摘要:$$\Large\displaystyle \int_{0}^{1}\frac{\sqrt[4]{x\left ( 1 x \right )^{3}}}{\left ( 1+x \right )^{3}}\mathrm{d}x~~,~~\int_{0}^{1}\frac{\sqrt[3]{x\lef 阅读全文
posted @ 2016-05-11 19:33 Renascence_5 阅读(832) 评论(0) 推荐(0) 编辑
摘要:如何计算 $\displaystyle \zeta \left ( 2 \right )=\frac{1}{1^{2}}+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\cdots =~?$ 这个问题是在1644年由意大利数学家蒙哥利(Pietro Mengoli)提出的,而大数学 阅读全文
posted @ 2016-05-11 16:43 Renascence_5 阅读(626) 评论(0) 推荐(0) 编辑
摘要:1.Irresistible Integrals http://pan.baidu.com/s/1c2z7QWo 2.Inside Interesting Integrals http://pan.baidu.com/s/1jI6Nkf8 window._bd_share_config={"comm 阅读全文
posted @ 2016-05-11 16:19 Renascence_5 阅读(374) 评论(0) 推荐(0) 编辑
摘要:$$\Large\displaystyle \int_{0}^{1}x\sqrt{1+x^{3}}\mathrm{d}x$$ $\Large\mathbf{Solution:}$ 易知 $$\int_{0}^{1}x\sqrt{1+x^{3}}\mathrm{d}x=\frac{1}{3}\int_ 阅读全文
posted @ 2016-05-10 20:46 Renascence_5 阅读(1244) 评论(0) 推荐(0) 编辑
摘要:$$\Large\displaystyle \int_{0}^{[x]}\left ( t \left [ t \right ] \right )\mathrm{d}t=\frac{[x]}{2}$$ $\Large\mathbf{Proof:}$ 我们来看更一般的形式,令$m=\left \lfl 阅读全文
posted @ 2016-05-10 20:41 Renascence_5 阅读(393) 评论(0) 推荐(0) 编辑
摘要:$$\Large\displaystyle \int_{0}^{1}\left [ \frac{1+\sqrt{1 x}}{x} +\frac{2}{\ln\left ( 1 x \right )}\right ]\mathrm{d}x$$ $\Large\mathbf{Solution:}$ 方法 阅读全文
posted @ 2016-05-10 20:30 Renascence_5 阅读(297) 评论(0) 推荐(0) 编辑
摘要:Since the Clausen functions are intimately related to a number of other important special functions, such as Inverse Tangent Integrals, Polylogarithms 阅读全文
posted @ 2016-05-10 19:36 Renascence_5 阅读(358) 评论(0) 推荐(0) 编辑
摘要:$$\Large\displaystyle \int_0^{1} \frac{\arccos^4 \left(x^2\right)}{\sqrt{1 x^2}}\,\mathrm{d}x$$ $\Large\mathbf{Solution:}$ Let $I$ denote the integral 阅读全文
posted @ 2016-05-09 20:29 Renascence_5 阅读(440) 评论(0) 推荐(0) 编辑
摘要:$$\Large\displaystyle \int_0^1 \dfrac{\operatorname{Li}_2\left(\dfrac{x}{4}\right)}{4 x}\ln\left(\dfrac{1+\sqrt{1 x}}{1 \sqrt{1 x}}\right)\mathrm{d}x= 阅读全文
posted @ 2016-05-09 20:10 Renascence_5 阅读(272) 评论(0) 推荐(0) 编辑
摘要:$$\Large\displaystyle \int_{0}^{1}\left \{ \frac{1}{x} \right \}\mathrm{d}x~,~\int_{0}^{1}\left \{ \frac{1}{x} \right \}^{2}\mathrm{d}x~,~\int_{0}^{1} 阅读全文
posted @ 2016-05-09 20:04 Renascence_5 阅读(252) 评论(0) 推荐(0) 编辑
摘要:$$\Large\displaystyle \int_0^1\frac{\ln^3(1+x)\,\ln^3x}x\mathrm{d}x$$ $\Large\mathbf{Solution:}$ Using the following known results: $$\begin{align } \ 阅读全文
posted @ 2016-05-09 17:16 Renascence_5 阅读(263) 评论(0) 推荐(0) 编辑
摘要:$$\Large\displaystyle \int_0^1\frac{\ln^3(1+x)\,\ln^2x}x\mathrm{d}x$$ $\Large\mathbf{Solution:}$ I will be using the following results: $$2\sum^\infty 阅读全文
posted @ 2016-05-05 17:11 Renascence_5 阅读(262) 评论(0) 推荐(0) 编辑
摘要:$$\Large\displaystyle {\;}_3F_2\left(\frac{1}{2},\frac{1}{2},\frac{1}{2};1,\frac{3}{2};1 \right)=\frac{4\mathbf{G}}{\pi}$$ $\Large\mathbf{Proof:}$ Wel 阅读全文
posted @ 2016-05-05 11:09 Renascence_5 阅读(353) 评论(0) 推荐(0) 编辑
摘要:$$\Large\displaystyle \int_0^\infty \frac{\ln \left(1+\dfrac{\pi^2}{4x} \right)}{e^{\sqrt{x}} 1}\mathrm{d}x$$ $\Large\mathbf{Solution:}$ Step 1 Split 阅读全文
posted @ 2016-05-04 20:08 Renascence_5 阅读(563) 评论(0) 推荐(0) 编辑
摘要:$$\Large\displaystyle \sum_{n=1}^\infty\frac{\Gamma\left(n+\dfrac{1}{2}\right)}{(2n+1)^4\,4^n\,n!}$$ $\Large\mathbf{Solution:}$ First, in view of Legr 阅读全文
posted @ 2016-05-04 16:30 Renascence_5 阅读(299) 评论(0) 推荐(0) 编辑