西西爆难积分题求解
在这里，主要展示西西12年7月在百度贴吧数学吧中贴出的30个积分题的求解，本文中主要参考的是自己以前的摘录，不知何故，与西哥的版本有些出入，但基本包含了西哥的所有问题，来源于网友的解答均会注明出处。
1. \[\int_0^1 {\frac{{\ln \left( {1 + {x^{2 + \sqrt 3 }}} \right)}}{{1 + x}}dx} = \frac{{{\pi ^2}}}{{12}}\left( {1 - \sqrt 3 } \right) + \ln 2 \cdot \ln \left( {1 + \sqrt 3 } \right)\]
2. \[\int_0^1 {\frac{{\ln \left( {1 + {x^{4 + \sqrt {15} }}} \right)}}{{1 + x}}dx} = \frac{{{\pi ^2}}}{{12}}\left( {2 - \sqrt {15} } \right) + \ln \frac{{1 + \sqrt 5 }}{2} \cdot \ln \left( {2 + \sqrt 3 } \right) + + \ln 2 \cdot \ln \left( {\sqrt 3 + \sqrt 5 } \right)\]
3. \[\int_0^1 {\frac{{\ln \left( {1 + {x^{6 + \sqrt {35} }}} \right)}}{{1 + x}}dx} = \frac{{{\pi ^2}}}{{12}}\left( {3 - \sqrt {35} } \right) + \ln \frac{{1 + \sqrt 5 }}{2} \cdot \ln \left( {8 + 3\sqrt 7 } \right) + + \ln 2 \cdot \ln \left( {\sqrt 5 + \sqrt 7 } \right)\]
4. \[\int_0^1 {\frac{{ar\tanh x\ln x}}{{x\left( {1 - x} \right)\left( {1 + x} \right)}}dx} \]
5. \[\int_0^1 {\frac{{\arctan {x^{3 + \sqrt 8 }}}}{{1 + {x^2}}}dx} \]
6. \[\int_0^{\frac{\pi }{3}} {x{{\ln }^2}\left( {2\sin \frac{x}{2}} \right)dx} \]
7. \[\int_0^{\frac{\pi }{2}} {{{\ln }^n}\sin xdx} \]
8. \[\int_0^\infty {\frac{{\sin x}}{{\cosh x - \cos x}} \cdot \frac{{{x^n}}}{{n!}}dx} \]
9. \[\int_0^\infty {\frac{{\sin x}}{{\cosh x + \cos x}} \cdot \frac{{{x^n}}}{{n!}}dx} \]
10. \[\int_{\frac{\pi }{4}}^{\frac{\pi }{2}} {\ln \ln \left( {\tan x} \right)dx} \]
11. \[\int_0^\infty {\frac{{x - \sin x}}{{\left( {{\pi ^2} + {x^2}} \right){x^3}}}dx} = \frac{1}{{2{\pi ^3}}}\left( {1 + \frac{{{\pi ^2}}}{2} - \pi - \frac{1}{{{e^\pi }}}} \right)\]
12. \[\int_0^\infty {\frac{1}{{{a^2} + {x^2}}}\frac{{x - \sin x}}{{{x^3}}}dx} = \frac{{{a^2} - 2a + 2 - 2{e^{ - a}}}}{{4{a^3}}},a > 0\]
13. \[\prod\limits_{k = 4}^\infty {\left[ {1 - {{\left( {\frac{3}{k}} \right)}^3}} \right]} = \frac{8}{{15561}} \cdot \frac{{\cos \left( {\frac{{3\sqrt 3 }}{2}\pi } \right)}}{\pi }\]
14. \[\sum\limits_{m = 1}^\infty {\sum\limits_{n = {2^{m - 1}}}^{{2^m} - 1} {\frac{m}{{\left( {2n + 1} \right)\left( {2n + 2} \right)}}} } = 1 - \gamma \]
15. \[\int_0^1 {\frac{{\left( {1 - x} \right)\ln x \cdot {e^{ - x}}}}{{\pi - x}}dx} \]
16. \[\mathop {\lim }\limits_{n \to \infty } \left[ { - \frac{1}{{2m}} + \ln \left( {\frac{e}{m}} \right) + \sum\limits_{n = 2}^m {\left( {\frac{1}{n} - \frac{{\varsigma \left( {1 - n} \right)}}{{{m^n}}}} \right)} } \right] = \gamma \]
17. \[\int_0^\infty {\frac{1}{{x{e^x}\left( {{\pi ^2} + {{\ln }^2}x} \right)}}dx} \]
18. \[\int_0^\infty {\frac{{\ln \left( {1 + {x^2}} \right)}}{{{e^{2\pi x}} - 1}}dx} \]
19. \[\sum\limits_{n = 1}^\infty {\frac{{\sum\limits_{k = 1}^n {\frac{1}{{{k^4}}}} }}{{{n^2}}}} \]
20. \[\int_0^1 {\int_0^1 {\int_0^1 {\int_0^1 {\frac{{dwdxdydz}}{{\left( {wxy - 1} \right)\left( {zwx - 1} \right)\left( {yzw - 1} \right)\left( {xyz - 1} \right)}}} } } } \]
21. \[\int_0^\infty {\frac{{{e^{ - {x^2}}}}}{{{\pi ^2} + {{\left( {\gamma + x} \right)}^2}}}dx} \]
22. \[\sqrt[3]{{1 + \sqrt[3]{{1 + \sqrt[3]{{1 + \sqrt[3]{{1 + \sqrt[3]{{1 + \cdots }}}}}}}}}} = \frac{2}{{\sqrt 3 }}\cos \left( {\frac{1}{3}\arccos \frac{{3\sqrt 3 }}{2}} \right)\]
23. \[\int_0^{2ar\cosh \pi } {\frac{{dx}}{{1 + \frac{{{{\sinh }^2}x}}{{{\pi ^4}}}}}} \]
24. \[\int_0^\infty {\frac{{x{e^x}}}{{\sqrt {4{e^x} - 3} \left( {1 + 2{e^x} - \sqrt {4{e^x} - 3} } \right)}}dx} \]
25. \[\int_0^{\frac{\pi }{2}} {\frac{{{{\ln }^2}\left( {2\cos x} \right)}}{{{{\ln }^2}\left( {2\cos x} \right) + {x^2}}}dx} \]
26. \[\mathop {\lim }\limits_{n \to \infty } \frac{{n!}}{{{n^n}}}\left( {\sum\limits_{k = 0}^n {\frac{{{n^k}}}{{k!}}} - \sum\limits_{k = n + 1}^\infty {\frac{{{n^k}}}{{k!}}} } \right)\]
27. \[\sum\limits_{k = 1}^\infty {\frac{1}{{{k^2}}}\cos \left( {\frac{9}{{k\pi + \sqrt {{k^2}{\pi ^2} - 9} }}} \right)} \]
28. \[\sum\limits_{n = - \infty }^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{{\left( {n\pi + \phi } \right)}^2}}}\cos \left( {\sqrt {{n^2}{\pi ^2} + {a^2} - {\phi ^2}} } \right)} = \frac{{a\cos a\cot \phi + \phi \sin a}}{{a\sin \phi }}\]
29. \[\sum\limits_{n = 0}^\infty {\frac{{{n^2}{\pi ^2} + {\phi ^2}}}{{{{\left( {{n^2}{\pi ^2} - {\phi ^2}} \right)}^2}}}{{\left( { - 1} \right)}^n}\cos \left( {\sqrt {{n^2}{\pi ^2} + {a^2} - {\phi ^2}} } \right)} = \frac{{\cos \sqrt {{a^2} - {\phi ^2}} }}{{2{\phi ^2}}} + \frac{{a\cos a\cot \phi + \phi \sin a}}{{2a\sin \phi }}\]
30. \[\sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}\left( {n + \frac{1}{2}} \right)}}{{{{\left( {n + \frac{1}{3}} \right)}^2}{{\left( {n + \frac{2}{3}} \right)}^2}}}} \cos \left[ {\pi \sqrt {\left( {n + \frac{1}{6}} \right)\left( {n + \frac{5}{6}} \right)} } \right] = {\pi ^2}{e^{\frac{{\pi \sqrt 3 }}{6}}}\]
31. \[\int_{ - 1}^1 {\frac{{\arctan x}}{{1 + x}}\ln \left( {\frac{{1 + {x^2}}}{2}} \right)dx} \]
32. \[\int_0^1 {\sin \left( {\pi x} \right){x^x}{{\left( {1 - x} \right)}^{1 - x}}dx} \]
33. \[\int_0^{\frac{\pi }{2}} {{x^2}{{\ln }^2}\left( {2\cos x} \right)dx} \]
34. \[\int_0^{\frac{\pi }{2}} {\frac{{{x^2}}}{{{x^2} + {{\ln }^2}\left( {2\cos x} \right)}}dx} \]
35. \[\int_0^{\frac{\pi }{2}} {\frac{{{x^2}{{\ln }^2}\left( {2\cos x} \right)}}{{{x^2} + {{\ln }^2}\left( {2\cos x} \right)}}dx}\]
36. \[\mathop {\lim }\limits_{n \to \infty } \frac{n}{{\ln n}}\left[ {\frac{1}{\pi }{{\left( {\sum\limits_{k = 1}^n {\sin \frac{\pi }{{\sqrt {{n^2} + k} }}} } \right)}^n} - \frac{1}{{\sqrt[4]{e}}}} \right]\]
37. \[\int_0^1 {{x^{ - x}}{{\left( {1 - x} \right)}^{ - 1 + x}}\sin \left( {\pi x} \right)dx = \frac{\pi }{e}} = 1.15573\]
38. \[\int_0^{ + \infty } {\frac{{dx}}{{1 + x\left| {\sin x} \right|}}} \]
39. \[\int_0^{ + \infty } {\frac{{dx}}{{{x^2} + {{\left( {{n^2}{x^2} - 1} \right)}^2}}}}\]
参阅：[1]http://tieba.baidu.com/p/2114477017
[2]http://tieba.baidu.com/p/3148596990
[3]http://tieba.baidu.com/p/2121721064
西西11年送的几个CMC培训题整理
如何掌握好数学系的两门基础课
西西云：
数学分析先刷完周民强系列，谢惠民系列,徐森林系列.提高:波利亚数学分析定理，国外一些analysis problem ,高等代数刷张贤科系列，王品超系列，提高:李炯生系列。这只是分析和代数一套书籍，差不多就可以了。
有关反三角和三角函数组合的虐心积分求解
昨天在数学爱好者群有位自称准粗二的网友发来两道有关反三角和三角函数组合的虐心积分题，顿感无措！
\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*}
一些亟待解决的问题
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！
Euler积分的一个延伸
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：
Here is a completely different way to approach this integral, which relies on some elementary complex analysis (Cauchy's theorem). It is based on an approach which I have seen several times employed to compute $\int_0^{\pi/2}\log{(\sin{x})}\,dx$ (particularly, in Ahlfors's book).
The idea is to integrate the principal branch of $f(z) := \log^2{(1 - e^{2iz})} = \log^2{(-2ie^{iz}\sin{z})}$ over the contour below, and then let $R \to \infty$ and $\epsilon \to 0$.
First of all, $1 - e^{2iz} \leq 0$ only when $z = k\pi + iy$, where $k$ is integral and $y \leq 0$. Thus, in the region of the plane which is obtained by omitting the lines $\{k\pi + iy: y\leq 0\}$ for $k \in \mathbb Z$, the principal branch of $\log{(1-e^{2iz})}$ is defined and analytic. Note that, for each fixed $R$ and $\epsilon$, the contour we wish to integrate over is contained entirely within this region.
By Cauchy's theorem, the integral over the contour vanishes for each fixed $R$. Since $f(x + iR) = \log^2{(1 - e^{2ix}e^{-2R})} \to 0$ uniformly as $R \to \infty$, the integral over the segment $[iR,\pi/2 + iR]$ vanishes in the limit. Similarly, since $1 - e^{2iz} = O(z)$ as $z \to 0$, we have $f(z) = O(\log^2{|z|})$ for small enough $z$, which, since $\epsilon \log^2{\epsilon} \to 0$ with $\epsilon$, means that the the integral over the circular arc from $i\epsilon$ to $\epsilon$ vanishes as $\epsilon \to 0$.
From the vertical sides of the contour, we get the contribution
\begin{align*}\int_{[\pi/2,\pi/2 + iR]} + \int_{[iR,i\epsilon]} f(z)\,dz & = i\int_0^R f(\pi/2 + iy)\,dy -i\int_\epsilon^R f(iy)\,dy.\end{align*}
Since $f(iy)$ and $f(\pi/2 + iy)$ are real, this contribution is purely imaginary.
Finally, the contribution from the bottom side of the contour, after letting $\epsilon \to 0$, is
\begin{align*}\int_0^{\pi/2} f(x)\,dx = \int_0^{\pi/2} \log^2{(-2ie^{ix}\sin{x})}\,dx,\end{align*}
and we know from the preceding remarks that the real part of this integral must vanish. For $x$ between $0$ and $\pi/2$, the quantity $2\sin{x}$ is positive. Writing $-ie^{ix} = e^{i(x - \pi/2)}$, we see that $x - \pi/2$ is the unique value of $\arg{(-2ie^{ix}\sin{x})}$ which lies in $(-\pi,\pi)$. Since we have chosen the principal branch of $\log{z}$, it follows from these considerations that $\log{(-2ie^{ix}\sin{x})} = \log{(2\sin{x})} + i(x-\pi/2)$, and therefore that
\begin{align*}\text{Re}{f(x)} &= \log^2{(2\sin{x})} - (x-\pi/2)^2 \\&= \log^2{(\sin{x})} + 2\log{2}\log{(\sin{x})} + \log^2{2} - (x - \pi/2)^2.\end{align*}
By setting $\int_0^{\pi/2} \text{Re}f(x)\,dx = 0$ we get
\begin{align*}\int_0^{\pi/2} \log^2{(\sin{x})}\,dx &= \int_0^{\pi/2}(x-\pi/2)^2\,dx - 2\log{2}\int_0^{\pi/2} \log{(\sin{x})}\,dx -\frac{\pi}{2}\log^2{2} \\& = \frac{1}{3}\left(\frac{\pi}{2}\right)^3 + \frac{\pi}{2} \log^2{2}\end{align*}
as expected
By similar methods, one can compute a variety of integrals of this form with little difficulty. Here are some examples I have computed for fun. All are proved by the same method, with the same contour, but different functions $f$.
1. Take $f(z) = \log{(1 + e^{2iz})} = \log{(2e^{iz}\cos{z})}$ and compare imaginary parts to get
$$\int_0^\infty \log{(\coth{y})}\,dy = \frac{1}{2}\left(\frac{\pi}{2}\right)^2.$$
2. Related to this question of yours (which incidentally led me here), one can show by taking $f(z) = \log^4(1 + e^{2iz})$ and comparing real parts that
$$\int_0^{\pi/2} x^2\log^2{(2\cos{x})}\,dx = \frac{1}{30}\left(\frac{\pi}{2}\right)^5 + \frac{1}{6}\int_0^{\pi/2} \log^4{(2\cos{x})}\,dx.$$Assuming the result of the other question, we then get
$$\int_0^{\pi/2} \log^4{(2\cos{x})}\,dx = \frac{19}{15}\left(\frac{\pi}{2}\right)^5.$$
3.Also related to the question cited in 2., taking $f(z) = z^2\log^2{(1 + e^{2iz})}$ and comparing real parts gives
$$\int_0^{\pi/2}x^2\log^2{(2\cos{x})}\,dx = \frac{1}{5}\left(\frac{\pi}{2}\right)^5 + \pi \int_0^\infty y\log^2{(1- e^{-2y})}\,dy.$$
Once more, assuming the result of the other question, we get
$$\int_0^\infty y\log^2{(1- e^{-2y})}\,dy = \frac{1}{45}\left(\frac{\pi}{2}\right)^4.$$
Actually, the integral in 3. has several interesting series expansions, and I would be very interested if someone could compute it without using the result from the question I cited. For one thing, that would give us a different proof of that result (which is why I started investigating it in the first place).
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
Use generating series. Consider the function $$f(z)=\sum_{k=0}^{\infty}S_{k}\frac{z^{k}}{k!}.$$ Then $$f(z)=\int_0^1 \left(\sum_{k=0}^\infty (-1)^k \log^k(\sin(\pi x)) z^k \right)dx=\int_{0}^{1}\frac{1}{\sin\left(\pi x\right)^{z}}dx=\frac{\Gamma\left(\frac{1-z}{2}\right)}{\sqrt{\pi}\Gamma\left(1-\frac{z}{2}\right)}.$$ The last equality follows from an identity of the Beta Function and then applying the Duplication Formula. From here, differentiating and plugging in $z=0$ allows you to conclude.
关于此类问题还可参阅：
无穷套根式的一些特殊情形（Nested Radical）
情形一：(Vieta)\[\frac{2}{\pi } = \sqrt {\frac{1}{2}} \sqrt {\frac{1}{2} + \frac{1}{2}\sqrt {\frac{1}{2}} } \sqrt {\frac{1}{2} + \frac{1}{2}\sqrt {\frac{1}{2} + \frac{1}{2}\sqrt {\frac{1}{2}} } } \cdots .\]
情形二：\[x = \sqrt[n]{{\left( {1 - q} \right){x^n} + q{x^{n - 1}}\sqrt[n]{{\left( {1 - q} \right){x^n} + q{x^{n - 1}}\sqrt[n]{ \cdots }}}}}.\]
有以下几种特殊情形：
\begin{align*}\frac{{b + \sqrt {{b^2} + 4a} }}{2} &= \sqrt {a + b\sqrt {a + b\sqrt {a + b\sqrt \cdots } } } \left( {n = 2,q = 1 - \frac{a}{{{x^2}}},x = \frac{b}{q}} \right)\\x &= \sqrt[n]{{{x^{n - 1}}\sqrt[n]{{{x^{n - 1}}\sqrt[n]{{{x^{n - 1}}\sqrt[n]{ \cdots }}}}}}}\left( {q = 1} \right)\\x &= \sqrt {x\sqrt {x\sqrt {x\sqrt {x\sqrt \cdots } } } } \left( {q = 1,n = 2} \right).\end{align*}
情形二：由以上情形可推得：\[{q^{\frac{{{n^k} - 1}}{{n - 1}}}}{x^{{n^j}}} = \sqrt[n]{{{q^{\frac{{{n^{k + 1}} - n}}{{n - 1}}}}\left( {1 - q} \right){x^{{n^{j + 1}}}} + \sqrt[n]{{{q^{\frac{{{n^{k + 2}} - n}}{{n - 1}}}}\left( {1 - q} \right){x^{{n^{j + 2}}}} + \sqrt[n]{ \cdots }}}}}.\]
特殊情形有：
\[\sqrt 2 = \sqrt {\frac{2}{{{2^{{2^0}}}}} + \sqrt {\frac{2}{{{2^{{2^1}}}}} + \sqrt {\frac{2}{{{2^{{2^2}}}}} + \sqrt {\frac{2}{{{2^{{2^3}}}}} + \sqrt {\frac{2}{{{2^{{2^4}}}}} + \cdots } } } } } \left( {q = \frac{1}{2},n = 2,x = 1,k = - 1} \right).\]
情形三：\[\sqrt[{n - 1}]{x} = \sqrt[n]{{x\sqrt[n]{{x\sqrt[n]{{x \cdots }}}}}}.\]
由于
\[\left\{ \begin{array}{l}1 + \frac{1}{n} + \frac{1}{{{n^2}}} + \cdots = 1 + \frac{1}{{n - 1}}\\\frac{1}{n} + \frac{1}{{{n^2}}} + \frac{1}{{{n^3}}} + \cdots = \frac{1}{{n - 1}}\\\frac{1}{n}\left( {1 + \frac{1}{n}\left( {1 + \frac{1}{n}\left( {1 + \cdots } \right)} \right)} \right) = \frac{1}{{n - 1}}\end{array} \right.\left( {n \ge 2,n \in {N_ + }} \right).\]
令$n=3$，我们有\[\sqrt x = \sqrt[3]{{x\sqrt[3]{{x\sqrt[3]{{x \cdots }}}}}}.\]
情形四： (Ramanujan)
\[x + n + a = \sqrt {ax + {{\left( {n + a} \right)}^2} + x\sqrt {a\left( {x + n} \right) + {{\left( {n + a} \right)}^2} + \left( {x + n} \right)\sqrt {a\left( {x + 2n} \right) + {{\left( {n + a} \right)}^2} + \left( {x + 2n} \right)\sqrt \cdots }}}.\]
特殊地，有\[x + 1 = \sqrt {1 + x\sqrt {1 + \left( {x + 1} \right)\sqrt {1 + \left( {x + 2} \right)\sqrt {1 + \cdots } } } } \left( {a = 0,n = 1} \right).\]由此有我们熟知的
\[3 = \sqrt {1 + 2\sqrt {1 + 3\sqrt {1 + 4\sqrt {1 + 5\sqrt \cdots } } } } \left( {a = 0,n = 1,x = 2} \right).\]
The justification of this process both in general and in the particular example of $\ln\sigma$, where $\sigma$ is Somos's quadratic recurrence constant in given by Vijayaraghavan (in Ramanujan 2000, p. 348).
情形五：
由下面两式
\[\left\{ \begin{array}{l}e = 1 + \frac{1}{{1!}} + \frac{1}{{2!}} + \frac{1}{{3!}} + \cdots \\e = 1 + 1 + \frac{1}{2}\left( {1 + \frac{1}{3}\left( {1 + \frac{1}{4}\left( {1 + \frac{1}{5}\left( {1 + \cdots } \right)} \right)}\right)} \right)\end{array} \right.\]
得
\[{x^{e - 2}} = \sqrt {x\sqrt[3]{{x\sqrt[4]{{x\sqrt[5]{{x \cdots }}}}}}}. \]
参考资料
$\sum{\frac{n}{{{e^{2\pi n}} - 1}}}$型的级数求解
两个结论：\[\left\{ \begin{array}{l}\sum\limits_{n = 1}^\infty {\frac{n}{{{e^{2\pi n}} - 1}}} = \frac{1}{{24}} - \frac{1}{{8\pi }}\\\sum\limits_{n = 0}^\infty {\frac{{2n + 1}}{{{e^{\pi \left( {2n + 1} \right)}} + 1}}} = \frac{1}{{24}}\end{array} \right..\]
Proof.What you require here are the Eisenstein series. In particular the evaluation of
$$E_2(\tau) = 1 – 24\sum_{n=1}^\infty \frac{ne^{2\pi i n \tau} }{1 - e^{2\pi i n \tau}},$$
at $\tau = i. $ Rearrange to get
$$\sum_{n=1}^\infty \frac{ne^{2\pi i n \tau} }{1 - e^{2\pi i n \tau} } = \frac{1}{24}(1 – E_2(i) ).$$
See Lambert series for additional information.
The function
$$G_ 2(\tau) = \zeta(2) \left(1 – 24\sum_{n=1}^\infty \frac{ne^{2\pi i n \tau} }{1 - e^{2\pi i n \tau}} \right)=\zeta(2)E_2(\tau)$$
satisfies the quasimodular transformation
$$G_ 2\left( \frac{a\tau+b}{c\tau+d} \right) =(c\tau+d)^2G_ 2(\tau) - \pi i c (c\tau + d).$$
And so with $a=d=0,$ $c=1$ and $b=-1$ we find $G_ 2(i) = \pi/2.$ Therefore
$$E_2(i) = \frac{ G_ 2( i)}{ \zeta(2)} = \frac{\pi}{2}\frac{6}{\pi^2} = \frac{3}{\pi}.$$
Hence we obtain
$$\sum_{n=1}^\infty \frac{n}{e^{2\pi n} – 1} = \frac{1}{24} - \frac{1}{8\pi},$$
as given in the comment to the question by Slowsolver.
There is a very nice generalisation of the sum in the question.
For odd $ m > 1 $ we have
$$\sum_{n=1}^\infty \frac{n^{2m-1} }{ e^{2\pi n} -1 } = \frac{B_{2m}}{4m},$$
where $B_k$ are the Bernoulli numbers defined by
\[\frac{z}{{{e^z} - 1}} = \sum\limits_{k = 0}^\infty {\frac{{{B_k}}}{{k!}}} {z^k}\;\;\; for |z|<2\pi .\]
另解：
$$ \displaystyle\mathcal{M}\Big[\frac{x}{e^{2\pi x}-1}] = \int_{0}^{\infty} \frac{x^{s}}{e^{2 \pi x}-1} \ dx = (2\pi)^{-(s+1)}\Gamma(s+1)\zeta(s+1) $$
So $$\displaystyle \frac{x}{e^{2\pi x}-1}=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}(2\pi)^{-(s+1)}\Gamma(s+1)\zeta(s+1) x^{-s}\ ds $$
which implies $$ \displaystyle \sum_{n=1}^{\infty}\frac{n}{e^{2\pi n}-1}=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}(2\pi)^{-(s+1)}\Gamma(s+1)\zeta(s+1)\zeta(s)\ ds $$
The integrand has poles at $s=-1, s=0$, and $s=1$ (and removable singularities at $s= -2, -3, -4, \ldots$)
I'm going to close the contour with a rectangle that has vertices at $$-i \infty, \frac{3}{2} - i \infty, \frac{3}{2} + i \infty, and \;i \infty$$ and is indented at the origin
$\Gamma(s)$ decays rapidly as $\text{Im} (s) \to \pm \infty$. So the integral goes to zero along the top and bottom of the rectangle.
And on the imaginary axis, the integrand is odd.
So $$\displaystyle \int_{\frac{3}{2}-i\infty}^{\frac{3}{2}+i\infty}(2\pi)^{-s}\Gamma(s+1)\zeta(s+1)\zeta(s)\ ds-\pi i \text{Res}[f,0] = 2\pi i\text{Res}[f,1]$$
where $$\displaystyle f(s) = (2\pi)^{-(s+1)}\Gamma(s+1)\zeta(s+1)\zeta(s) $$.
$$ \displaystyle \text{Res}[f,0] = \lim_{s \to 0} s (2 \pi)^{-(s+1)} \Gamma(s+1) \zeta(s+1) \zeta(s) = \lim_{s\to 0} s\zeta(s+1) (2\pi)^{-(s+1)}\Gamma(s+1)\zeta(s)$$
$$= 1\Big(\frac{1}{2 \pi} \Big)(1)\zeta(0)=-\frac{1}{4 \pi} $$
$$ \displaystyle\text{Res}[f,1] = \lim_{s \to 1} (s-1) (2 \pi)^{-(s+1)} \Gamma(s+1) \zeta(s+1) \zeta(s) $$
$$= \displaystyle \lim_{s\to 1}(s-1)\zeta(s) (2\pi)^{-(s+1)}\Gamma(s+1)\zeta(s+1)= 1\Big(\frac{1}{4 \pi^{2}}\Big)(1)\Big(\frac{\pi^{2}}{6}\Big) =\frac{1}{24} $$
Therefore, $$\displaystyle \sum_{n=1}^{\infty} \frac{n}{e^{2 \pi n}-1} = \frac{1}{2 \pi i} \Big( 2 \pi i (\frac{1}{24}) + \pi i (\frac{-1}{4 \pi}) \Big) = \frac{1}{24} - \frac{1}{8 \pi}.$$
Poof.We will use the Mellin transform technique. Recalling the Mellin transform and its inverse
$$ F(s) =\int_0^{\infty} x^{s-1} f(x)dx, \quad\quad f(x)=\frac{1}{2 \pi i} \int_{c-i \infty}^{c+i \infty} x^{-s} F(s)\, ds. $$
Now, let's consider the function
$$ f(x)= \frac{x}{e^{\pi x}+1}. $$
Taking the Mellin transform of $f(x)$, we get
$$ F(s)={\pi }^{-s-1}\Gamma \left( s+1 \right) \left(1- {2}^{-s} \right) \zeta \left( s+1 \right),$$
where $\zeta(s)$ is the zeta function. Representing the function in terms of the inverse Mellin Transform, we have
$$ \frac{x}{e^{\pi x}+1}=\frac{1}{2\pi i}\int_{C}{\pi }^{-s-1}\Gamma \left( s+1 \right) \left( 1-{2}^{-s} \right) \zeta \left( s+1 \right) x^{-s}ds. $$
Substituting $x=2n+1$ and summing yields
$$\sum_{n=0}^{\infty}\frac{2n+1}{e^{\pi (2n+1)}+1}=\frac{1}{2\pi i}\int_{C}{\pi}^{-s-1}\Gamma \left( s+1 \right)\left(1-{2}^{-s} \right) \zeta\left( s+1 \right) \sum_{n=0}^{\infty}(2n+1)^{-s}ds$$
$$ = \frac{1}{2\pi i}\int_{C}{\pi }^{-s-1}\Gamma \left( s+1 \right) \left(1-{2}^{-s} \right)^2\zeta\left( s+1 \right) \zeta(s)ds.$$
Now, the only contribution of the poles comes from the simple pole $s=1$ of $\zeta(s)$ and the residue equals to $\frac{1}{24}$. So, the sum is given by
$$ \sum_{n=0}^{\infty}\frac{2n+1}{e^{\pi (2n+1)}+1}=\frac{1}{24} $$
Notes: 1)
$$ \sum_{n=0}^{\infty}(2n+1)^{-s}= \left(1- {2}^{-s} \right) \zeta \left( s \right). $$
2) The residue of the simple pole $s=1$, which is the pole of the zeta function, can be calculated as
$$ r = \lim_{s=1}(s-1)({\pi }^{-s-1}\Gamma \left( s+1 \right) \left({2}^{-s}-1 \right)^2\zeta\left( s+1 \right) \zeta(s))$$
$$ = \lim_{s\to 1}(s-1)\zeta(s)\lim_{s\to 1} {\pi }^{-s-1}\Gamma \left( s+1 \right) \left({2}^{-s}-1 \right)^2\zeta\left( s+1 \right) = \frac{1}{24}. $$
For calculating the above limit, we used the facts
$$ \lim_{s\to 1}(s-1)\zeta(s)=1, \quad \zeta(2)=\frac{\pi^2}{6}. $$
3) Here is the technique for computing the Mellin transform of $f(x)$.
Using the change of variables $u=-\ln(x)$ and the identity
$$ \int_{0}^{\infty}\frac{u^{s-1}}{e^u -1}=\zeta{(s)}\Gamma{(s)} $$
we reach to the deisred result
$$ \int_0^1 \frac{\ln x }{x-1}= \int_{0}^{\infty}\frac{u}{e^u -1}=\zeta{(2)}\Gamma{(2)} =\sum_{n=1}^\infty \frac{1}{n^2}. $$
Note that,
$$ \int_{0}^{\infty}\frac{u^{s-1}}{e^u - 1}=\int_{0}^{\infty}\frac{u^{s-1}}{e^u}(1-e^{-u})^{-1}= \sum_{n=0}^{\infty} \int_{0}^{\infty}{u^{s-1}e^{-(n+1)u}}$$
$$= \sum_{n=0}^{\infty}\frac{1}{(n+1)^s} \int_{0}^{\infty}{y^{s-1}e^{-y}}= \sum_{n=1}^{\infty}\frac{1}{n^s} \Gamma(s)= \zeta(s) \Gamma(s).$$
另一个积分公式：
If $Re(s)>1,Re(q)>0$,define\[\zeta(s,q)=\sum_{n=0}^{\infty}\frac1{(q+n)^s}.\],then the function has an integral representation in terms of the Mellin transform as \[\zeta(s,q)=\frac{1}{\Gamma(s)} \int_0^\infty \frac{t^{s-1}e^{-qt}}{1-e^{-t}}dt.\]
Ramanujan相关问题研究
问题一：
\[\int_0^\infty {\frac{{dx}}{{\left( {1 + {x^2}} \right)\left( {1 + {r^2}{x^2}} \right)\left( {1 + {r^4}{x^2}} \right)\left( {1 + {r^6}{x^2}} \right) \cdots }}} = \frac{\pi }{{2\left( {1 + r + {r^3} + {r^6} + {r^{10}} + \cdots } \right)}}.\]
Proof.If we set
$$ f(x)=\prod_{n=0}^{+\infty}(1+r^{2n}x^2) $$
we have:
$$\int_{0}^{+\infty}\frac{dx}{f(x)}=\pi i\sum_{m=0}^{+\infty}\operatorname{Res}\left(f(z),z=\frac{i}{r^m}\right)=\frac{\pi}{2}\sum_{m=0}^{+\infty}\frac{1}{r^m}\prod_{n\neq m}(1-r^{2n-2m})^{-1}\tag{1}$$
but since
$$ \prod_{n=0}^{+\infty}(1-x^n z)^{-1}=\sum_{n=0}^{+\infty}\frac{z^n}{(1-x)\cdot\ldots\cdot(1-x^n)}$$
is one of the Euler's partitions identities, and:
$$\frac{\pi}{2}\sum_{m=0}^{+\infty}\frac{1}{r^m}\prod_{n\neq m}(1-r^{2n-2m})^{-1}=\frac{\pi}{2}\prod_{n=1}^{+\infty}(1-r^{2n})^{-1}\sum_{m=0}^{+\infty}\frac{(1/r)^m}{(1-(1/r^2))\cdot\ldots\cdot(1-(1/r^2)^m)}$$
we have:
$$\int_{0}^{+\infty}\frac{dz}{f(z)}=\frac{\pi}{2}\prod_{n=1}^{+\infty}(1-r^{2n})^{-1}\prod_{m=0}^{+\infty}\left(1-\frac{1}{r^{2m+1}}\right)^{-1}\tag{2}$$
and the claim follows from the Jacobi triple product identity:
$$\sum_{k=-\infty}^{+\infty}s^k q^{\binom{k+1}{2}}=\prod_{m\geq 1}(1-q^m)(1+s q^m)(1+s^{-1}q^{m-1}).$$
问题二：Ramanujan Log-Trigonometric Integrals
\begin{align*}R_n^ - &= \frac{2}{\pi }\int_0^{\frac{\pi }{2}} {{{\left( {{\theta ^2} + {{\ln }^2}\cos \theta } \right)}^{ - 2\left( { - n - 1} \right)}}\sqrt {\frac{1}{2} + \frac{1}{2}\sqrt {\frac{1}{2} + \cdots +\frac{1}{2}\sqrt {\frac{{{{\ln }^2}\cos \theta }}{{{\theta ^2} + {{\ln }^2}\cos \theta }}} } } d\theta } = {\left( {\ln 2} \right)^{ - {2^{ - n}}}}\\R_n^ + &= \frac{2}{\pi }\int_0^{\frac{\pi }{2}} {{{\left( {{\theta ^2} + {{\ln }^2}\cos \theta } \right)}^{2\left( { - n - 1} \right)}}\sqrt {\frac{1}{2} + \frac{1}{2}\sqrt {\frac{1}{2} + \cdots +\frac{1}{2}\sqrt {\frac{{{{\ln }^2}\cos \theta }}{{{\theta ^2} + {{\ln }^2}\cos \theta }}} } } d\theta } = {\left( {\ln 2} \right)^{{2^{ - n}}}}.\end{align*}
问题二：Ramanujan's eccentric Integral formula
\[\int_0^\infty {\frac{{1 + \frac{{{x^2}}}{{{{\left( {b + 1} \right)}^2}}}}}{{1 + \frac{{{x^2}}}{{{a^2}}}}} \times \frac{{1 + \frac{{{x^2}}}{{{{\left( {b + 2} \right)}^2}}}}}{{1 + \frac{{{x^2}}}{{{{\left( {a + 1} \right)}^2}}}}} \times \cdots dx} = \frac{{\sqrt \pi }}{2} \times \frac{{\Gamma \left( {a + \frac{1}{2}} \right)\Gamma \left( {b + 1} \right)\Gamma \left( {b - a + \frac{1}{2}} \right)}}{{\Gamma \left( a \right)\Gamma \left( {b + \frac{1}{2}} \right)\Gamma \left( {b - a + 1} \right)}}.\]
Proof.Andrey Rekalo:This is one of those precious cases when Ramanujan himself provided (a sketch of) a proof. The identity was published in his paper "Some definite integrals" (Mess. Math. 44 (1915), pp. 10-18) together with several related formulae.
It might be instructive to look first at the simpler identity (i.e. the limiting case when $b\to\infty$; the identity mentioned in the original question can be obtained by a similar approach):
$$\int\limits_{0}^{\infty} \prod_{k=0}^{\infty}\frac{1}{ 1 + x^{2}/(a+k)^{2}}dx = \frac{\sqrt{\pi}}{2} \frac{ \Gamma(a+\frac{1}{2})}{\Gamma(a)},\quad a>0.\qquad\qquad\qquad(1)$$
Ramanujan derives (1) by using a partial fraction decomposition of the product $\prod_{k=0}^{n}\frac{1}{ 1 + x^{2}/(a+k)^{2}}$, integrating term-wise, and passing to the limit $n\to\infty$. He also indicates that alternatively (1) is implied by the factorization
$$\prod_{k=0}^{\infty}\left[1+\frac{x^2}{(a+k)^2}\right] = \frac{ [\Gamma(a)]^2}{\Gamma(a+ix)\Gamma(a-ix)},$$
which follows readily from Euler's product formula for the gamma function. Thus (1) is equivalent to the formula
$$\int\limits_{0}^{\infty}\Gamma(a+ix)\Gamma(a-ix)dx=\frac{\sqrt{\pi}}{2} \Gamma(a)\Gamma\left(a+\frac{1}{2}\right).$$
------------------------------------------------------------------
There is a nice paper "Wallis-Ramanujan-Schur-Feynman" by Amdeberhan et al (American Mathematical Monthly 117 (2010), pp. 618-632) that discusses interesting combinatorial aspects of formula (1) and its generalizations.
David Hansen:Here is a proof of Ramanujan's identity (thanks to Todd Trimble for encouraging me to post this!). As Andrey Rekalo notes, we have the identity $$\prod_{k=0}^{\infty}(1+\frac{x^2}{(k+a)^2})=\frac{\Gamma(a)^2}{|\Gamma(a+ix)|^2}$$. In particular, the integrand in Ramanujan's integral is $\frac{\Gamma(b+1)^2 |\Gamma(a+ix)|^2}{\Gamma(a)^2 |\Gamma(b+1+ix)|^2}$. Hence, after a little algebra (and also changing $b$ to $b-1$; I personally think Ramanujan made the wrong aesthetic choice here), we need to prove the integral evaluation $$I=\int_{-\infty}^{\infty} \frac{|\Gamma(a+ix)|^2}{|\Gamma(b+ix)|^2}dx=\sqrt{\pi}\frac{\Gamma(a)\Gamma(a+1/2)\Gamma(b-a-1/2)}{\Gamma(b-1/2)\Gamma(b)\Gamma(b-a)}$$.
Now, if $f(x)$ has Mellin transform $F(s)$, then one form of Parseval's theorem for Mellin transforms is the identity $$\int_{0}^{\infty}f(x)x^{-1}dx=\frac{1}{2\pi}\int_{-\infty}^{\infty}|F(it)|^2 dt$$ (under suitable conditions of course). Applying this with the Mellin pair $$f(x)=\Gamma(b-a)^{-1}x^{a}(1-x)^{b-a-1} \; \mathrm{if} \; 0\leq x \leq 1$$ (and $f=0$ otherwise), $$F(s)=\frac{\Gamma(s+a)}{\Gamma(s+b)}$$ gives
\begin{align*}I&=2\pi \Gamma(b-a)^{-2} \int_{0}^{\infty}x^{2a-1}(1-x)^{2b-2a-2}dx\\&=2\pi \Gamma(b-a)^{-2} \frac{\Gamma(2a) \Gamma(2b-2a-1)}{\Gamma(2b-1)}.\end{align*}
Next, apply the formula $$\Gamma(2z)=2^{2z-1}\pi^{-1/2}\Gamma(z)\Gamma(z+1/2)$$ to each of the $\Gamma$-functions in the quotient here, getting
$$I=\sqrt{\pi} \frac{\Gamma(a)\Gamma(a+1/2)\Gamma(b-a-1/2)\Gamma(b-a)}{\Gamma(b-a)^2 \Gamma(b-1/2) \Gamma(b)}$$, and cancelling a $\Gamma(b-a)$ concludes the proof.
Exercise: Give a proof, along similar lines, of the formula
$$\int_{-\infty}^{\infty} |\Gamma(a+ix)\Gamma(b+ix)|^2 dx=\sqrt{\pi}\frac{\Gamma(a)\Gamma(a+1/2)\Gamma(b)\Gamma(b+1/2)\Gamma(a+b)}{\Gamma(a+b+1/2)}$$,
and determine for what range of $a,b$ it holds.
问题二：Ramanujan's eccentric Integral formula
If $\alpha$ and $\beta$ are positive numbers such that $\alpha\dots \beta=\pi^2$ then,
\[\alpha \cdot \sum\limits_{n = 1}^\infty {\frac{n}{{{e^{2n\alpha }} - 1}}} + \beta \sum\limits_{n = 1}^\infty {\frac{n}{{{e^{2n\beta }} - 1}}} = \frac{{\alpha + \beta }}{{24}} - \frac{1}{4}.\]
Proof.I believe this formula is true, provided the $\alpha$ in the second sum is changed to a $\beta$, as suggested by Todd Trimble's comment. Let
$$ P(x) = \prod_{n=1}^\infty \frac{1}{1-x^n} $$
be the generating function for the number of partitions of a non-negative integer $n$. Dedekind proved that $P$ satisfies the transformation formula
$$ \log P(e^{-2\pi t}) - \log P(e^{-2\pi /t}) = \frac{\pi}{12} \Bigl( \frac{1}{t} - t \Bigr)+ \frac{1}{2} \log t $$
for $t > 0$.
Differentiating this formula with respect to $t$ gives
$$ -\sum_{n=1}^\infty \frac{2\pi n}{e^{2\pi n t}-1} - \frac{1}{t^2} \sum_{n=1}^\infty \frac{2\pi n}{e^{2\pi n/t} -1} = \frac{\pi}{12} \Bigl( -\frac{1}{t^2} - 1\Bigr) + \frac{1}{2t} $$
Now multiply through by $-t/2$ and substitute $\alpha = \pi t$, $\beta = \pi /t$ to get
$$ \sum_{n=1}^\infty \frac{\alpha n}{e^{2n\alpha}-1} + \sum_{n=1}^\infty \frac{\beta n}{e^{2n\beta}-1} = \frac{1}{24}(\beta+\alpha) - \frac{1}{4}$$
which is Ramanujan's formula.
The transformation formula for $P$ is related to the theory of modular forms, of which
the Eisenstein series mentioned in Derek Jennings' answer to your question on math.stackexchange are important examples. Briefly, if we define
$$ \eta(\tau) = \frac{e^{2\pi i \tau/24}}{P(e^{2\pi i \tau})} = e^{2\pi i \tau/24} \prod_{n=1}^\infty (1-e^{2\pi i n \tau}), $$
then $\eta(\tau)^{24}$ is a modular form of weight $12$. As such, $\eta$ satisfies the identity
$$ \eta(-1/\tau) = \sqrt{-i \tau}\; \eta(\tau). $$
The transformation formula for $P$ follows by setting $\tau = it$ and taking logs.
Euler Sum的若干研究
本文主要展示并分享有关Euler Sum的若干求解问题。
问题一：求\[\sum_{n=1}^\infty{\frac{H_n}{n^q}}\]
Solution.
\begin{align}&\sum_{j=0}^k\zeta(k+2-j)\zeta(j+2)\\&=\sum_{m=1}^\infty\sum_{n=1}^\infty\sum_{j=0}^k\frac1{m^{k+2-j}n^{j+2}}\tag{1}\\&=(k+1)\zeta(k+4)+\sum_{\substack{m,n=1\\m\ne n}}^\infty\frac1{m^2n^2}\frac{\frac1{m^{k+1}}-\frac1{n^{k+1}}}{\frac1m-\frac1n}\tag{2}\\&=(k+1)\zeta(k+4)+\sum_{\substack{m,n=1\\m\ne n}}^\infty\frac1{nm^{k+2}(n-m)}-\frac1{mn^{k+2}(n-m)}\tag{3}\\&=(k+1)\zeta(k+4)+2\sum_{m=1}^\infty\sum_{n=m+1}^\infty\frac1{nm^{k+2}(n-m)}-\frac1{mn^{k+2}(n-m)}\tag{4}\\&=(k+1)\zeta(k+4)+2\sum_{m=1}^\infty\sum_{n=1}^\infty\frac1{(n+m)m^{k+2}n}-\frac1{m(n+m)^{k+2}n}\tag{5}\\&=(k+1)\zeta(k+4)\\&+2\sum_{m=1}^\infty\sum_{n=1}^\infty\frac1{m^{k+3}n}-\frac1{(m+n)m^{k+3}}\\&-2\sum_{m=1}^\infty\sum_{n=1}^\infty\frac1{m(n+m)^{k+3}}+\frac1{n(n+m)^{k+3}}\tag{6}\\&=(k+1)\zeta(k+4)+2\sum_{m=1}^\infty\frac{H_m}{m^{k+3}}-4\sum_{n=1}^\infty\sum_{m=1}^\infty\frac1{n(n+m)^{k+3}}\tag{7}\\&=(k+1)\zeta(k+4)+2\sum_{m=1}^\infty\frac{H_m}{m^{k+3}}-4\sum_{n=1}^\infty\sum_{m=n+1}^\infty\frac1{nm^{k+3}}\tag{8}\\&=(k+1)\zeta(k+4)+2\sum_{m=1}^\infty\frac{H_m}{m^{k+3}}-4\sum_{n=1}^\infty\sum_{m=n}^\infty\frac1{nm^{k+3}}+4\zeta(k+4)\tag{9}\\&=(k+5)\zeta(k+4)+2\sum_{m=1}^\infty\frac{H_m}{m^{k+3}}-4\sum_{m=1}^\infty\sum_{n=1}^m\frac1{nm^{k+3}}\tag{10}\\&=(k+5)\zeta(k+4)+2\sum_{m=1}^\infty\frac{H_m}{m^{k+3}}-4\sum_{m=1}^\infty\frac{H_m}{m^{k+3}}\tag{11}\\&=(k+5)\zeta(k+4)-2\sum_{m=1}^\infty\frac{H_m}{m^{k+3}}\tag{12}\end{align}
Letting $q=k+3$ and reindexing $j\mapsto j-1$ yields
$$\sum_{j=1}^{q-2}\zeta(q-j)\zeta(j+1)=(q+2)\zeta(q+1)-2\sum_{m=1}^\infty\frac{H_m}{m^q}\tag{13}$$
and finally
$$\sum_{m=1}^\infty\frac{H_m}{m^q}=\frac{q+2}{2}\zeta(q+1)-\frac12\sum_{j=1}^{q-2}\zeta(q-j)\zeta(j+1)\tag{14}$$
Explanation
$\hphantom{0}(1)$ expand $\zeta$
$\hphantom{0}(2)$ pull out the terms for $m=n$ and use the formula for finite geometric sums on the rest
$\hphantom{0}(3)$ simplify terms
$\hphantom{0}(4)$ utilize the symmetry of $\frac1{nm^{k+2}(n-m)}+\frac1{mn^{k+2}(m-n)}$
$\hphantom{0}(5)$ $n\mapsto n+m$ and change the order of summation
$\hphantom{0}(6)$ $\frac1{mn}=\frac1{m(m+n)}+\frac1{n(m+n)}$
$\hphantom{0}(7)$ $H_m=\sum_{n=1}^\infty\frac1n-\frac1{n+m}$ and use the symmetry of $\frac1{m(n+m)^{k+3}}+\frac1{n(n+m)^{k+3}}$
$\hphantom{0}(8)$ $m\mapsto m-n$
$\hphantom{0}(9)$ subtract and add the terms for $m=n$
$(10)$ combine $\zeta(k+4)$ and change the order of summation
$(11)$ $H_m=\sum_{n=1}^m\frac1n$
$(12)$ combine sums
参考文献
[1]
[2]
