Perron's formula

Perron's formula recovers a summatory arithmetic function from its Dirichlet series by the inverse Mellin integralwith the usual convergence and endpoint conventions.

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.

A truncated Perron integral over $c-iT$ to $c+iT$ recovers a summatory function with an error controlled by

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 $T\ge 2$.