摘要: $$\Large\int_{0}^{z}x^{t}\ln\Gamma \left ( 1+x \right )\mathrm{d}x~,~z 0\, ,\, t\in N^{ }$$ $\Large\mathbf{Solution:}$ Notice that $$\begin{align } \i 阅读全文
posted @ 2016-04-27 19:46 Renascence_5 阅读(1040) 评论(0) 推荐(1)
摘要: $$\Large\sum_{k=1}^{\infty}\frac{(2^{2k 1} 2)(4^{2k+1} 3^{2k+1})}{144^k\,k\,(2k+1)}\zeta(2k)$$ $\Large\mathbf{Solution:}$ Within the interval $\displa 阅读全文
posted @ 2016-04-27 19:12 Renascence_5 阅读(529) 评论(0) 推荐(0)
摘要: $$\Large\sum_{n=1}^{\infty}\frac{\left(H_{n}^{(2)}\right)^{2}}{n^{2}}=\frac{19}{24}\zeta(6)+\zeta^{2}(3)$$ $\Large\mathbf{Proof:}$ We use the Abel's r 阅读全文
posted @ 2016-04-27 18:58 Renascence_5 阅读(472) 评论(0) 推荐(0)
摘要: $$\Large\sum_{n=0}^\infty \frac{H_{2n+1}}{(2n+1)^2}=\frac{21}{16}\zeta(3)$$ $\Large\mathbf{Proof:}$ Let $\displaystyle S_1=\sum_{n=1}^\infty \frac{H_n 阅读全文
posted @ 2016-04-27 17:25 Renascence_5 阅读(516) 评论(0) 推荐(0)
摘要: $$\Large\int_{0}^{1}\frac{\arctan x}{\sqrt{1 x^{2}}}\mathrm{d}x$$ $\Large\mathbf{Solution:}$ 首先第一种做法,含参积分.不多说直接上图 第二种方法则是利用级数,易知 $$\begin{align } \int 阅读全文
posted @ 2016-04-27 12:43 Renascence_5 阅读(6295) 评论(0) 推荐(0)