As an integral
出典: フリー百科事典『ウィキペディア（Wikipedia）』 (2018/10/27 20:57 UTC 版)
「Mertens 関数」の記事における「As an integral」の解説
オイラー積を使うと、 1 ζ ( s ) = ∏ p ( 1 − p − s ) = ∑ n = 1 ∞ μ ( n ) n s {\displaystyle {\frac {1}{\zeta (s)}}=\prod _{p}(1-p^{-s})=\sum _{n=1}^{\infty }{\frac {\mu (n)}{n^{s}}}} ζ ( s ) {\displaystyle \zeta (s)} はリーマンゼータ関数 and the product is taken over primes. 次に,ディリクレ級数とヘロンの公式を使い,次の式が得られる。 1 2 π i ∫ c − i ∞ c + i ∞ x s s ζ ( s ) d s = M ( x ) {\displaystyle {\frac {1}{2\pi i}}\int _{c-i\infty }^{c+i\infty }{\frac {x^{s}}{s\zeta (s)}}\,ds=M(x)} (c > 1.) Conversely, one has the メリン変換 1 ζ ( s ) = s ∫ 1 ∞ M ( x ) x s + 1 d x {\displaystyle {\frac {1}{\zeta (s)}}=s\int _{1}^{\infty }{\frac {M(x)}{x^{s+1}}}\,dx} ( R e ( s ) > 1 {\displaystyle \mathrm {Re} (s)>1} .) 第二チェビシェフ関数を使い以下のようにあらわせる ψ ( x ) = M ( x 2 ) log ( 2 ) + M ( x 3 ) log ( 3 ) + M ( x 4 ) log ( 4 ) + ⋯ . {\displaystyle \psi (x)=M\left({\frac {x}{2}}\right)\log(2)+M\left({\frac {x}{3}}\right)\log(3)+M\left({\frac {x}{4}}\right)\log(4)+\cdots .} Assuming that there are not multiple non-trivial roots of ζ ( ρ ) {\displaystyle \zeta (\rho )} we have the "exact formula" by the residue theorem: M ( x ) = ∑ ρ x ρ ρ ζ ′ ( ρ ) − 2 + ∑ n = 1 ∞ ( − 1 ) n − 1 ( 2 π ) 2 n ( 2 n ) ! n ζ ( 2 n + 1 ) x 2 n . {\displaystyle M(x)=\sum _{\rho }{\frac {x^{\rho }}{\rho \zeta '(\rho )}}-2+\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}(2\pi )^{2n}}{(2n)!n\zeta (2n+1)x^{2n}}}.} Weyl conjectured that the Mertens function satisfied the approximate functional-differential equation y ( x ) 2 − ∑ r = 1 N B 2 r ( 2 r ) ! D t 2 r − 1 y ( x t + 1 ) + x ∫ 0 x y ( u ) u 2 d u = x − 1 H ( log x ) {\displaystyle {\frac {y(x)}{2}}-\sum _{r=1}^{N}{\frac {B_{2r}}{(2r)!}}D_{t}^{2r-1}y\left({\frac {x}{t+1}}\right)+x\int _{0}^{x}{\frac {y(u)}{u^{2}}}\,du=x^{-1}H(\log x)} where H(x) is the Heaviside step function, B are Bernoulli numbers and all derivatives with respect to t are evaluated at t = 0. There is also a trace formula involving a sum over the Möbius function and zeros of Riemann Zeta in the form ∑ n = 1 ∞ μ ( n ) n g ( log n ) = ∑ γ h ( γ ) ζ ′ ( 1 / 2 + i γ ) + 2 ∑ n = 1 ∞ ( − 1 ) n ( 2 π ) 2 n ( 2 n ) ! ζ ( 2 n + 1 ) ∫ − ∞ ∞ g ( x ) e − x ( 2 n + 1 / 2 ) d x , {\displaystyle \sum _{n=1}^{\infty }{\frac {\mu (n)}{\sqrt {n}}}g(\log n)=\sum _{\gamma }{\frac {h(\gamma )}{\zeta '(1/2+i\gamma )}}+2\sum _{n=1}^{\infty }{\frac {(-1)^{n}(2\pi )^{2n}}{(2n)!\zeta (2n+1)}}\int _{-\infty }^{\infty }g(x)e^{-x(2n+1/2)}\,dx,} where the first sum on the right-hand side is over the nontrivial zeros of the Riemann zeta function, and (g,h) are related by a Fourier transform, such that 2 π g ( x ) = ∫ − ∞ ∞ h ( u ) e i u x d u . {\displaystyle 2\pi g(x)=\int _{-\infty }^{\infty }h(u)e^{iux}\,du.}
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