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计算：

\[ \lim_{n\rightarrow\infty}{\left(1-\frac{1}{1\cdot 2}\right)\left(1-\frac{1}{2\cdot 3}\right)\cdots\left(1-\frac{1}{n(n+1)}\right)} \]

Solution：

We write

$ P=\prod_{n=1}^\infty \left(1-\frac{1}{n(n+1)}\right)=\prod_{n=1}^\infty \left(\frac{n^2+n-1}{n^2+n}\right)=\prod_{n=1}^\infty\frac{(n-a_1)(n-a_2)}{(n-b_1)(n-b_2)}  $,

where  $a_1=\frac{-1+\sqrt{5}}{2},\;a_2=\frac{-1-\sqrt{5}}{2},\;b_1=0$, and $b_2=-1.$

Using the [i]Weierstrass product[/i] for the gamma function, $\frac{1}{\Gamma(x)}=xe^\gamma x\prod_{n=1}^\infty \left(1+\frac{x}{n} \right)e^{-x/n}$, where $\gamma$ denotes Euler's constant, and the fundamental relation $\Gamma(x+1)=x\Gamma(x),$ we can deduce that

 $P= \prod_{j=1}^2 \frac{\Gamma(1-b_j)}{\Gamma(1-a_j)}=\frac{\Gamma(1)\Gamma(2)}{\Gamma\left(\frac{3-\sqrt{5}}{2}\right)\Gamma\left(\frac{3+\sqrt{5}}{2}\right)}=\frac{1}{\Gamma\left(\frac{3-\sqrt{5}}{2}\right)\Gamma\left(\frac{3+\sqrt{5}}{2}\right)}\approx 0.2966751356.$

[This standard method of determining certain infinite products is explained, for example, in Section 12.13 of [i]A Course of Modern Analysis [/i]([i]Fourth Edition[/i]) by E. T. Whittaker and G. N. Watson.]

Using the relation $\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin\pi x},$ for $x$ not an integer, it should be possible to deduce a closed form for the infinite product given by the computer algebra system [i]Maple[/i]:  $\; - \frac{\sin \pi\left(\frac{1+\sqrt{5}}{2}\right)}{\pi}.$

\[ \int_{0}^{1}{\frac{\arctan{\sqrt{x^{2}+2}}}{(x^{2}+1)\sqrt{x^{2}+2}}dx} \]

(Ahmed's integral)

Solution:

\[ \frac{\pi^{2}}{16}=\int_{0}^{1}{\int_{0}^{1}{\frac{dxdy}{(1+x^2)(1+y^2)}}} \]
\[=\int_{0}^{1}{\int_{0}^{1}{\frac{1}{(1+x^{2})(2+x^2+y^2)}+\frac{1}{(1+y^{2})(2+x^2+y^2)}dxdy}}\]
\[=2\int_{0}^{1}{\int_{0}^{1}{\frac{1}{(1+x^2)(2+x^2+y^2)}dydx}} \]
\[=2\int_{0}^{1}{\frac{1}{(1+x^2)\sqrt{2+x^2}}\arctan{\frac{1}{\sqrt{2+x^2}}}dx} \]
\[ =2\int_{0}^{1}{\left(\frac{\pi}{2(1+x^2)\sqrt{2+x^2}}-\frac{\arctan{\sqrt{2+x^2}}}{(1+x^2)\sqrt{2+x^2}} \right)dx} \]
\[ =\frac{\pi^{2}}{6}-2\int_{0}^{1}{\frac{\arctan{\sqrt{2+x^2}}}{(1+x^2)\sqrt{2+x^2}}dx} \]
\[ \Rightarrow \int_{0}^{1}{\frac{\arctan{\sqrt{2+x^2}}}{(1+x^2)\sqrt{2+x^2}}dx}=\frac{5}{96}\pi^{2}\]
 

1.设$f(x)$在$ \left[0,\frac{\pi}{2}\right]$上连续可导，$f(0)=0,f(\frac{\pi}{2})=1,$求证：
\[ \int_{0}^{\frac{\pi}{2}}{|f(x)\cdot\sin{x}+f'(x)|dx}\geq 1 \]
证明：设\[ F(x)=e^{-\cos{x}}f(x) \]
则\[F'(x)=e^{-\cos{x}}\left(f(x)\sin{x}+f'(x)\right) \]
\[ \Rightarrow F'(x)e^{\cos{x}}=f(x)\cdot\sin{x}+f'(x) \]
\[ \Rightarrow \int_{0}^{\frac{\pi}{2}}{|f(x)\cdot\sin{x}+f'(x)|dx}=\int_{0}^{\frac{\pi}{2}}{|e^{\cos{x}}F'(x)|dx} \]
注意到 \[ e^{\cos{x}}\geq 1, x\in\left[0,\frac{\pi}{2}\right] \]
所以\[ \int_{0}^{\frac{\pi}{2}}{|e^{\cos{x}}F'(x)|dx}\geq \left|\int_{0}^{\frac{\pi}{2}}{F'(x)dx}\right|=1 \]

2.若曲面$z=f(x,y)$的所有切平面均过原点，证明该曲面是锥面。
证明：由于曲面$z=f(x,y)$处处有切平面，故处处可微。
设$ (x_{0},y_{0},z_{0})$是曲面上任意一点,则有：
\[ z_{0}=f(x_{0},y_{0})\]
过这个点的切平面为
\[ z-z_{0}=f'_{x}(x_{0},y_{0})(x-x_{0})+f'_{y}(x_{0},y_{0})(y-y_{0}) \]
而该平面过原点，
\[ \Rightarrow z_{0}=x_{0}\cdot f'_{x}(x_{0},y_{0})+y_{0}\cdot f'_{y}(x_{0},y_{0}) \]
故对$z=f(x,y)$上任意的点$(x,y,z)$有
\[  \Rightarrow z=x\cdot f'_{x}(x,y)+y\cdot f'_{y}(x,y) \]
$z=f(x,y)$是一次齐次函数
\[ \Rightarrow f(tx,ty)=tf(x,y) \]
用极坐标代换$ x=r\cos{\theta}, y=r\sin{\theta} $
\[ z=f(x,y)=f(r\cos{\theta},r\sin{\theta})=rf(\cos{\theta},\sin{\theta})=\sqrt{x^2+y^2}f(\cos{\theta},\sin{\theta}) \]

3.设
\[ H=\{(x,y,z):x^2+y^2+z^2=1,z\geq 0 \} \]
\[ C=\{(x,y,0):x^2+y^2=1\} \]
而$P$是$C$的内接正五边形，求$H$位于$P$的上方部分面积。
我们考虑旋转体表面积公式，$ y=f(x) x\in[a,b] $
\[ P=2\pi \int_{a}^{b}{y\sqrt{1+(y'_{x})^{2}}dx} \]
设$H$位于$P$上部分面积为$S_{H}$
\[ S_{H}+5I=\frac{1}{2}4\pi=2\pi \]
\[ I=\frac{1}{2}\cdot 2\pi \int_{\sin{54^{o}}}^{1}{\sqrt{1-x^2}\sqrt{1+\frac{x^2}{1-x^2}}dx}=\pi(1-\sin{54^{o}}) \]
\[ \Rightarrow S_{H}=5\pi\cdot\sin{54^{o}}-3\pi \]