Eufisky - The lost book

## 几个恒等式证明

1.证明:

$\sum_{j=1}^{n-1}{\frac{1}{1-\exp \left\{ \frac{2\pi ij}{n} \right\}}}=\frac{n-1}{2}.$

2.设$N$为自然数, $\{x\}$表示$x$的小数部分.证明$\sum_{n=1}^N{\left( \left\{ x+\frac{n}{N} \right\} -\frac{1}{2} \right)}=\left\{ Nx \right\} -\frac{1}{2}.$

$\int_0^1{\left( \left\{ ax \right\} -\frac{1}{2} \right) \left( \left\{ bx \right\} -\frac{1}{2} \right) dx}=\frac{1}{12ab}.$

3.求极限
$\lim_{m\rightarrow \infty}\lim_{n\rightarrow \infty}\int_0^1{\int_0^1{\cdots \int_0^1{\sum_{i=0}^m{\exp \left\{ -\frac{n}{\sum_{k=1}^n{x_{k}^{m-1}}} \right\} \frac{\prod_{j=1}^i{\sum_{k=1}^n{x_{k}^{j-1}}}}{\left( \sum_{k=1}^n{x_{k}^{m-1}} \right) ^i}}dx_1dx_2\cdots dx_n}}}.$
4.证明
\begin{align*}f\left( x \right) &=\frac{1}{a}+\frac{x}{a\left( a+d \right)}+\cdots +\frac{x^n}{a\left( a+d \right) \cdots \left( a+nd \right)}+\cdots\\&=\frac{e^{x/d}}{dx^{a/d}}\int_0^x{e^{-t/d}t^{a/d-1}dt}.\end{align*}

5.证明Ramanujan的恒等式
$\int_0^{\infty}{e^{-3\pi x^2}\frac{\sin\text{h}\pi x}{\sin\text{h}3\pi x}dx}=\frac{1}{\sqrt{3}e^{2\pi /3}}\sum_{n=0}^{\infty}{\frac{e^{-2n\left( n+1 \right) \pi}}{\left( 1+e^{-\pi} \right) ^2\left( 1+e^{-3\pi} \right) ^2\cdots \left( 1+e^{-\left( 2n+1 \right) \pi} \right) ^2}}.$

6.Compute
$$\sum_{n=1}^{\infty}{\left( \frac{n^n}{n!e^n}-\frac{1}{\sqrt{2\pi n}} \right)}.$$
By the software Mathematica, I find
$$\sum_{n=1}^{\infty}{\left( \frac{n^n}{n!e^n}-\frac{1}{\sqrt{2\pi n}} \right)}=-\frac{2}{3}-\frac{\zeta \left( 1/2 \right)}{\sqrt{2\pi}}.$$

Well, $-\frac{1}{\sqrt{2\pi}}\zeta\left(\tfrac{1}{2}\right)$ is the $\zeta$-regularization of the divergent series $\sum_{n\geq 1}\frac{1}{\sqrt{2\pi n}}$, hence the problem boils down to finding the $\zeta$-regularization of the divergent series $\sum_{n\geq 1}\frac{n^n}{n!e^n}$. As pointed out in the comments,

$$W(x) = \sum_{n\geq 1}\frac{n^{n-1}(-1)^{n-1}}{n!}x^n$$
holds for any $x\in\left(-\frac{1}{e},\frac{1}{e}\right)$ by Lagrange inversion theorem, hence
$$ze^{-z} W'(-ze^{-z})=\sum_{n\geq 1}\frac{n^n}{n!e^{nz}}z^{n}=\frac{z}{1-z} =\sum_{n\geq 1}z^n\tag{1}$$
holds for any $z\in(-W(e^{-1}),1)$. Pretty strange identity, I can give you that.<br>
Similarly, over the same interval
$$-W(-z e^{-z})=\sum_{n\geq 1}\frac{n^{n-1}}{n!e^{nz}}z^n = z \tag{2}$$
$$1=\sum_{n\geq 1}\frac{n^{n}}{n!e^{nz}}z^{n-1}-\sum_{n\geq 1}\frac{n^{n}}{n!e^{nz}}z^{n}=\frac{1}{1-z}-\frac{z}{1-z}.\tag{3}$$
Since $\zeta(0)=-\frac{1}{2}$, it should not be difficult to prove from $(1)$ and $(2)$ that the $\zeta$-regularization of $\sum_{n\geq 1}\frac{n^n}{n!e^n}$ equals $-\frac{2}{3}$ as wanted, for instance by computing $\sum_{n\geq 1}\frac{n^{n-1-k}}{n!e^n}$ for any $k\in\mathbb{N}$:
$$\sum_{n\geq 1}\frac{n^{n-2}}{n!e^n}=\int_{-1/e}^{1}\frac{W(x)}{x}\,dx = \frac{1}{2},\qquad \sum_{n\geq 1}\frac{n^{n-3}}{n!e^n}=-\int_{-1/e}^{1}\frac{W(x)}{x}(1+\log(-x))\,dx=\frac{5}{12}$$
$$\sum_{n\geq 1}\frac{n^{n-4}}{n!e^n}=\frac{7}{18},\qquad \sum_{n\geq 1}\frac{n^{n-4}}{n!e^n}=\frac{1631}{4320},$$
$$\sum_{n\geq 1}\frac{n^{n-1-k}}{n!e^n}= \frac{1}{\Gamma(k)}\int_{0}^{1}(1-x)(x-1-\log x)^{k-1}\,dx.\tag{4}$$
Indeed the substitution $x=e^{-s}$ in $(4)$ and the integral representation for the $\zeta$ function complete the proof.
Taking $$F\left(x\right)=\sum_{n\geq1}\frac{n^{n-1}}{n!e^{n}}x^{n}-\frac{1}{\sqrt{2\pi}}\sum_{n\geq1}\frac{x^{n}}{n^{3/2}}=-W\left(-\frac{x}{e}\right)-\frac{\mathrm{Li}_{3/2}\left(x\right)}{\sqrt{2\pi}},\,\left|x\right|<1$$ where $W\left(x\right)$ is the [Lambert $W$ function][1] and $\mathrm{Li}_{3/2}\left(x\right)$ is the [Polylogarithm function][2], we obtain, differentiating both sides,that $$\sum_{n\geq1}\left(\frac{n^{n}}{n!e^{n}}-\frac{1}{\sqrt{2\pi n}}\right)x^{n-1}=-\frac{W\left(-\frac{x}{e}\right)}{x\left(W\left(-\frac{x}{e}\right)+1\right)}-\frac{\mathrm{Li}_{1/2}\left(x\right)}{x\sqrt{2\pi}}$$ so $$\sum_{n\geq1}\left(\frac{n^{n}}{n!e^{n}}-\frac{1}{\sqrt{2\pi n}}\right)=\lim_{x\rightarrow1^{-}}\left(-\frac{W\left(-\frac{x}{e}\right)}{x\left(W\left(-\frac{x}{e}\right)+1\right)}-\frac{\mathrm{Li}_{1/2}\left(x\right)}{x\sqrt{2\pi}}\right).$$ Now, [we know][3] that $$\mathrm{Li}_{v}\left(z\right)=\left(\Gamma\left(1-v\right)\left(1-z\right)^{v-1}+\zeta\left(v\right)\right)\left(1+O\left(\left|1-z\right|\right)\right),v\neq1,\,z\rightarrow1$$  and now we claim $$-\frac{W\left(-\frac{x}{e}\right)}{x\left(W\left(-\frac{x}{e}\right)+1\right)}\sim\frac{1}{\sqrt{2\left(1-x\right)}}-\frac{2}{3}$$ as $x\rightarrow1^{-}$. This is true because, [since][5] $$W\left(z\right)\sim-1+\sqrt{2ze+2}-\frac{2}{3}e\left(z+\frac{1}{e}\right)$$ as $z\rightarrow-1/e$, we have $$-\frac{W\left(-\frac{x}{e}\right)}{x\left(W\left(-\frac{x}{e}\right)+1\right)}\sim\frac{1-\sqrt{2\left(1-x\right)}+\frac{2}{3}\left(1-x\right)}{x\sqrt{2\left(1-x\right)}-\frac{2}{3}\left(1-x\right)x}$$ $$=\frac{1}{x}\left(-1+\frac{1}{\sqrt{2\left(1-x\right)}}\left(\frac{1}{1-\sqrt{2-2x}/3}\right)\right)=\frac{1}{x}\left(-1+\frac{1}{\sqrt{2\left(1-x\right)}}\sum_{k\geq0}\left(\frac{\sqrt{2-2x}}{3}\right)^{k}\right)$$ $$=\frac{1}{x}\left(-\frac{2}{3}+\frac{1}{\sqrt{2\left(1-x\right)}}+O\left(\sqrt{1-x}\right)\right)$$ then the claim.

[1]:https://en.wikipedia.org/wiki/Lambert_W_function
[2]:https://en.wikipedia.org/wiki/Polylogarithm
[3]:http://functions.wolfram.com/ZetaFunctionsandPolylogarithms/PolyLog/06/01/02/01/01/
[4]:http://mathworld.wolfram.com/StirlingsSeries.html
[5]:http://functions.wolfram.com/ElementaryFunctions/ProductLog/06/01/02/
This is a general answer to the followup question by Jack D'Aurizio.

**Proposition**
>Let $n\in\mathbb{N}$. We have the asymptotic expansion
$$n!\sim \sqrt{2\pi n} \frac{n^n}{e^n} \left[ 1+ \frac1{12n} +\frac1{288n^2}-\frac{139}{51840n^3}-\frac{157}{2488320n^4}+\cdots \right]$$

Note that this is not a convergent series, but an asymptotic expansion. The error in the truncated series is asymptotically equal to the first omitted term. Regard the series on the right as an element of the ring of power series over rational numbers $\mathbb{Q}[[T]]$.
$$S(T)=1+ \frac1{12}T+\frac1{288}T^2-\frac{139}{51840}T^3-\frac{157}{2488320}T^4 + \cdots.$$
Consider the multiplicative inverse of $S(T)$ in $\mathbb{Q}[[T]]$.
$$S^{-1}(T)=1-\frac1{12}T+ g_2 T^2 + g_3 T^3 + g_4 T^4 + \cdots.$$
Let $Y_s(T)=\sum_{n=0}^{\infty} h_n(s) T^n \in\mathbb{Q}[s][[T]]$ be defined by
$$\left(\frac12 T^2\right)^{s-1}\sum_{n=0}^{\infty} h_n(s) T^n = \left[ \frac12 T^2 + \frac13 T^3 + \frac14 T^4+\cdots \right]^{s-1}.$$
Then we have

**Theorem**
>$$\sum_{n=1}^{\infty} n^p\left[ \frac{n^n}{n!e^n}- \frac1{\sqrt{2\pi n}} \sum_{k=0}^p \frac{g_k}{n^k}\right]=(-2)^p p!h_{2p+1}(-p) - \frac1{\sqrt{2\pi}}\sum_{k=0}^p g_k \zeta\left(k+\frac12-p\right).$$

With $p=0$, it is the original series
$$\sum_{n=1}^{\infty} \left[\frac{n^n}{n!e^n} - \frac1{\sqrt{2\pi n}}\right]=-\frac23 - \frac{\zeta\left(\frac12\right)}{\sqrt{2\pi}}$$

With $p=1$, it gives the value of
$$\sum_{n=1}^{\infty} \left[ \frac{n^{n+1}}{n!e^n} - \sqrt{\frac{n}{2\pi}} + \frac{1}{12\sqrt{2\pi n}}\right] = -\frac 4{135} - \frac{\zeta\left(-\frac12\right)}{\sqrt{2\pi}} + \frac{\zeta\left(\frac12\right)}{12\sqrt{2\pi}}.$$

7.Prove that $$\sum_{m\leqslant x}\sum_{n\leqslant x}\Big\{\frac{x}{m+n}\Big\}=\Big(2\log2-\frac{\pi^2}{12}\Big)x^2+O(x\log x),$$

where $\{x\}$ is the fractional part of the real number $x$.

I know
$$\sum_{n\leqslant x} \Big\{\frac{x}{n} \Big\}=(1-\gamma)x+O\big(x^{1/2}\big),$$

where $\gamma$ is Euler's constant. But I don't know whether it is useful. Can you help me?

Write $\{x\} = x - \lfloor x\rfloor$,s so that
$\sum_{m \leq x} \sum_{n \leq x} \left\{ \frac{x}{m + n}\right\} = \sum_{m \leq x} \sum_{n \leq x} \frac{x}{m + n} - \sum_{m \leq x} \sum_{n \leq x} \left\lfloor \frac{x}{m + n}\right\rfloor.$
Then note that
$\sum_{m \leq x} \sum_{n \leq x} \left\lfloor \frac{x}{m + n}\right\rfloor = \sum_{m \leq x} \sum_{n \leq x} \sum_{\ell \leq \frac{x}{m + n}} 1 = \sum_{\ell \leq \frac{x}{2}} \sum_{n \leq \frac{x}{\ell} - 1} \sum_{m \leq \frac{x}{\ell} - n} 1.$
The sum over $m$ is $\lfloor x/\ell\rfloor - n$. The ensuing sum over $n$ is
$\left\lfloor \frac{x}{\ell}\right\rfloor \left(\left\lfloor \frac{x}{\ell}\right\rfloor - 1\right) - \sum_{n \leq \frac{x}{\ell} - 1} n.$
Via partial summation,
$\sum_{n \leq \frac{x}{\ell} - 1} n = \left(\frac{x}{\ell} - 1\right) \left(\left\lfloor \frac{x}{\ell} \right\rfloor - 1\right) - \int_{1}^{\frac{x}{\ell} - 1} \lfloor t\rfloor \\, dt,$
and this integral is equal to $\frac{1}{2} \left(\frac{x}{\ell} - 1\right)^2 + O(\frac{x}{\ell})$. So the sum over $n$ and $m$ simplifies to
$\frac{x^2}{2 \ell^2} + O\left(\frac{x}{\ell}\right).$
The ensuing sum over $\ell$ is
$\frac{x^2}{2} \sum_{\ell = 1}^{\infty} \frac{1}{\ell^2} - \frac{x^2}{2} \sum_{\ell > \frac{x}{2}} \frac{1}{\ell^2} + O(x \log x).$
The first sum over $\ell$ is $\zeta(2) = \pi^2/6$. The second is $O(1/x)$. So this simplifies to
$\frac{\pi^2 x^2}{12} + O(x \log x).$

Now we deal with
$\sum_{m \leq x} \sum_{n \leq x} \frac{x}{m + n} = x \sum_{m \leq x} \sum_{n \leq x} \frac{1}{m + n}.$
We deal with the sum over $n$ via partial summation: it is equal to
$\frac{\lfloor x\rfloor}{m + x} + \int_{1}^{x} \frac{\lfloor t \rfloor}{(m + t)^2} \\, dt = \log \frac{m + x}{m + 1} + O\left(\frac{1}{m}\right),$
where we have used the fact that $\lfloor x \rfloor = x - \{x\} = x + O(1)$, the fact that the antiderivative of $t/(m + t)^2$ is $m/(m + t) + \log(m + t)$, and the fact that the antiderivative of $1/(m + t)^2$ is $-1/(m + t)$.

So it remains to evaluate
$\sum_{m \leq x} \log \frac{m + x}{m + 1} = \lfloor x\rfloor \log \frac{2x}{x + 1} + (x + 1) \int_{1}^{x} \frac{\lfloor t\rfloor}{(t + x)(t + 1)} \\, dt.$
The antiderivative of $\frac{t}{(t + x)(t + 1)}$ is $\frac{x}{x - 1} \log \frac{t + x}{t + 1}$, and so after some simplification, we arrive at $(2\log 2) x + O(\log x)$.

## 几个重要定理

1.Mittag-Leffler's theorem.

\begin{align*}\frac{1}{\sin \left( z \right)}&=\sum_{n\in \mathbb{Z}}{\frac{\left( -1 \right) ^n}{z-n\pi}}=\frac{1}{z}+2z\sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^n}{z^2-\left( n\,\pi \right) ^2}},\\\cot \left( z \right) &\equiv \frac{\cos \left( z \right)}{\sin \left( z \right)}=\sum_{n\in \mathbb{Z}}{\frac{1}{z-n\pi}}=\frac{1}{z}+2z\sum_{k=1}^{\infty}{\frac{1}{z^2-\left( k\,\pi \right) ^2}},\\\frac{1}{\sin ^2\left( z \right)}&=\sum_{n\in \mathbb{Z}}{\frac{1}{\left( z-n\,\pi \right) ^2}},\\\frac{1}{z\sin \left( z \right)}&=\frac{1}{z^2}+\sum_{n\ne 0}{\frac{\left( -1 \right) ^n}{\pi n\left( z-\pi n \right)}}=\frac{1}{z^2}+\sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^n}{n\,\pi}}\frac{2z}{z^2-\left( n\,\pi \right) ^2}.\end{align*}
2.Ramanujan's Master Theorem

Ramanujan's Proof
This proof was given by none other than Ramanujan.
Recall Euler's integral representation of the Gamma Function -
$$\int_0^\infty e^{-mx}x^{n-1}dx = m^{-n}\Gamma(n).$$
where $m,n>0$. Let $m=r^k$ with $r>0$, multiply both sides by $\frac{f^{(k)}h^k}{k!}$ and sum on $k, 0 \leq k <\infty$, to obtain
$$\sum_{k=0}^\infty \frac{f^{(k)}(a)h^k}{k!}\int_0^\infty e^{-r^k x}x^{n-1}dx=\Gamma(n) \sum_{k=0}^\infty \frac{f^{(k)}(a)(hr^{-n})^k}{k!}.$$

Next, expand $e^{-r^k x}, 0\leq k<\infty$, in its Maclaurin Series, invert the order of summation and integration, and apply Taylor's Theorem to deduce that
$$\int_0^\infty x^{n-1}\sum_{j=0}^\infty \frac{f(h r^j+a)}{j!}(-x)^j dx = \Gamma(n)f(hr^{-n}+a).$$

Now define $f(hr^m+a)=\varphi(m)$, where $m$ is real and $a,h$ and $r$ are regarded as constants. Then
$$\int_0^\infty x^{n-1}\sum_{j=0}^\infty \frac{\varphi(j) (-x)^j}{j!}dx= \Gamma(n) \varphi(-n).$$
This completes Ramanujan's proof. Ramanujan was very fond of this clever, original technique and he used it many contexts.

3.Glasser's Master Theorem

$$\phi(x)=|a|x-\sum_{n=1}^N\frac{|\alpha_n|}{x-\beta_n}$$

$$I=\int_{0}^{\infty} \left[\left(\frac{2015}{2015+x}+\cdots +\frac{2}{2+x}+\frac{1}{1+x}-x\right)^{2016}+1 \right] ^{-1}\mathrm{d}x,$$
$$I=\frac{1}{2}\int_{-\infty}^{\infty} \left[\left(\frac{2015}{2015+x}+\cdots +\frac{2}{2+x}+\frac{1}{1+x}-x\right)^{2016}+1 \right] ^{-1}\mathrm{d}x,$$
$$I=\frac{1}{2}\int_{-\infty}^{\infty} \left[\left(\sum^{2015}_{i=1}\frac{i}{x+i}-x\right)^{2016}+1 \right] ^{-1}\mathrm{d}x.$$

Now, letting $f(x)=\frac{1}{x^{2016}+1}$, and noting that $f(x)=f(-x)$,
$$I=\frac{1}{2}\int_{-\infty}^{\infty} f\left(\sum^{2015}_{i=1}\frac{i}{x+i}-x\right)\mathrm{d}x=\frac{1}{2}\int_{-\infty}^{\infty} f\left(-\left(\sum^{2015}_{i=1}\frac{i}{x+i}-x\right)\right)\mathrm{d}x$$
$$I=\frac{1}{2}\int_{-\infty}^{\infty} f\left(x-\sum^{2015}_{i=1}\frac{i}{x-(-i)}\right)\mathrm{d}x \tag {1}$$
Using Glasser's Master Theorem,
$$I=\frac{1}{2}\int^{\infty}_{-\infty} f(x)\ \mathrm{d}x=\frac{1}{2}\int^{\infty}_{-\infty} \frac{1}{x^{2016}+1}\ \mathrm{d}x=\int^{\infty}_{0} \frac{1}{x^{2016}+1}\ \mathrm{d}x \tag {2}$$

Now we know that $$B(a,b)=\int^{\infty}_0\frac{t^{a-1}}{(1+t)^{a+b}}dt$$
From $(2)$,after substituting $x^{2016} =t$,
$$I=\frac{1}{2016}\int^{\infty}_{0}\frac{t^{\frac{1}{2016}-1}}{(1+t)^{\frac{1}{2016}+\frac{2015}{2016}}}dt=\frac{1}{2016}B(\frac{1}{2016},\frac{2015}{2016})$$
Therefore $$\color{red}{I=\frac{1}{2016}\frac{\Gamma(\frac{1}{2016})\Gamma(\frac{2015}{2016})}{\Gamma(1)}=\frac{\pi}{2016\sin(\frac{\pi}{2016})}\approx1.0000004047320180811575}$$

## 几个逼格稍高的积分级数题

\begin{align}\int_0^{ + \infty } {\frac{{\sin nx}}{{x + \frac{1}{{x + \frac{2}{{x + \frac{3}{{x +  \cdots }}}}}}}}dx}  &= \frac{{\sqrt {\frac{\pi }{2}} }}{{n + \frac{1}{{n + \frac{2}{{n + \frac{3}{{n +  \cdots }}}}}}}}\\\int_0^{ + \infty } {\frac{{\sin \frac{{n\pi x}}{2}}}{{x + \frac{{{1^2}}}{{x + \frac{{{2^2}}}{{x + \frac{{{3^2}}}{{x +  \cdots }}}}}}}}dx}  &= \frac{1}{{n + \frac{{{1^2}}}{{n + \frac{{{2^2}}}{{n + \frac{{{3^2}}}{{n +  \cdots }}}}}}}}.\end{align}

\begin{align}\sum\limits_{n = 0}^{ + \infty } {\left[ {\left( {1 + \frac{1}{3} + \frac{1}{5} +  \cdots  + \frac{1}{{2n + 1}}} \right) \cdot \frac{1}{{{5^n}\left( {2n + 1} \right)}}} \right]}  &= \frac{{{\pi ^2}}}{{4\sqrt 5 }} - \frac{{\sqrt 5 }}{{24}}{\left( {\ln \left( {2 + \sqrt 5 } \right)} \right)^2}\\\sum\limits_{n = 0}^{ + \infty } {\left[ {\left( {1 + \frac{1}{3} + \frac{1}{5} +  \cdots  + \frac{1}{{2n + 1}}} \right) \cdot \frac{1}{{{9^n}\left( {2n + 1} \right)}}} \right]}  &= \frac{{{\pi ^2}}}{8} - \frac{3}{8}{\left( {\ln 2} \right)^2}\end{align}

\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).

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

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

## 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.