Eufisky - The lost book

## 有关反三角和三角函数组合的虐心积分求解

\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*}

# 最近经常在群里见到一些棘手的角度求解问题，俺到今天才知道原来这类问题都属于三角形中的角格点问题，以前一直不专业地称为角度求解问题，此外，田永海老师近来还出了一本专著三角形中的角格点问题。下面汇集了自己在群里对某几个此类问题的解答。关于此类问题求解的方法，还可参阅：

[1]三角形的格点

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## 一些亟待解决的问题

\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*}

\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*}

## 涉及到一个有意思的方程的积分

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

## Euler积分的一个延伸

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

\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*}

\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*}

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

\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*}

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

$$\frac{\partial }{\partial x} \beta(x,y) = 4\int_0^{\frac{\pi}{2}} \sin^{2x-1}(t)\ln(\sin t) \cos^{2y-1}(t) \ dt$$
$$\frac{\partial}{\partial y} \left( \frac{\partial}{\partial x} \beta(x,y) \right) = 8\int_0^{\frac{\pi}{2}} \sin^{2x-1}(t) \ln(\sin t ) \ln(\cos t) \cos^{2y-1}(t) \ dt$$
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 $= \frac{\pi}{8} \left(4\ln(2)^2 - \frac{\pi^2}{6} \right)$

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.

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).

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

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

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.

## 来自西哥的一道解几题

(1)求$\triangle ABF_2$内心的轨迹；

(2)当$\triangle ABF_2$内切圆面积最大时，求直线$AB$的方程.
(1)根据题意有$F_1(-3,0),F_2(3,0)$，不妨设$A(x_1,y_1),B(x_2,y_2)$，且
$\left\{ \begin{array}{l}{x_1} = 5\cos \alpha \\{y_1} = 4\sin \alpha\end{array} \right.\text{及}\left\{ \begin{array}{l}{x_2} = 5\cos \beta \\{y_2} = 4\sin \beta\end{array} \right.$

$\frac{{4\sin \beta - 4\sin \alpha }}{{5\cos \beta - 5\cos \alpha }} = \frac{{4\sin \beta }}{{5\cos \beta + 3}} \Leftrightarrow 3\left( {\sin \beta - \sin \alpha } \right) = 5\sin \left( {\alpha - \beta } \right).$
$5\cos \frac{{\alpha - \beta }}{2} = - 3\cos \frac{{\alpha + \beta }}{2}\Leftrightarrow \tan \frac{\alpha }{2}\tan \frac{\beta }{2} = - 4.$

\begin{align*}\left| {A{F_2}} \right| &= 5 - 3\cos \alpha \\\left| {B{F_2}} \right| &= 5 - 3\cos \beta \\\left| {AB} \right| &= \left| {A{F_1}} \right| + \left| {A{F_1}} \right| = 3\left( {\cos \beta  + \cos \alpha }\right) + 10.\end{align*}

$\left\{ \begin{array}{l}x = \frac{{\left| {B{F_2}} \right|{x_1} + \left| {A{F_2}} \right|{x_2} + 3\left| {AB} \right|}}{{20}}\\y = \frac{{\left| {B{F_2}} \right|{y_1} + \left| {A{F_2}} \right|{y_2}}}{{20}}\end{array} \right.$

$\left\{ \begin{array}{l}x = \frac{{34\left( {\cos \beta + \cos \alpha } \right) - 30\cos \alpha \cos \beta + 30}}{{20}}\\y = \frac{{5\left( {\sin \beta + \sin \alpha } \right) - 3\sin \left( {\alpha + \beta } \right)}}{5}\end{array} \right.$

\begin{align*}\sin \beta  + \sin \alpha  &= 2\sin \frac{{\alpha  + \beta }}{2}\cos \frac{{\alpha  - \beta }}{2} =  - \frac{6}{5}\sin \frac{{\alpha  + \beta }}{2}\cos \frac{{\alpha  + \beta }}{2}\\&=  - \frac{3}{5}\sin \left( {\alpha  + \beta } \right)\\\cos \beta  + \cos \alpha  &= 2\cos \frac{{\alpha  + \beta }}{2}\cos \frac{{\alpha  - \beta }}{2} =  - \frac{6}{5}{\cos ^2}\frac{{\alpha  + \beta }}{2}\\&=  - \frac{3}{5}\cos \left( {\alpha  + \beta } \right) - \frac{3}{5}\\\cos \alpha \cos \beta  &= \frac{1}{2}\cos \left( {\alpha  + \beta } \right) + \frac{1}{2}\cos \left( {\alpha  - \beta } \right)\\&= \frac{1}{2}\cos \left( {\alpha  + \beta } \right) + \frac{1}{2}\left( {2{{\cos }^2}\frac{{\alpha  - \beta }}{2} - 1} \right)\\&= \frac{1}{2}\cos \left( {\alpha  + \beta } \right) - \frac{1}{2} + \frac{9}{{25}}{\cos ^2}\frac{{\alpha  + \beta }}{2}\\&= \frac{1}{2}\cos \left( {\alpha  + \beta } \right) - \frac{1}{2} + \frac{9}{{25}}\frac{{\cos \left( {\alpha  + \beta } \right) + 1}}{2}\\&= \frac{{17}}{{25}}\cos \left( {\alpha  + \beta } \right) - \frac{8}{{25}}.\end{align*}

$\left\{ \begin{array}{l}x = - \frac{{51}}{{25}}\cos \left( {\alpha + \beta } \right) + \frac{{24}}{{25}}\\y = - \frac{6}{5}\sin \left( {\alpha + \beta } \right)\end{array} \right. \Rightarrow {\left( {\frac{{25x - 24}}{{51}}} \right)^2} + \frac{{25}}{{36}}{y^2} = 1\left( {x = \frac{{27}}{{25}},y = 0} \right).$
(2)由题意
$S = \frac{1}{2} \times 20r = \frac{1}{2} \times \left| {{F_1}{F_2}} \right| \times 4\left| {\sin \beta - \sin \alpha } \right| = \frac{1}{2} \times 24\left| {\sin \beta - \sin \alpha } \right|.$
\begin{align*}\Rightarrow r &= \frac{6}{5}\left| {\sin \beta  - \sin \alpha } \right| = 2\left| {\sin \left( {\alpha  - \beta } \right)} \right|\\&= 4\left| {\frac{{\sin \frac{{\alpha  - \beta }}{2}\cos \frac{{\alpha  - \beta }}{2}}}{{{{\sin }^2}\frac{{\alpha  -\beta }}{2} + {{\cos }^2}\frac{{\alpha  - \beta }}{2}}}} \right|\\&= 4\left| {\frac{{\tan \frac{{\alpha  - \beta }}{2}}}{{{{\tan }^2}\frac{{\alpha  - \beta }}{2} + 1}}} \right|.\end{align*}

$r = \frac{4}{{\left| {t + \frac{1}{t}} \right|}} = \frac{4}{{\left| t \right| + \frac{1}{{\left| t \right|}}}} \le \frac{4}{{\frac{4}{3} + \frac{3}{4}}} = \frac{{48}}{{25}}.$

$\left\{ \begin{array}{l}{x_1} = 5\frac{{1 - {{\tan }^2}\frac{\alpha }{2}}}{{{{\tan }^2}\frac{\alpha }{2} + 1}} = - 3\\{y_1} = 4\frac{{2\tan \frac{\alpha }{2}}}{{{{\tan }^2}\frac{\alpha }{2} + 1}} = \frac{{16}}{5}\end{array} \right.,\left\{ \begin{array}{l}{x_2} = - 3\\{y_2} = - \frac{{16}}{5}\end{array} \right.\text{或}\left\{ \begin{array}{l}{x_1} = 5\frac{{1 - {{\tan }^2}\frac{\alpha }{2}}}{{{{\tan }^2}\frac{\alpha }{2} + 1}} = - 3\\{y_1} = 4\frac{{2\tan \frac{\alpha }{2}}}{{{{\tan }^2}\frac{\alpha }{2} + 1}} = - \frac{{16}}{5}\end{array} \right.,\left\{ \begin{array}{l}{x_2} = - 3\\{y_2} = \frac{{16}}{5}\end{array} \right.$

则$$x_{1}+x_{2}=-\dfrac{2a^2ck^2}{b^2+a^2k^2},x_{1}x_{2}=\dfrac{a^2c^2k^2-a^2b^2}{b^2+a^2k^2}$$

因为$$|AF_{1}|=a+ex_{1},|AF_{2}|=a-ex_{1},|BF_{1}|=a+ex_{2},|BF_{2}|=a-ex_{2}$$

\begin{align*}y&=\dfrac{(a-ex_{2})y_{1}+(a-ex_{1})y_{2}}{4a}=\dfrac{a(y_{1}+y_{2})-e(x_{1}y_{2}+x_{2}y_{1})}{4a}\\&=\dfrac{a\cdot\dfrac{2b^2ck}{b^2+a^2k^2}-\dfrac{c}{a}\left(-\dfrac{2a^2b^2k}{b^2+a^2k^2}\right)}{4a}\\&=\dfrac{b^2ck}{b^2+a^2k^2}\end{align*}

\begin{align*}y^2&=\dfrac{b^4c^2k^2}{(b^2+a^2k^2)^2}=\dfrac{b^2c^2\cdot\dfrac{b^2c-b^2x}{a^2x+c^3}}{\left(b^2+a^2\cdot\dfrac{b^2c-b^2x}{a^2x+c^3}\right)^2}=\dfrac{\dfrac{b^6c^2(c-x)}{a^2x+c^3}}{\left[\dfrac{(a^2x+c^3)b^2+a^2(b^2c-b^2x)}{a^2x+c^3}\right]^2}\\&=\dfrac{\dfrac{b^6c^2(c-x)}{a^2x+c^3}}{\left[\dfrac{b^2c(a^2+c^2)}{a^2x+c^3}\right]^2}=\dfrac{b^2(c-x)(a^2x+c^3)}{(a^2+c^2)^2}\end{align*}

$$(a^2+c^2)^2y^2=(b^2c-b^2x)(a^2x+c^3)=b^2(cb^2x+c^4-a^2x)$$
$$a^2b^2x^2+(a^2+c^2)^2y^2-cb^4x-b^2c^4=0$$

(2)设直线$AB$方程为$x=ky-3$与椭圆联立得$$(16k^2+25)y^2-96ky-256=0$$

\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*}

$\sqrt 2 = \sqrt {\frac{2}{{{2^{{2^0}}}}} + \sqrt {\frac{2}{{{2^{{2^1}}}}} + \sqrt {\frac{2}{{{2^{{2^2}}}}} + \sqrt {\frac{2}{{{2^{{2^3}}}}} + \sqrt {\frac{2}{{{2^{{2^4}}}}} + \cdots } } } } } \left( {q = \frac{1}{2},n = 2,x = 1,k = - 1} \right).$

$\left\{ \begin{array}{l}1 + \frac{1}{n} + \frac{1}{{{n^2}}} + \cdots = 1 + \frac{1}{{n - 1}}\\\frac{1}{n} + \frac{1}{{{n^2}}} + \frac{1}{{{n^3}}} + \cdots = \frac{1}{{n - 1}}\\\frac{1}{n}\left( {1 + \frac{1}{n}\left( {1 + \frac{1}{n}\left( {1 + \cdots } \right)} \right)} \right) = \frac{1}{{n - 1}}\end{array} \right.\left( {n \ge 2,n \in {N_ + }} \right).$

$x + n + a = \sqrt {ax + {{\left( {n + a} \right)}^2} + x\sqrt {a\left( {x + n} \right) + {{\left( {n + a} \right)}^2} + \left( {x + n} \right)\sqrt {a\left( {x + 2n} \right) + {{\left( {n + a} \right)}^2} + \left( {x + 2n} \right)\sqrt \cdots }}}.$

$3 = \sqrt {1 + 2\sqrt {1 + 3\sqrt {1 + 4\sqrt {1 + 5\sqrt \cdots } } } } \left( {a = 0,n = 1,x = 2} \right).$
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).

$\left\{ \begin{array}{l}e = 1 + \frac{1}{{1!}} + \frac{1}{{2!}} + \frac{1}{{3!}} + \cdots \\e = 1 + 1 + \frac{1}{2}\left( {1 + \frac{1}{3}\left( {1 + \frac{1}{4}\left( {1 + \frac{1}{5}\left( {1 + \cdots } \right)} \right)}\right)} \right)\end{array} \right.$

${x^{e - 2}} = \sqrt {x\sqrt[3]{{x\sqrt[4]{{x\sqrt[5]{{x \cdots }}}}}}}.$

## $\sum{\frac{n}{{{e^{2\pi n}} - 1}}}$型的级数求解

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

$\frac{z}{{{e^z} - 1}} = \sum\limits_{k = 0}^\infty {\frac{{{B_k}}}{{k!}}} {z^k}\;\;\; for |z|<2\pi .$

$$\displaystyle\mathcal{M}\Big[\frac{x}{e^{2\pi x}-1}] = \int_{0}^{\infty} \frac{x^{s}}{e^{2 \pi x}-1} \ dx = (2\pi)^{-(s+1)}\Gamma(s+1)\zeta(s+1)$$

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,

$$\int_{0}^{\infty}\frac{u^{s-1}}{e^u - 1}=\int_{0}^{\infty}\frac{u^{s-1}}{e^u}(1-e^{-u})^{-1}= \sum_{n=0}^{\infty} \int_{0}^{\infty}{u^{s-1}e^{-(n+1)u}}$$

$$= \sum_{n=0}^{\infty}\frac{1}{(n+1)^s} \int_{0}^{\infty}{y^{s-1}e^{-y}}= \sum_{n=1}^{\infty}\frac{1}{n^s} \Gamma(s)= \zeta(s) \Gamma(s).$$

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

$\int_0^\infty {\frac{{dx}}{{\left( {1 + {x^2}} \right)\left( {1 + {r^2}{x^2}} \right)\left( {1 + {r^4}{x^2}} \right)\left( {1 + {r^6}{x^2}} \right) \cdots }}} = \frac{\pi }{{2\left( {1 + r + {r^3} + {r^6} + {r^{10}} + \cdots } \right)}}.$

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:
$$\sum_{k=-\infty}^{+\infty}s^k q^{\binom{k+1}{2}}=\prod_{m\geq 1}(1-q^m)(1+s q^m)(1+s^{-1}q^{m-1}).$$

\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*}

$\int_0^\infty {\frac{{1 + \frac{{{x^2}}}{{{{\left( {b + 1} \right)}^2}}}}}{{1 + \frac{{{x^2}}}{{{a^2}}}}} \times \frac{{1 + \frac{{{x^2}}}{{{{\left( {b + 2} \right)}^2}}}}}{{1 + \frac{{{x^2}}}{{{{\left( {a + 1} \right)}^2}}}}} \times \cdots dx} = \frac{{\sqrt \pi }}{2} \times \frac{{\Gamma \left( {a + \frac{1}{2}} \right)\Gamma \left( {b + 1} \right)\Gamma \left( {b - a + \frac{1}{2}} \right)}}{{\Gamma \left( a \right)\Gamma \left( {b + \frac{1}{2}} \right)\Gamma \left( {b - a + 1} \right)}}.$
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
$$\int\limits_{0}^{\infty}\Gamma(a+ix)\Gamma(a-ix)dx=\frac{\sqrt{\pi}}{2} \Gamma(a)\Gamma\left(a+\frac{1}{2}\right).$$

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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.

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.