求极限

[TheMatrix]

lim _N->∞ (Σ₂^N 1/ln(n) - ∫₂^N 1/ln(x) dx)

再加一问：

lim _N->∞ (Σ ₂^N ln(n)^-s - ∫₂^N ln(x)^-s dx)
是不是s的解析函数？

[changbaihou]

Direct calculation gives

\sum_{2\leq n\leq N}1/(\log n)=\int_{2^{-}}^{N}1/(\log x)d[x]=\int_2^Ndx/(\log x)-int_{2^{-}}^N1/(\log x)d{x}=\int_2^Ndx/(\log x)-{x}/\logx|_{2^{-}}^N-int_2^Ndx/(x\log^2x)

where {x}=x-[x] is the fractional part of x. Thus, the limit wanted is equal to
1/\log2-int_2^{\infty}{x}/(x\log^2x)dx

The same argument shows that the second limit gives a function analytic on Re(s)>0.

[TheMatrix]

Direct calculation gives

\sum_{2\leq n\leq N}1/(\log n)=\int_{2^{-}}^{N}1/(\log x)d[x]=\int_2^Ndx/(\log x)-int_{2^{-}}^N1/(\log x)d{x}=\int_2^Ndx/(\log x)-{x}/\logx|_{2^{-}}^N-int_2^Ndx/(x\log^2x)

where {x}=x-[x] is the fractional part of x. Thus, the limit wanted is equal to
1/\log2-int_2^{\infty}{x}/(x\log^2x)dx

The same argument shows that the second limit gives a function analytic on Re(s)>0.

format 一下：

[changbaihou]

format 一下：

谢谢。刚发现一个typo，在"where {x}=..."之前最后一个积分中，integrand分子missed了一个{x}。

[(ッ)]

发散的

[TheMatrix]

发散的

题目中那个积分积的不是1/x，而是积1/ln(x)。