摘要: 1.Table of Integrals,Series and Products , Eighth Edition , I.S.Gradshteyn, I.M.Ryzhik 这就是众所周知的"积分大典",也是最新版,包含了巨量的积分公式,绝对是值得拥有的,遗憾的是只有公式,没有证明,当然,在书的开头阅读全文
posted @ 2016-04-26 21:00 Renascence_5 阅读(341) 评论(0) 编辑
摘要: $$\Large\displaystyle \int_{0}^{1}\frac{\ln\left ( x+\sqrt{x^{2}+1} \right )}{x}\mathrm{d}x$$ $\Large\mathbf{Solution:}$ 注意到: $$\int_{0}^{1}\frac{\ln\阅读全文
posted @ 2016-06-06 14:16 Renascence_5 阅读(205) 评论(0) 编辑
摘要: 原文地址:http://spaces.ac.cn/archives/3154/ 原文作者:苏剑林 标准思路 简单来说,$n$维球体积就是如下$n$重积分 $$V_n(r)=\int_{x_1^2+x_2^2+\dots+x_n^2\leq r^2}\mathrm{d}x_1 \mathrm{d}x_阅读全文
posted @ 2016-05-27 09:09 Renascence_5 阅读(882) 评论(0) 编辑
摘要: $$\Large\displaystyle \int^{\infty}_{0}\frac{\tanh\left(\, x\,\right)} {x\left[\, 1 2\cosh\left(\, 2x\,\right)\,\right]^{2}}\,{\rm d}x$$ $\Large\mathb阅读全文
posted @ 2016-05-15 21:06 Renascence_5 阅读(620) 评论(0) 编辑
摘要: $$\Large\displaystyle \int_0^\infty{_3F_2}\left(\begin{array}c\dfrac58,\dfrac58,\dfrac98\\\dfrac12,\dfrac{{13}}8\end{array}\middle|\ { x}\right)^2\fra阅读全文
posted @ 2016-05-15 20:51 Renascence_5 阅读(204) 评论(0) 编辑
摘要: $$\Large\displaystyle \int_0^\infty\frac{\ln\left(1+x+\sqrt{x^2+2\,x}\right)\,\ln\left(1+\sqrt{x^2+2\,x+2}\right)}{x^2+2x+1}\mathrm dx$$ $\Large\mathb阅读全文
posted @ 2016-05-14 21:52 Renascence_5 阅读(126) 评论(0) 编辑
摘要: $$\Large\displaystyle\int_0^\infty\frac{\ln\left(\displaystyle\frac{1+x^{4+\sqrt{15\vphantom{\large A}}}}{1+x^{2+\sqrt{3\vphantom{\large A}}}}\right)}阅读全文
posted @ 2016-05-14 21:43 Renascence_5 阅读(220) 评论(0) 编辑
摘要: $$\Large\displaystyle \int_0^1\frac{\ln x\ln^2(1+x)}{x}\ \mathrm dx$$ $\Large\mathbf{Solution:}$ $$\begin{align } \int_0^1\frac{\ln x\ln^2(1+x)}{x}\ \阅读全文
posted @ 2016-05-14 20:44 Renascence_5 阅读(528) 评论(0) 编辑
摘要: $$\Large\displaystyle \int_0^{\pi/2}\ln^2(\sin x)\ln(\cos x)\tan x \,{\rm d}x$$ $\Large\mathbf{Solution:}$ Let $J$ donates the integral and it is easy阅读全文
posted @ 2016-05-13 10:58 Renascence_5 阅读(145) 评论(0) 编辑
摘要: $$\Large\displaystyle \int_{0}^{\frac{\pi }{2}}x^{2}\ln\left ( \sin x \right )\ln\left ( \cos x \right )\mathrm{d}x$$ $\Large\mathbf{Solution:}$ Tools阅读全文
posted @ 2016-05-13 10:48 Renascence_5 阅读(149) 评论(0) 编辑
摘要: 链接:http://pan.baidu.com/s/1eSNkz4Y window._bd_share_config={"common":{"bdSnsKey":{},"bdText":"","bdMini":"2","bdMiniList":false,"bdPic":"","bdStyle":"阅读全文
posted @ 2016-05-13 10:21 Renascence_5 阅读(116) 评论(0) 编辑
摘要: $$\Large\displaystyle \sum_{n=1}^{\infty}\frac{H_{2n}}{n(6n+1)}$$ $\Large\mathbf{Solution:}$ Let $S$ denote the sum. Then $$\begin{align } S=\sum_{n=1阅读全文
posted @ 2016-05-12 16:24 Renascence_5 阅读(121) 评论(0) 编辑
摘要: $$\Large\displaystyle \sum_{n=1}^{\infty} \frac{\widetilde{H_n}}{n^{3}}$$ where $\widetilde{H_n}$ is the alternating harmonic number. $\Large\mathbf{S阅读全文
posted @ 2016-05-12 16:09 Renascence_5 阅读(114) 评论(0) 编辑
摘要: $$\sum_{n = 1}^\infty {\frac{1}{{{n^3}}}} \left( {\sum\limits_{k = 1}^n {\frac{{{{\left( { 1} \right)}^{k 1}}}}{k}} } \right) = \frac{7}{4}\zeta \left阅读全文
posted @ 2016-05-12 15:40 Renascence_5 阅读(106) 评论(0) 编辑
摘要: 计算下面两个积分的比值: $$\Large\displaystyle \int_{0}^{1}\frac{1}{\sqrt{1+t^{4}}}\mathrm{d}t~,~\int_{0}^{1}\frac{1}{\sqrt{1 t^{4}}}\mathrm{d}t$$ $\Large\mathbf{阅读全文
posted @ 2016-05-11 20:00 Renascence_5 阅读(168) 评论(0) 编辑
摘要: $$\Large\displaystyle \int_{0}^{1}\frac{\ln\left ( 1+x^{2} \right )}{1+x^{2}}\mathrm{d}x$$ $\Large\mathbf{Solution:}$ 方法一: 考虑含参积分 $$\mathcal{I}\left (阅读全文
posted @ 2016-05-11 19:52 Renascence_5 阅读(138) 评论(0) 编辑