logarithmic denominator explains why coefficients near $n=x$ require separate control: truncating the contour cannot instantly distinguish the two sides of the endpoint. This estimate is a starting point for explicit Prime number theorem error bounds.

For $c=1+1/\log x$, $1\le y\le x$ and $2\le T\le x$, apply the truncated Perron formula with Von Mangoldt function coefficients at both endpoints on the same vertical line. The difference kernel is $((x+y{)}^s-x^s)/s={\int }_x^{x+y}{u}^{s-1}du$, of modulus $O(y)$. The usual near-integer error bound contributes $O(x{\log }^{2}x/T)$. The logarithmic derivative on this line is finite because of the Euler product.

For $v\ne 1$, $K_T(v)$ differs from $1_{v>1}$ by $O(v^c\min (1,(T|\log v|{)}^{-1}))$. Close the contour to the left for $v>1$ and to the right for $v<1$; horizontal sides give the logarithmic denominator and the nearby transition is bounded. At $v=1$, direct integration gives $K_T(1)={\pi }^{-1}\arctan (T/c)=1/2+O(c/T)$. This explains the half-weight endpoint in the truncated Perron formula. The constants may be taken uniform for $1\le c\le 2$ and $T\ge 2$.