实在没办法呀
15 个赞
·······
13 个赞
好的
13 个赞
不会错过了
13 个赞
额，6
12 个赞
\log\prod(N)=(N+2)\log N-N+A-\int_{N}^{\infty}\frac{\overline{B}_{1}(x)dx}{x},A=1+\int_{1}^{\infty}\frac{\overline{B}_{1}(x)dx}{x}\
\log\prod(s)=(s+\frac{1}{2})\log s-s+A-\int_{\infty}^{0}\frac{\overline{B}_{1}(t)dt}{t+s}\
\log\prod(s)=\lim_{n \to \infty}[s\log (N+1)\sum_{N}^{n=1}\log n-\sum_{N}^{n=1}\log (s+n)]
=\lim_{n \to \infty}[s\log (N+1)+\int_{1}^{N}\log xdx-\frac{1}{2}\log N+\int_{1}^{N}\frac{\overline{B}_{1}dx}{x}
-\int_{1}^{N}\log(s+x)dx-\frac{1}{2}[\log (s+1)+\log (s+N)]
-\int_{1}^{N}\frac{\overline{B}_{1}(x)dx}{s+x}]
=\lim_{n \to \infty}[s\log (N+1)+N \log N-N+1+\frac{1}{2}\log N+\int_{1}^{N}\frac{\overline{B}_{1}(x)dx}{x}
-(s+N)\log (s+N)+(s+N)+(s+1)\log(s+1)
-(s+1)-\frac{1}{2}\log (s+1)-\log (s+N)-\int_{1}^{N}\frac{\overline{B}_{1}(x)dx}{s+x}]
=(s+\frac{1}{2})\log (s+1)+\int_{1}^{\infty}\frac{\overline{B}_{1}(x)dx}{x}-\int_{1}^{N}\frac{\overline{B}_{1}(x)dx}{s+x}
+\lim_{n \to \infty}[s\log (N+1)+(N\frac{1}{2})\log N
-(s+N+\frac{1}{2})\log (s+N)]
=(s+\frac{1}{2})\log (s+1)+(A-1)-\int_{1}^{\infty}\frac{\overline{B}_{1}(x)dx}{s+x}
+\lim[s\log\frac{N+1}{2}-(N+\frac{1}{2})\log(1+\frac{s}{2})]
终于写完力（喜
14 个赞
哈哈哈，可以啊！
12 个赞
你这加入日期属于论坛的元老级人物了
12 个赞
13 个赞
膜拜大佬
12 个赞
洛谷：https://www.luogu.com.cn/
信友队论坛：https://discourse.xinyoudui.com/
ASCLL码：https://www.asciim.cn/
bilibili(b站)：https://www.bilibili.com/
15 个赞
…
8 个赞
可以再加一个网站
8 个赞
8 个赞
8 个赞
我也是
7 个赞