$$\begin{align*}&\int_0^{\frac{\pi }{3}} {\arccos \frac{{1 - \cos x}}{{2\cos x}}dx} \\&\int_0^{\frac{\pi }{2}} {\arccos \sqrt {\frac{{\cos x}}{{1 + 2\cos x}}} dx} .\end{align*}$$

今天回家，11:00的车，临走前写几个要解决的问题，虽然第三个积分在潘承洞的素数定理的初等证明里看到过.

$$\begin{align*}\gamma&=- \int_0^1 {\ln \ln \frac{1}{x}dx}  = \int_0^1 {\left( {\frac{1}{{1 - x}} + \frac{1}{{\ln x}}} \right)dx}  = \int_0^\infty  {{e^{ - x}}\ln xdx} \\\beta \left( 3 \right) &= \frac{{{{\left( { - 1} \right)}^{n - 1}}}}{{{{\left( {2n - 1} \right)}^3}}} = \frac{{{\pi ^3}}}{{32}}.\end{align*}$$

接下来是一个恒等式的证明：

$$\begin{align*}&\frac{1}{{\left( {{x_1} - {x_2}} \right)\left( {{x_1} - {x_3}} \right) \cdots \left( {{x_1} - {x_n}} \right)}} = \frac{1}{{{x_1} - {x_2}}}\left[ {\frac{1}{{\left( {{x_2} - {x_3}} \right)\left( {{x_2} - {x_4}} \right) \cdots \left( {{x_2} - {x_n}} \right)}}} \right] \\&+ \frac{1}{{{x_1} - {x_3}}}\left[ {\frac{1}{{\left( {{x_3} - {x_2}} \right)\left( {{x_3} - {x_4}} \right) \cdots \left( {{x_3} - {x_n}} \right)}}} \right] +  \cdots \frac{1}{{{x_1} - {x_n}}}\left[ {\frac{1}{{\left( {{x_n} - {x_2}} \right)\left( {{x_n} - {x_3}} \right) \cdots \left( {{x_n} - {x_{n - 1}}} \right)}}} \right].\end{align*}$$

也祝愿自己和亲们假期愉快！

$$f(z) = {1\over\log{(i(1-e^{i2z}))}},$$

the logarithm being the principal branch. As can be deduced from the comments, the original integral is equal to

$$\int_0^\pi \operatorname{Re}{f(x)}\,dx.$$

Next consider, for $R>\epsilon > 0$, the following contour:

 

 

It is straightforward to check that for each fixed $\epsilon$ and $R$, the function $f$ is analytic on an open set containing this contour (just consider where $i(1-e^{i2z})\leq0$; this can only happen when $\operatorname{Im}{z}<0$ or when $\operatorname{Im}{z} = 0$ and $\operatorname{Re}{z}$ is an integer multiple of $\pi$). It then follows from Cauchy's theorem that $f$ integrates to zero over it. First let's see that the integrals over the quarter-circular portions of the contour vanish in the limit $\epsilon \to 0$. I'll look at the quarter circle near $\pi$ (near the bottom right corner of the contour) but the one near zero is similar, if not easier. Writing $i(1-e^{i2z}) = i(1-e^{i2(z-\pi)})$, it is clear that $i(1-e^{i2z}) = O(z-\pi)$ as $z\to\pi$. It follows from this that $f(z) = O(1/\log{|z-\pi|})$ as $z\to0$, and therefore that the integral of $f$ over the bottom right quarter circle is $O(\epsilon/\log{\epsilon})$ as $\epsilon \to0$, hence it vanishes in the limit as claimed.

 

Thus for fixed $R$, we can let $\epsilon \to 0$ to see that $f$ integrates to zero over the rectangle with corners $0,\pi, \pi +iR,$ and $iR$. Now the vertical sides of this rectangle give the contribution

$$\begin{align*}-\int_0^R f(iy)\,idy + \int_0^R f(iy+\pi)\,idy = 0,\end{align*}$$

since $f(iy) = f(iy+\pi)$. It follows at once that for each $R$, we can set the contribution from the horizontal sides equal to zero, giving

$$\begin{align*}\int_0^{\pi} f(x)\,dx - \int_0^\pi f(x+iR)\,dx=0, \qquad R>0. \tag{1}\end{align*}$$

(Note that the above equation implies that the integral in $x$ of $f(x+iR)$ over the interval $[0,\pi]$ is constant as a function of $R$; another way to evaluate the integral is to prove this directly, which I'll add below in a moment.) Now $f(x+iR) \to 1/\log{i} = 2/\pi$ as $R\to\infty$, uniformly for $x\in[0,\pi]$. Thus

$$\begin{align*}\lim_{R\to\infty} \int_0^\pi f(x+iR)\,dx = \pi\cdot {2\over \pi} = 2,\end{align*}$$

and it follows from $(1)$ that the original integral is equal to $2$ as well.

(By the way, the above idea is basically a replication of the technique I used here and here, and which I originally got from Ahlfors' book on complex analysis.)

 

选自：http://math.stackexchange.com/questions/625456/what-is-int-0-pi-fracxx2-ln22-sin-x-mathrmdx?lq=1

求解 $$\int_0^{\frac{\pi }{2}} {{{\ln }^2}\sin xdx}  = \int_0^{\frac{\pi }{2}} {{{\ln }^2}\cos xdx}$$ 的值.

显然 $$\int_0^{\frac{\pi }{2}} {{{\ln }^2}\sin xdx} \underline{\underline {{\text{令}x = \frac{\pi }{2} - t}}} \int_0^{\frac{\pi }{2}} {{{\ln }^2}\cos xdx}.$$

记待求积分为$I$，注意到

 

(1)$\int_0^\pi  {{{\ln }^2}\sin \left( x \right)dx}  = 2\int_0^{\frac{\pi }{2}} {{{\ln }^2}\sin xdx}  = 2I$;

 

(2)$\int_0^\pi  {\ln \sin xdx}  = 2\int_0^{\frac{\pi }{2}} {\ln \sin xdx} \text{(Euler 积分)} =  - \pi \ln 2$;

 

(3)$\sum\limits_{n = 1}^\infty  {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{{{{\left( {2n - 1} \right)}^3}}}}=\beta(3)  = \frac{{{\pi ^3}}}{32}$,其中$\beta(s)$是Dirichlet beta 函数;

 

(4)$\int_0^{\frac{\pi }{2}} {{{\ln }^2}\tan xdx}  = \int_0^{\frac{\pi }{2}} {{{\ln }^2}\cot xdx}  = \frac{{{\pi ^3}}}{8}.$

 

事实上，我们有

$$\begin{align*}\int_0^{\frac{\pi }{2}} {{{\ln }^2}\tan xdx}  &= \int_0^\infty  {\frac{{{{\ln }^2}x}}{{1 + {x^2}}}dx}  = 2\int_0^1 {\frac{{{{\ln }^2}x}}{{1 + {x^2}}}dx}  = 2\int_0^1 {{{\ln }^2}x\sum\limits_{n = 1}^\infty  {{{\left( { - {x^2}} \right)}^{n - 1}}} dx} \\&= 2\sum\limits_{n = 1}^\infty  {{{\left( { - 1} \right)}^{n - 1}}\int_0^1 {{x^{2n - 2}}{{\ln }^2}xdx} }  = 4\sum\limits_{n = 1}^\infty  {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{{{{\left( {2n - 1} \right)}^3}}}}  = \frac{{{\pi ^3}}}{8}.\end{align*}$$

 

又

$$\begin{align*}2I &= \int_0^{\frac{\pi }{2}} {{{\left[ {\left( {\ln \sin \left( {2x} \right) - \ln 2} \right)} \right]}^2}dx}  - \int_0^{\frac{\pi }{2}} {2\ln \sin x\ln \cos xdx} \\&{ = \int_0^{\frac{\pi }{2}} {{{\ln }^2}\sin \left( {2x} \right)dx}  - 2\ln 2\int_0^{\frac{\pi }{2}} {{\rm{lnsin}}\left( {2x} \right)dx}  + \frac{{{{\ln }^2}2}}{2}\pi  - \int_0^{\frac{\pi }{2}} {2\ln \sin x\ln \cos xdx} }\\&{ = \frac{1}{2}\int_0^\pi  {{{\ln }^2}\sin \left( x \right)dx}  - \ln 2\int_0^\pi  {\ln \sin xdx}  + \frac{{{{\ln }^2}2}}{2}\pi  - \int_0^{\frac{\pi }{2}} {2\ln \sin x\ln \cos xdx} }\\&{ = I + \frac{{3{{\ln }^2}2}}{2}\pi  - 2\int_0^{\frac{\pi }{2}} {\ln \sin x\ln \cos xdx} }.\end{align*}$$

 

 

$$2I = \int_0^{\frac{\pi }{2}} {{{\ln }^2}\tan xdx}  + 2\int_0^{\frac{\pi }{2}} {\ln \sin x\ln \cos xdx} .$$

因此

$$\begin{align*}3I &= \int_0^{\frac{\pi }{2}} {{{\ln }^2}\tan xdx}  + \frac{{3{{\ln }^2}2}}{2}\pi  = \frac{{{\pi ^3}}}{8} + \frac{{3{{\ln }^2}2}}{2}\pi \\I &= \frac{{{\pi ^3}}}{{24}} + \frac{{{{\ln }^2}2}}{2}\pi.\end{align*}$$

同时我们有

$$\int_0^{\frac{\pi }{2}} {\ln \sin x\ln \cos xdx}  = \frac{{{{\ln }^2}2}}{2}\pi  - \frac{{{\pi ^3}}}{{48}}.$$

这题挺好的，哈哈，是Euler积分的一个延伸吧.

关于Dirichlet beta 函数可参阅：

http://mathworld.wolfram.com/DirichletBetaFunction.html或是http://en.wikipedia.org/wiki/Dirichlet_beta_function

 

解法二： $$\beta(x,y) = 2\int_0^{\frac{\pi}{2}} \sin^{2x-1}(t) \cos^{2y-1}(t) \ dt$$

$$\frac{\partial }{\partial x} \beta(x,y) = 4\int_0^{\frac{\pi}{2}} \sin^{2x-1}(t)\ln(\sin t) \cos^{2y-1}(t) \ dt$$

$$\frac{\partial}{\partial y} \left( \frac{\partial}{\partial x} \beta(x,y) \right) = 8\int_0^{\frac{\pi}{2}} \sin^{2x-1}(t) \ln(\sin t ) \ln(\cos t) \cos^{2y-1}(t) \ dt$$

and we have $$\beta(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$

so differentiate and put $x =\frac{1}{2} , y = \frac{1}{2}$

$$\psi \left(\frac{1}{2} \right) = -2\ln 2 - \gamma$$

$$\psi(1) = -\gamma$$

$$\psi^{(1)}(1) = \frac{\pi^2}{6}$$

$$\beta \left( \frac{1}{2} , \frac{1}{2} \right) = \frac{\Gamma \left( \frac{1}{2} \right)^2}{\Gamma(1)} = \pi$$  

thus you'll have the integral $= \frac{\pi}{8} \left(4\ln(2)^2 - \frac{\pi^2}{6} \right)$

参阅：http://math.stackexchange.com/questions/492878/find-int-0-frac-pi2-ln-sinx-ln-cosx-mathrm-dx?rq=1

解法三：A third approach would be the Fourier series:   

Namely, consider  

$$\ln \left (2\sin \frac{x}{2}\right )=-\sum_{n=1}^{\infty}\frac{\cos nx}{n};(0<x<2\pi)$$   

After squared:  

$$\ln^2\left (2\sin \frac{x}{2}\right )=\sum_{n=1}^{\infty} \sum_{k=1}^{\infty}\frac{\cos kx\cos nx}{kn}$$     

Now, integrate the last equation from $x=0$ to $x=\pi$  

On the right side, we get:  

$$\frac{\pi}{2}\sum_{k=1}^{\infty}\frac{1}{k^2}=\frac{\pi}{2}\frac{\pi^2}{6}=\frac{\pi^3}{12}$$    

because $$I=\int_{0}^{\pi}\cos kx\cos nx\,dx=0;k\neq n$$ $$I=\frac{\pi}{2};k=n$$

On the left side:  

$$\int_{0}^{\pi}\ln^2\left (2\sin \frac{x}{2}\right )\,dx= \ln^22 \int_{0}^{\pi}dx + 4\ln 2 \int_{0}^{\frac{\pi}{2}} \ln \left (\sin x\right )dx+2 \int_{0}^{\frac{\pi}{2}} \ln^2 \left (\sin x\right )dx$$  

Since we  know that $\int_{0}^{\frac{\pi}{2}} \ln(\sin x)dx=-\frac{\pi}{2}\ln(2)$ then we get $\int_{0}^{\frac{\pi}{2}}\ln^{2}(\sin x)dx$ from the equation.

解法四：

参阅：http://math.stackexchange.com/questions/58654/integrate-square-of-the-log-sine-integral-int-0-frac-pi2-ln2-sinx/58672#58672

 

再看一下推广：

 

$$\int_0^{\frac{\pi }{2}} {{{\ln }^n}\sin xdx}  = \frac{{\sqrt \pi  }}{{{2^{n + 1}}}}\frac{{{d^k}}}{{d{\alpha ^k}}}{\left( {\frac{{\Gamma \left( {\frac{{2\alpha  + 1}}{2}} \right)}}{{\Gamma \left( {\frac{{2\alpha  + 2}}{2}} \right)}}} \right)_{\alpha  = 0}}.$$

证明.注意到 $$\int_0^{\frac{\pi }{2}} {{{\ln }^n}\sin xdx}  = \int_0^{\frac{\pi }{2}} {\frac{{{{\ln }^n}t}}{{\sqrt {1 - {t^2}} }}dt}.$$

令 $$I\left( \alpha  \right) = \int_0^1 {\frac{{{t^{2\alpha }}}}{{\sqrt {1 - {t^2}} }}dt}  = \frac{1}{2}B\left( {\frac{1}{2},\frac{{2\alpha  + 1}}{2}} \right) = \frac{{\sqrt \pi  }}{2}\frac{{\Gamma \left( {\frac{{2\alpha  + 1}}{2}} \right)}}{{\Gamma \left( {\frac{{2\alpha  + 2}}{2}} \right)}},$$

则有 $${I^{\left( n \right)}}\left( \alpha  \right) = \int_0^1 {\frac{{{t^{2\alpha }}{2^n}{{\ln }^n}t}}{{\sqrt {1 - {t^2}} }}dt}  .$$

故有

$$\int_0^{\frac{\pi }{2}} {{{\ln }^n}\sin xdx}  = \int_0^{\frac{\pi }{2}} {\frac{{{{\ln }^n}t}}{{\sqrt {1 - {t^2}} }}dt}  = \frac{{{I^{\left( n \right)}}\left( 0 \right)}}{{{2^n}}} = \frac{{\sqrt \pi  }}{{{2^{n + 1}}}}\frac{{{d^k}}}{{d{\alpha ^k}}}{\left( {\frac{{\Gamma \left( {\frac{{2\alpha  + 1}}{2}} \right)}}{{\Gamma \left( {\frac{{2\alpha  + 2}}{2}} \right)}}} \right)_{\alpha  = 0}}.$$

另一种解法

参阅：http://math.stackexchange.com/questions/121473/solve-the-integral-s-k-1k-int-01-log-sin-pi-xk-dx?rq=1

 

关于此类问题还可参阅：

http://math.stackexchange.com/questions/307593/a-hard-log-definite-integral-int-0-pi-4-ln3-sin-x-mathrm-dx?lq=1

或是http://math.stackexchange.com/questions/289587/log-sin-and-log-cos-integral-maybe-relate-to-fourier-series/309781#309781

http://math.stackexchange.com/questions/330057/how-to-evaluate-i-displaystyle-int-0-pi-2x2-ln-sin-x-ln-cos-xdx