积分习题XVI:超重积分(2)

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我觉得我上次讲的不太友好,题目太难了.

我们应该从简单点的开始讲,比如:

$$\int_{{{\left[ {a,b} \right]}^n}} {{\text{d}}{x_1}{\text{d}}{x_2} \ldots {\text{d}}{x_n}}  = {\left( {b - a} \right)^n}$$

嗯,就是求超立方体体积...

然后我们来求超球体的体积,当然是用球坐标咯.

$$\begin{aligned}
dV&=r^{n-1}\sin ^{n-2}(\varphi _{1})\sin ^{n-3}(\varphi _{2})\cdots \sin(\varphi _{n-2})\,dr\,d\varphi _{1}\,d\varphi _{2}\cdots d\varphi _{n-1}\\
V_{n}(R)&=\int _{0}^{R}\int _{0}^{\pi }\cdots \int _{0}^{2\pi }r^{n-1}\sin ^{n-2}(\varphi _{1})\cdots \sin(\varphi _{n-2})\,d\varphi _{n-1}\cdots d\varphi _{1}\,dr\\
&=\left(\int _{0}^{R}r^{n-1}\,dr\right)\left(\int _{0}^{\pi }\sin ^{n-2}(\varphi _{1})\,d\varphi _{1}\right)\cdots \left(\int _{0}^{2\pi }d\varphi _{n-1}\right)\\
&={\frac {R^{n}}{n}}\left(2\int _{0}^{\frac {\pi }{2}}\sin ^{n-2}(\varphi _{1})\,d\varphi _{1}\right)\cdots \left(4\int _{0}^{\frac {\pi }{2}}d\varphi _{n-1}\right)\\
&={\frac {R^{n}}{n}}\textstyle \mathrm {B} \left({\frac {n-1}{2}},{\frac {1}{2}}\right)\mathrm {B} \left({\frac {n-2}{2}},{\frac {1}{2}}\right)\cdots \mathrm {B} \left(1,{\frac {1}{2}}\right)\cdot 2\mathrm {B} \left({\frac {1}{2}},{\frac {1}{2}}\right)\\
&={\frac {R^{n}}{n}}{\frac {\Gamma \left({\frac {n-1}{2}}\right)\Gamma \left({\frac {1}{2}}\right)}{\Gamma \left({\frac {n}{2}}\right)}}{\frac {\Gamma \left({\frac {n-2}{2}}\right)\Gamma \left({\frac {1}{2}}\right)}{\Gamma \left({\frac {n-1}{2}}\right)}}\cdots {\frac {\Gamma \left(1\right)\Gamma \left({\frac {1}{2}}\right)}{\Gamma \left({\frac {3}{2}}\right)}}\cdot 2{\frac {\Gamma \left({\frac {1}{2}}\right)\Gamma \left({\frac {1}{2}}\right)}{\Gamma \left(1\right)}}\\
&={\frac {2\pi ^{\frac {n}{2}}R^{n}}{n\Gamma \left({\frac {n}{2}}\right)}}={\frac {\pi ^{\frac {n}{2}}R^{n}}{\Gamma \left({\frac {n}{2}}+1\right)}}
\end{aligned}$$Read More

积分习题XV:超重积分

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超重积分就是那种看上去很重很重的积分......嗯?这个字读Chóng来着....那就是n重积分的意思了....

发明一个约定:

$$I(n) = \int_{{{[0,1]}^n}} {\mathop f\limits_{i \in 1\sim n} [{x_i}]{\text{d}}{x_i}}  = \int_0^1 {\int_0^1 . } ..\int_0^1 {f({x_1},{x_2},...,{x_n}){\text{d}}{x_1}{\text{d}}{x_2}...{\text{d}}{x_n}}$$

Well....你以为打一大堆积分号不累啊...

来试验一下,求证:$$\boxed{\bigstar I(n) = \int_{{{[0,1]}^n}} {\mathop {\min }\limits_{i \in 1\sim n} [{x_i}]{\text{d}}{x_i}}  = \frac{1}{{n + 1}}}$$

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级数习题V

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上次我展示了一个极为惊艳的积分:

\[\int_0^1 {{e^{i\pi x}}} \; {x^x}{(1 - x)^{1 - x}}\; dx = \frac{e}{2}\frac{\pi }{3}\frac{i}{4}\]

这个积分神奇的把$0,1,2,3,4,i,e,\pi,x^x$结合在了一起.

Well,可是有很多吃瓜群众仍不满足,执意要把欧拉常数$\gamma$塞进去...

这个积分很精密的,禁不住这么魔改.不过还真有能塞一起的式子,不但能把$e,\pi,\gamma$一网打尽还能塞个Glaisher常数$\rm A$进去.

\[ - \zeta'(2) = \sum\limits_{k = 1}^\infty  {\frac{{\ln k}}{{{k^2}}}}  = \ln \prod\limits_{k = 1}^\infty  {\sqrt[{{k^2}}]{k}}  = \zeta (2)\ln \left( {\frac{{{{\rm A}^{12}}}}{{2\pi {e^\gamma }}}} \right)\]

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积分习题XI:孪生积分

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所谓孪生积分就是一种积分技巧,就是一次性算两个积分.唔,这是很常用的积分技巧啦....教科书上基本都有.

一般用于三角函数积分.....就是用的多了会养成习惯....然后连简单的积分都用这个.....然后就没有然后了,因为简单的考试题用基本的积分技巧就能秒了,用这个不是找事情么....


我们来研究这样一个积分:

$$F(x) = \int {\frac{{c + d\;{\text{Tri}}(x)}}{{a + b\;{\text{Tri}}(x)}}{\text{d}}x}$$

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积分习题X:格拉瑟主定理

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格拉瑟主定理(Glasser's Master Theorem)的标准表述如下:

恒等式$PV\int_{ - \infty }^{ + \infty } {F(\phi (x)){\rm{dx}}}  = PV\int_{ - \infty }^{ + \infty } {F(x){\rm{dx}}} $对任意可积函数$F(x)$和$\phi (x) = |a|x - \sum\limits_{n = 1}^N {\frac{{|{\alpha _n}|}}{{x - {\beta _n}}}} $成立.

其中$a,\left\{ {{\alpha _n}} \right\}_{n = 1}^N,\left\{ {{\beta _n}} \right\}_{n = 1}^N$可以是任意常数,PV是柯西主值(Cauchy Principal Value)不可以两边约掉的啊魂淡可我怎么总是有种两边划掉的强烈冲动怎么办才好啊好烦啊要控制不住了的说....

So....这玩意儿有什么用呢?

叫你证仨积分,汝敢答应否?

$$\begin{aligned}
{I_1} &= \int_{ - \infty }^{ + \infty } {\frac{{{x^8} + 12{x^7} + 58{x^6} + 144{x^5} + 193{x^4} + 132{x^3} + 36{x^2}}}{{{x^{10}} + 12{x^9} + 51{x^8} + 72{x^7} - 81{x^6} - 300{x^5} - 43{x^4} + 576{x^3} + 664{x^2} + 264x + 36}}{\rm{d}}x} = \pi \\
{I_2} &= \int_{ - \infty }^{ + \infty } {\frac{{{x^8} + 4{x^7} + 6{x^6} + 4{x^5} + {x^4}}}{{{x^{12}} + 4{x^{11}} - 2{x^{10}} - 24{x^9} - 10{x^8} + 56{x^7} + 48{x^6} - 40{x^5} - 49{x^4} + 4{x^3} + 20{x^2} + 8x + 1}}{\rm{d}}x = \frac{\pi }{{\sqrt 2 }}} \\
{I_3} &=\int_0^\infty \frac{x^{14}-15x^{12}+82x^{10}-190x^8+184x^6-60x^4+16x^2}{x^{16}-20x^{14}+156x^{12}-616x^{10}+1388x^8-1792x^6+1152x^4-224x^2+16}\; dx = \frac{\pi}{2}
\end{aligned}$$

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积分习题IX

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又到了喜闻乐见的杂题时间了,不过今天到场的有个重量级人物:

\[\int_0^1 {{e^{i\pi x}}} {\mkern 1mu} {x^x}{(1 - x)^{1 - x}}{\mkern 1mu} dx = \frac{e}{2}\frac{\pi }{3}\frac{i}{4}\]

哈,牛逼的木有吧,神奇的把$0,1,2,3,4,i,e,\pi,x^x$结合在了一起.

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积分习题IV

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最近碰到一些积分,都是有关开n次根的,那就整理一下好了.

第一个Supercircles的面积积分涉及到的这个:$\int_0^1 {\sqrt[n]{{1 - {x^n}}}} dx$

复习一下欧拉Gamma函数和欧拉Beta函数.

$$\begin{aligned}
\Gamma \left( z \right) &= \int_0^\infty {{t^{z - 1}}{e^{ - t}}dt} \\
{\rm B}\left( {a,b} \right) &= \frac{{\Gamma \left( a \right)\Gamma \left( b \right)}}{{\Gamma \left( {a + b} \right)}}\\
&= \int_0^1 {{t^{a - 1}}{{\left( {1 - t} \right)}^{b - 1}}dt}
\end{aligned}$$

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