Euler's work VII

Series expansion of the complete beta function

We show the proof of the following series posted by @infseriesbot $$\frac{1}{x} + \frac{x}{1!(x+1)}+\frac{x(x+1)}{2!(x+2)}+... = \frac{\pi}{\sin(\pi x)}$$

Proof:

This series is just a special case of the series expansion of the complete beta function: $$B(x,y) = \sum_{j=0}^{\infty}\frac{(1-y)_{j}}{j!(x+j)}$$ To prove this, consider the definition of the beta function: $$B(x,y) = \int_{0}^{1} t^{x-1}(1-t)^{y-1}dt$$ and recall the series expansion of $(1-t)^{y-1}$ $$(1-t)^{y-1} = \sum_{j=0}^{\infty} (-1)^{j} \binom{y-1}{j}t^{j}$$ Hence $$\begin{align*} \int_{0}^{1} t^{x-1}(1-x)^{y-1}dt =& \int_{0}^{1} t^{x-1}\sum_{j=0}^{\infty} (-1)^{j} \binom{y-1}{j}t^{j}dt\\ =& \sum_{j=0}^{\infty} (-1)^{j} \binom{y-1}{j} \int_{0}^{\infty} t^{x+j-1}dt\\ =& \sum_{j=0}^{\infty} (-1)^{j} \binom{y-1}{j} \frac{1}{(x+j)} \end{align*}$$ Recall the reflection formula for the upper indice $$\binom{-v}{m} = (-1)^{m}\binom{v+m-1}{m}\tag{1}$$ and the fact that the Pochhammer polynomials can be expressed as the quotient of two gamma functions: $$(x)_{n} = \frac{\Gamma(n+x)}{\Gamma(x)} \tag{2}$$ Then we have $$\begin{align*} \sum_{j=0}^{\infty} (-1)^{j} \binom{y-1}{j}\frac{1}{(x+j)} =& \sum_{j=0}^{\infty}\binom{j-y}{j}\frac{1}{(x+j)}\quad (\textrm{ from } (1))\\ =& \sum_{j=0}^{\infty}\frac{(j-y)!}{j!(-y)!}\frac{1}{(x+j)}\\ =& \sum_{j=0}^{\infty}\frac{\Gamma(j-y+1)}{j!\Gamma(1-y)} \frac{1}{(x+j)}\\ =& \sum_{j=0}^{\infty}\frac{(1-y)_{j}}{j!(x+j)} \quad (\textrm{ from } (2)) \end{align*}$$ Therefore $$B(x,y) = \sum_{j=0}^{\infty}\frac{(1-y)_{j}}{j!(x+j)}$$ Now, if we put $y = 1-x$ we have: $$\sum_{j=0}^{\infty}\frac{(x)_{j}}{j!(x+j)} = B(x,1-x) = \frac{\Gamma(x)\Gamma(1-x)}{\Gamma(1)}$$ Recall the Euler's reflection formula $$\Gamma(x)\Gamma(1-x) = \frac{\pi}{\sin(\pi x)} \quad x\notin \mathbb{N}$$ Hence $$\boxed{\sum_{j=0}^{\infty}\frac{(x)_{j}}{j!(x+j)} = \frac{1}{x} + \frac{x}{1!(x+1)}+\frac{x(x+1)}{2!(x+2)}+... = \frac{\pi}{\sin(\pi x)}}$$

Generalized hypergeometric functions IV

Another nice involving ${}_{3}F_{2}$

We show the proof of this nice series posted by @PDemichel : $$\sum_{n=0}^{\infty} \frac{(4n)!}{256^n n!^4} = \frac{\pi}{\Gamma^2\left(\frac{5}{8}\right)\Gamma^2\left(\frac{7}{8}\right)}$$

It turned out that we proved it earlier using different formulas to calculate ${}_{3}F_{2}$.

Proof: In the past we have used the duplication or triplication formula for Pochhamer polyonomials or rising factorials. It turned out that this formulas are particular cases for the general formula (Proof in the Appendix): $$(x)_{nm} = m^{nm} \prod_{j=1}^{m}\left(\frac{x+j-1}{m} \right)_{n}$$ With this formula, we can expand $(4n)!$ in the following way: $$(4n)! = \frac{\Gamma(4n+1)}{\Gamma(1)} = (1)_{4n} = 4^{4n}\left(\frac{1}{4}\right)_{n}\left(\frac{1}{2}\right)_{n}\left(\frac{3}{4}\right)_{n}\left(1\right)_{n}$$ Additionally, we know that $(1)_{n} = n!$. Hence, $$\sum_{n=0}^{\infty} \frac{(4n)!}{256^n n!^4} = \sum_{n=0}^{\infty} \frac{4^{4n}\left(\frac{1}{4}\right)_{n}\left(\frac{1}{2}\right)_{n}\left(\frac{3}{4}\right)_{n}\left(1\right)_{n}}{256^n (1)_{n}(1)_{n}(1)_{n}n!} = \sum_{n=0}^{\infty} \frac{\left(\frac{1}{4}\right)_{n}\left(\frac{3}{4}\right)_{n}\left(\frac{1}{2}\right)_{n}}{ (1)_{n}(1)_{n}n!}={}_{3}F_{2} \left(\frac{1}{4},\frac{3}{4},\frac{1}{2};1,1;1\right)$$ Now recall the Watson’s Sum formula for ${}_{3}F_{2}$: $${{}_{3}F_{2}}\left({a,b,c\atop\frac{1}{2}(a+b+1),2c};1\right)=\frac{\Gamma% \left(\frac{1}{2}\right)\Gamma\left(c+\frac{1}{2}\right)\Gamma\left(\frac{1}{2% }(a+b+1)\right)\Gamma\left(c+\frac{1}{2}(1-a-b)\right)}{\Gamma\left(\frac{1}{2% }(a+1)\right)\Gamma\left(\frac{1}{2}(b+1)\right)\Gamma\left(c+\frac{1}{2}(1-a)% \right)\Gamma\left(c+\frac{1}{2}(1-b)\right)},$$ Therefore $$\sum_{n=0}^{\infty} \frac{(4n)!}{256^n n!^4} ={}_{3}F_{2} \left(\frac{1}{4},\frac{3}{4},\frac{1}{2};1,1;1\right) = \frac{\Gamma\left(\frac{1}{2}\right)\Gamma(1)\Gamma(1)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{1}{2}\left(\frac{1}{4}+\frac{4}{4}\right)\right)\Gamma\left(\frac{1}{2}\left(\frac{3}{4}+\frac{4}{4}\right)\right)\Gamma\left(\frac{1}{2}+\frac{1}{2}\left(\frac{3}{4}\right)\right)\Gamma\left(\frac{1}{2}+\frac{1}{2}\left(\frac{1}{4}\right)\right)}$$ Hence $$\sum_{n=0}^{\infty} \frac{(4n)!}{256^n n!^4} = \frac{\pi}{\Gamma^2\left(\frac{5}{8}\right)\Gamma^2\left(\frac{7}{8}\right)}$$

Appendix

We show the proof of this result for Pochhammer polynomials:
$$(x)_{nm} = m^{nm} \prod_{j=1}^{m}\left(\frac{x+j-1}{m} \right)_{n}$$ Consider: $$(x)_{nm} = x(x+1)(x+2)\ldots (x+nm-1)$$ Factorize $(x)_{nm}$ in the following way: $$(x)_{nm} = A_{1}\cdot A_{2} \cdot A_{3}\ldots A_{n}$$ where $$\begin{matrix} A_{1} =& x& (x+1)&(x+2)&\ldots &(x+m-1)\\ A_{2} =& (x+m)& (x+m+1) &(x+m+2) & \ldots & (x+2m-1)\\ A_{3} =& (x+2m)&(x+2m+1)&(x+2m+2)& \ldots &(x+3m-1)\\ \vdots & \vdots & \vdots &\vdots & \ldots & \vdots\\ A_{n} =&(x+(n-1)m) &(x+(n-1)m+1)& (x+(n-1)m+2) & \ldots & (x+nm-1) \end{matrix}$$
Now factor m out of each factor in each $A_{j}$
$$\begin{matrix} A_{1} =& m^m & \frac{x}{m}& \left(\frac{x+1}{m}\right)& \left(\frac{x+2}{m}\right)&\ldots &\left(\frac{x+m-1}{m}\right)\\ A_{2} =& m^m & \left(\frac{x}{m}+1\right)&\left(\frac{x+1}{m}+1\right)&\left(\frac{x+2}{m}+1\right)& \ldots& \left(\frac{x+m-2}{m}\right)\\ A_{3} =& m^m & \left(\frac{x}{m}+2\right)&\left(\frac{x+1}{m}+2\right)&\left(\frac{x+2}{m}+2\right)&\ldots &\left(\frac{x+3m-1}{m}\right)\\ \vdots &\vdots & \vdots& \vdots &\vdots &\ldots &\vdots \\ A_{n} =& \underbrace{m^m}_{B_{0}} & \underbrace{\left(\frac{x}{m}+n-1\right)}_{B_{1}} & \underbrace{\left(\frac{x+1}{m}+n-1\right)}_{B_{2}} &\underbrace{\left(\frac{x+2}{m}+n-1\right)}_{B_{3}} &\ldots & \underbrace{\left(\frac{x+mn-1}{m}\right)}_{B_{m}} \end{matrix}$$ Now it touned out that column $B_{0}=m^{nm}$ and $$B_{j} = \left(\frac{x+j-1}{m} \right)_{n}$$ Therefore $$\boxed{(x)_{nm} = B_{0}\cdot B_{1} \cdot B_{2} \ldots B_{m} = m^{nm} \prod_{j=1}^{m}\left(\frac{x+j-1}{m} \right)_{n} }$$

Generalized hypergeometric functions III

Series involving ${}_{2}F_{1}$

Today we show the proof of this nice series posted by @infseriesbot: $$\sum_{n=0}^{\infty} \frac{(3n)!}{54^n n!^3} = \frac{\sqrt{ \pi}}{\Gamma\left(\frac{2}{3}\right)\Gamma\left(\frac{5}{6}\right)}$$

Proof

First recall that the Pochhammer polyonomial or rising factorial can be expressed as a quiotinent of two gamma functions: $$(x)_{n} = x(x+1)(x+2)\cdots(x+n-1) = \frac{\Gamma(n+x)}{\Gamma(x)}$$ Hence $$(3n)! = \frac{\Gamma(3n+1)}{\Gamma(1)} = (1)_{3n}$$ Now we will use the triplication formula for Pochhamer polynomials: $$(x)_{3n} = 3^{3n}\left(\frac{x}{3}\right)_{n}\left(\frac{x+1}{3}\right)_{n}\left(\frac{x+2}{3}\right)_{n}$$ Is easy to show the proof of this: By definition $$(x)_{3n} = x(x+1)(x+2)\cdots(x+3n-1)$$ If we factorize in this form: $$(x)_{3n} = \underbrace{\left[x(x+3)(x+6)...(x+3n-3)\right]}_{n-\textrm{terms}}\underbrace{\left[(x+1)(x+4)(x+7)...(x+3n-2)\right]}_{n-\textrm{terms}}\underbrace{\left[(x+2)(x+5)(x+8)...(x+3n-1)\right]}_{n-\textrm{terms}}$$ Consider $$A = x(x+3)(x+6)\cdots(x+3n-3) = 3^n\left(\frac{x}{3}\right)\left(\frac{x}{3}+1\right)\dots \left(\frac{x}{3}+n-1\right) = 3^n\left(\frac{x}{3}\right)_{n}$$ $$B = (x+1)(x+4)(x+7)\cdots(x+3n-2) = 3^n\left(\frac{x+1}{3}\right)\left(\frac{x+1}{3}+1\right)\dots \left(\frac{x+1}{3}+n-1\right) = 3^n\left(\frac{x+1}{3}\right)_{n}$$ $$C = (x+2)(x+5)(x+8)\cdots(x+3n-1)= 3^n\left(\frac{x+2}{3}\right)\left(\frac{x+2}{3}+1\right)\dots \left(\frac{x+2}{3}+n-1\right) = 3^n\left(\frac{x+2}{3}\right)_{n}$$ Hence $$(x)_{3n} = ABC = 3^{3n}\left(\frac{x}{3}\right)_{n}\left(\frac{x+1}{3}\right)_{n}\left(\frac{x+2}{3}\right)_{n}$$ Therefore $$(3n)! = (1)_{3n} = 3^{3n}\left(\frac{1}{3}\right)_{n}\left(\frac{2}{3}\right)_{n}\left(1\right)_{n}$$ Now, using the fact that $(1)_{n}=n!$ $$\sum_{n=0}^{\infty} \frac{(3n)!}{54^n n!^3} = \sum_{n=0}^{\infty} \frac{3^{3n}\left(\frac{1}{3}\right)_{n}\left(\frac{2}{3}\right)_{n}\left(1\right)_{n}}{54^n (1)_{n} (1)_{n} n!}= \sum_{n=0}^{\infty} \frac{\left(\frac{1}{3}\right)_{n}\left(\frac{2}{3}\right)_{n}}{ (1)_{n}}\left(\frac{1}{2^n}\right)\frac{1}{n!} = {}_{2}F_{1}\left(\frac{1}{3},\frac{2}{3};1;\frac{1}{2}\right)$$ Now using the following identity for ${}_{2}F_{1}$: $${}_{2}F_{1} \left(a,b;\tfrac{1}{2}a+\tfrac{1}{2}b+\tfrac{1}{2};\tfrac{1}{2}\right)=\sqrt{% \pi}\frac{\Gamma\left(\tfrac{1}{2}a+\tfrac{1}{2}b+\tfrac{1}{2}\right)}{\Gamma% \left(\tfrac{1}{2}a+\tfrac{1}{2}\right)\Gamma\left(\tfrac{1}{2}b+\tfrac{1}{2}% \right)}.$$ we have $$\sum_{n=0}^{\infty} \frac{(3n)!}{54^n n!^3} ={}_{2}F_{1}\left(\frac{1}{3},\frac{2}{3};1;\frac{1}{2}\right) = \frac{\sqrt{\pi}{\Gamma(1)}}{\Gamma\left(\frac{1}{6}+\frac{1}{2}\right)\left(\frac{2}{6}+\frac{1}{2}\right)} = \frac{\sqrt{\pi}}{\Gamma\left(\frac{2}{3}\right)\Gamma\left(\frac{5}{6}\right)}$$ $$\boxed{\sum_{n=0}^{\infty} \frac{(3n)!}{54^n n!^3} = \frac{\sqrt{\pi}}{\Gamma\left(\frac{2}{3}\right)\Gamma\left(\frac{5}{6}\right)}}$$

Generalized hypergeometric functions II

Series involving ${}_{3}F_{2}$

Today we show the proof of this nice series posted by @infseriesbot $$\sum_{n=0}^{\infty} \frac{(4n)!}{2^{8n}n!^4} = \frac{\pi}{\Gamma\left(\frac{5}{8}\right)^2\Gamma\left(\frac{7}{8}\right)^2}$$ The proof will rely on some properties of the Pochhammer polynomials (rising factorials) and the generalized hypergeometric functions

Proof

Consider the Pochhammer polynomial: $$(x)_{n} = x(x+1)(x+2)\cdots (x+n-1) = \prod_{j=0}^{n-1}(x+j) \quad n=1,2,3,...$$ also called rising factorial.

Pochhammer polynomials can be expressed as the quotient of two gamma functions: $$(x)_{n} = \frac{\Gamma(n+x)}{\Gamma(x)}$$ Therefore $$(4n)! = \Gamma(4n+1) = (1)_{4n}$$ Pochhammer polynomials also obey the following duplication formula: $$(x)_{2n} = 4^n\left(\frac{x}{2}\right)_{n}\left(\frac{1+x}{2}\right)_{n}$$ Therefore $$\begin{align*} (4n)! = (1)_{4n} =& 4^{2n}\left(\frac{1}{2}\right)_{2n}\left(1\right)_{2n}\\ =& 4^{2n}\left[4^{n}\left(\frac{1}{4}\right)_{n}\left(\frac{3}{4}\right)_{n}\right]\left[4^{n}\left(\frac{1}{2}\right)_{n}\left(1\right)_{n}\right]\\ =& 4^{4n} \left(\frac{1}{4}\right)_{n}\left(\frac{3}{4}\right)_{n}\left(\frac{1}{2}\right)_{n}\left(1\right)_{n} \end{align*}$$ From the definition of the Pochhammer polynomials is easy to show that $\displaystyle (1)_{n} = n!$
Therefore
$$\begin{align*} S = \sum_{n=0}^{\infty} \frac{(4n)!}{2^{8n}n!^4} =&\sum_{n=0}^{\infty} \frac{4^{4n} \left(\frac{1}{4}\right)_{n}\left(\frac{3}{4}\right)_{n}\left(\frac{1}{2}\right)_{n}\left(1\right)_{n}}{2^{8n}(1)_{n}(1)_{n}(1)_{n}}\frac{1}{n!} \\ =& \sum_{n=0}^{\infty} \frac{ \left(\frac{1}{4}\right)_{n}\left(\frac{3}{4}\right)_{n}\left(\frac{1}{2}\right)_{n}}{(1)_{n}(1)_{n}}\frac{1}{n!}\\ =& {}_3F_{2}\left({\frac{1}{4},\frac{3}{4},\frac{1}{2}\atop 1,1};1\right) \end{align*}$$ where $\displaystyle {}_3F_{2}$ is the Generalized Hypergeometric function

$\displaystyle {}_3F_{2}$ satisfy the following identity: $${{}_{3}F_{2}}\left({a,b,c\atop d,e};1\right)=\frac{\Gamma\left(e\right)\Gamma% \left(d+e-a-b-c\right)}{\Gamma\left(e-a\right)\Gamma\left(d+e-b-c\right)}{{}_{% 3}F_{2}}\left({a,d-b,d-c\atop d,d+e-b-c};1\right)$$ whenever $\operatorname{Re}(d+e-a-b-c)>0$ and $\operatorname{Re}(e-a)>0$

Therefore $$S = {}_3F_{2}\left({\frac{1}{4},\frac{3}{4},\frac{1}{2}\atop 1,1};1\right) = \frac{\sqrt{\pi}}{\Gamma\left(\frac{3}{4}\right)^2}{{}_{% 3}F_{2}}\left({\frac{1}{4},\frac{1}{4},\frac{1}{2}\atop 1,\frac{3}{4}};1\right)$$ ${}_{3}F_{2}$ also satisfy the Dixon’s Well-Poised Sum: $${{}_{3}F_{2}}\left({a,b,c\atop a-b+1,a-c+1};1\right)=\frac{\Gamma\left(\frac{1% }{2}a+1\right)\Gamma\left(a-b+1\right)\Gamma\left(a-c+1\right)\Gamma\left(% \frac{1}{2}a-b-c+1\right)}{\Gamma\left(a+1\right)\Gamma\left(\frac{1}{2}a-b+1% \right)\Gamma\left(\frac{1}{2}a-c+1\right)\Gamma\left(a-b-c+1\right)}$$ Hence $${}_{ 3}F_{2}\left({\frac{1}{4},\frac{1}{4},\frac{1}{2}\atop 1,\frac{3}{4}};1\right) = \frac{\Gamma\left(\frac{9}{8}\right)\Gamma\left(\frac{3}{4}\right)\Gamma\left(\frac{3}{8}\right)}{\Gamma\left(\frac{5}{4}\right)\Gamma\left(\frac{7}{8}\right)\Gamma\left(\frac{5}{8}\right)\sqrt{\pi} } = \frac{\Gamma\left(\frac{3}{8}\right)^2}{\sqrt{2\pi}\Gamma\left(\frac{5}{8}\right)^{2}}$$ Therefore $$S = \frac{\Gamma\left(\frac{3}{8}\right)^2}{\sqrt{2}\Gamma\left(\frac{3}{4}\right)^2\Gamma\left(\frac{5}{8}\right)^{2}} = \frac{\pi}{\Gamma^2\left(\frac{5}{8}\right)\Gamma^2\left(\frac{7}{8}\right)}$$ Hence, we can conclude $$\boxed{ \sum_{n=0}^{\infty} \frac{(4n)!}{2^{8n}n!^4} = \frac{\pi}{\Gamma\left(\frac{5}{8}\right)^2\Gamma\left(\frac{7}{8}\right)^2}}$$

Euler's work VI

Series related to the Catalan constant

We show te proof of the following result. $$\sum_{n=0}^{\infty} \frac{(2n)!!(n+1)(2n+5)}{(2n+1)!!(2n+1)(2n+3)^2}=1$$ The proof involves one of the series representation of the Catalan constant.

Proof

First note that $$\frac{(n+1)(2n+5)}{(2n+1)(2n+3)^2} = \frac{1}{2(2n+1)}+\frac{1}{2(2n+3)^2}$$ Hence $$\begin{align*} S=\sum_{n=0}^{\infty} \frac{(2n)!!(n+1)(2n+5)}{(2n+1)!!(2n+1)(2n+3)^2} =& \sum_{n=0}^{\infty} \frac{(2n)!!}{(2n+1)!!}\left[\frac{1}{2(2n+1)}+\frac{1}{2(2n+3)^2}\right] \\ =& \frac{1}{2}\underbrace{ \sum_{n=0}^{\infty} \frac{(2n)!!}{(2n+1)!!(2n+1)}}_{S_{1}} + \frac{1}{2}\underbrace{\sum_{n=0}^{\infty}\frac{(2n)!!}{(2n+1)!!(2n+3)^2}}_{S_{2}} \end{align*}$$ In a previous post we evaluated $S_{1}$: $$S_{1}= \sum_{n=0}^{\infty} \frac{(2n)!!}{(2n+1)!!(2n+1)} = 2\beta(2)$$ where $\beta(2)$ is the Calatan constant.

For $S_{2}$:

$$\begin{align*} S_{2} =& \sum_{n=0}^{\infty}\frac{(2n)!!}{(2n+1)!!(2n+3)^2}\\ =&\sum_{n=0}^{\infty}\frac{(2n+2)!!}{(2n+3)!!(2n+3)(2n+2)}\\ =&\sum_{j=1}^{\infty}\frac{(2j)!!}{(2j+1)!!(2j+1)(2j)} \quad (n=j-1) \end{align*}$$ Note that $$\underbrace{\sum_{n=1}^{\infty} \frac{(2j)!!}{(2j+1)!!(2j+1)}}_{2\beta(2)-1} = \underbrace{\sum_{j=1}^{\infty}\frac{(2j)!!}{(2j+1)!!2j}}_{S_{3}} - \underbrace{\sum_{j=1}^{\infty}\frac{(2j)!!}{(2j+1)!!(2j+1)(2j)}}_{S_{2}}$$ We just have to find $S_{3}$:

In another post we found the following series expansion: $$\frac{\arcsin(x)}{\sqrt{1-x^2}} = \sum_{j=0}^{\infty} \frac{(2j)!!}{(2j+1)!!}x^{2j+1}$$ Dividing by $x^2$ $$\frac{\arcsin(x)}{x^2\sqrt{1-x^2}} = \sum_{j=0}^{\infty} \frac{(2j)!!}{(2j+1)!!}x^{2j-1} = \frac{1}{x} + \sum_{j=1}^{\infty} \frac{(2j)!!}{(2j+1)!!}x^{2j-1}$$ $$\Longrightarrow \sum_{j=1}^{\infty} \frac{(2j)!!}{(2j+1)!!}x^{2j-1} = \frac{\arcsin(x)}{x^2\sqrt{1-x^2}}- \frac{1}{x}$$ Integrating from $0$ to $1$: $$\Longrightarrow S_{3} = \sum_{j=1}^{\infty} \frac{(2j)!!}{(2j+1)!!2j} = \int_{0}^{1}\frac{\arcsin(x)}{x^2\sqrt{1-x^2}}- \frac{1}{x} dx$$ $$I= \int \frac{\arcsin(x)}{x^2\sqrt{1-x^2}}- \frac{1}{x} dx = \int\frac{\arcsin(x)}{x^2\sqrt{1-x^2}}dx - \ln(x) + C$$ $$\begin{align*} \int\frac{\arcsin(x)}{x^2\sqrt{1-x^2}}dx =& \int \frac{w}{\sin^2(w)}dw \quad (w \mapsto \arcsin(x))\\ \stackrel{IBP}{=}& -w\cot(w) + \int \cot(w) dw \\ =& -w\cot(w) +\ln(\sin w) + C\\ =& -\frac{\sqrt{1-x^2}\arcsin(x)}{x} +\ln(x) + C \end{align*}$$ Therefore $$I = \int \frac{\arcsin(x)}{x^2\sqrt{1-x^2}}- \frac{1}{x} dx = -\frac{\sqrt{1-x^2}\arcsin(x)}{x}+ C$$ Hence $$S_{3} = \int_{0}^{1}\frac{\arcsin(x)}{x^2\sqrt{1-x^2}}- \frac{1}{x} dx = \left[-\frac{\sqrt{1-x^2}\arcsin(x)}{x}\right]_{0}^{1} = 1$$ Therefore $$S_{2} = S_{3} -\beta(2) +1 = 2-2\beta(2)$$ $$S = \frac{1}{2}S_{1} + \frac{1}{2}S_{2} = \beta(2) + 1 - \beta(2) = 1$$ We can conclude: $$\boxed{\sum_{n=0}^{\infty} \frac{(2n)!!(n+1)(2n+5)}{(2n+1)!!(2n+1)(2n+3)^2}=1}$$

Residue theorem V

Series involving hyperbolic cotangent

We show the proof following result about series involving hyperbolic functions posted by @infseriesbot: $$\pi \sum_{n=1}^{\infty} \frac{\coth \pi n}{n^{4m-1}} = \zeta(4m) + \sum_{n=1}^{2m-1}(-1)^{k-1}\zeta(2k)\zeta(4m-2k)$$ The proof will rely on the residue theorem.

Proof

In order to calculate the residue, we will need the expansion in zeta numbers of $\displaystyle \coth(\pi x)$ and $\cot(x)$ $$\coth(\pi x) = \frac{1}{\pi x} - \frac{2}{\pi } \sum_{n=1}^{\infty} (-1)^n\zeta(2n)x^{2n-1}$$ $$\cot(\pi x) = \frac{1}{\pi x} - \frac{2}{\pi } \sum_{n=1}^{\infty}\zeta(2n)x^{2n-1}$$

1. To obtain these expansions we need the Fourier series expansion of $\cos(ax)$

Since $\cos(ax)$ is defined for all $x$, and represents a continuous, piece-wise smooth, odd function is everywhere equal to its Fourier series, which is absolutely and uniformly convergent.

Therefore,
$$\cos ax = \sum_{n=-\infty}^{\infty} c_{n}e^{inx} \quad a\neq n$$ where $$c_{n} = \frac{1}{2\pi} \int_{-\pi}^{\pi} \cos(ax)e^{-inx}dx$$ Hence $$\begin{align*} c_{n} = \frac{1}{2\pi} \int_{-\pi}^{\pi} \cos(ax)e^{-inx}dx =& \frac{1}{2\pi} \int_{-\pi}^{\pi} \left[\frac{e^{iax}+ e^{-iax}}{2}\right]e^{-inx}dx \\ =& \frac{1}{4\pi} \int_{-\pi}^{\pi} e^{ix(a-n)}dx + \frac{1}{4\pi} \int_{-\pi}^{\pi} e^{ix(-a-n)}dx\\ =& \frac{1}{4\pi}\left[\frac{e^{ i(a-n)} -e^{- i(a-n)}}{i(a-n)}\right] + \frac{1}{4\pi}\left[\frac{e^{ i(a+n)} -e^{- i(a+n)}}{i(a+n)}\right]\\ =& \frac{1}{4\pi}\left[\frac{e^{ a} e^{-\pi in} -e^{-\pi i a}e^{- \pi i n}}{i(a-n)}\right] + \frac{1}{4\pi}\left[\frac{e^{\pi i a}e^{\pi in)} -e^{-\pi i a}e^{i\pi n}}{i(a+n)}\right]\\ =&\frac{e^{i\pi n} (e^{i \pi a}-e^{-i \pi a} )}{4\pi i}\left[\frac{1}{a-n}+\frac{1}{a+n}\right]\\ =& (-1)^n \frac{\sin(a\pi)}{2\pi}\left[ \frac{2a}{a^2-n^2}\right]\\ =& (-1)^n \frac{a\sin(a\pi)}{\pi(a^2-n^2)}\end{align*}$$ Therefore $$\begin{align*} \cos(ax) = \sum_{n=-\infty}^{\infty} (-1)^n \frac{a\sin(a\pi)}{\pi(a^2-n^2)}e^{inx} =&\frac{\sin(\pi a)}{\pi a}+ \sum_{j=-\infty}^{j=-1} (-1)^j \frac{a\sin(a\pi)}{\pi(a^2-j^2)}e^{ijx} + \sum_{n=1}^{\infty} (-1)^n \frac{a\sin(a\pi)}{\pi(a^2-n^2)}e^{inx} \\ =&\frac{\sin(\pi a)}{\pi a}+ \sum_{n=1}^{\infty} (-1)^n \frac{a\sin(a\pi)}{\pi(a^2-n^2)}e^{-inx} + \sum_{n=1}^{\infty} (-1)^n \frac{a\sin(a\pi)}{\pi(a^2-n^2)}e^{inx} \quad (n= -j)\\ =&\frac{\sin(\pi a)}{\pi a}+ \sum_{n=1}^{\infty} (-1)^n \frac{a\sin(a\pi)}{\pi(a^2-n^2)}(e^{-inx} +e^{inx})\\ =&\frac{\sin(\pi a)}{\pi a}+ \frac{2}{\pi}\sum_{n=1}^{\infty} (-1)^n \frac{a\sin(a\pi)\cos(nx)}{\pi(a^2-n^2)}\\ =&\frac{2}{\pi}\sin(\pi a)\left[\frac{1}{2 a}+ \sum_{n=1}^{\infty} (-1)^n \frac{a\cos(nx)}{(a^2-n^2)}\right] \end{align*}$$ Hence $$\boxed{\cos(ax) = \frac{2}{\pi}\sin(\pi a)\left[\frac{1}{2 a}+ \sum_{n=1}^{\infty} (-1)^n \frac{a\cos(nx)}{(a^2-n^2)}\right] } \tag{1}$$
2. With this result we can obtain the expansion in partial fraction of $\coth(x)$ and $\cot(x)$:

If we put $\displaystyle x=\pi \Longrightarrow \cos(n\pi) = (-1)^n$ in the Fourier series expansion of $\cot(ax)$ then $$\cos(a \pi) = \frac{2}{\pi}\sin(\pi a)\left[\frac{1}{2 a}+ \sum_{n=1}^{\infty} \frac{a}{(a^2-n^2)}\right]$$ If we divide by $\sin(a\pi)$ $$\cot(a \pi) = \frac{2}{\pi}\left[\frac{1}{2 a}+ \sum_{n=1}^{\infty} \frac{a}{(a^2-n^2)}\right]$$ Now, if we allow $a$ to take complex values: let $a= iv$ $$\cot(-iv \pi) = \frac{2}{\pi}\left[-\frac{i}{2 v} - \sum_{n=1}^{\infty} \frac{iv}{(v^2+n^2)}\right]$$ Multiplying by $i$ $$\coth(v \pi) = \frac{2}{\pi}\left[\frac{1}{2 v} + \sum_{n=1}^{\infty} \frac{v}{(v^2+n^2)}\right]$$ We have found the expansion in partial fractions of $\cot$ and $\coth$ $$\boxed{ \cot(x \pi) = \frac{2}{\pi}\left[\frac{1}{2 x}+ \sum_{n=1}^{\infty} \frac{x}{(x^2-n^2)}\right] \quad x\neq \pm 1, \pm 2, \pm 3,..} \tag{2}$$ $$\boxed{\coth(x \pi) = \frac{2}{\pi}\left[\frac{1}{2 x} + \sum_{n=1}^{\infty} \frac{x}{(x^2+n^2)}\right]\quad x\neq \pm i, \pm 2i, \pm 3i,..} \tag{3}$$
3. Another consequence is the expansion in zeta numbers of $\coth(x)$ and $\cot(x)$:

Suppose that $|x|\lt 1$, using (2) we have $$\begin{align*} \cot(x \pi) =& \frac{1}{\pi x}- \frac{2}{\pi}\sum_{n=1}^{\infty} \frac{x}{(n^2-x^2)} \\ = &\frac{1}{\pi x}- \frac{2}{\pi}\sum_{n=1}^{\infty} \frac{x}{n^2\left(1-\left(\frac{x}{n}\right)^2\right)} \\ = &\frac{1}{\pi x}- \frac{2}{\pi}\sum_{n=1}^{\infty} \frac{x}{n^2} \sum_{j=0}^{\infty} \left(\frac{x}{n}\right)^{2j}\\ =&\frac{1}{\pi x}- \frac{2}{\pi}\underbrace{\sum_{j=0}^{\infty}\sum_{n=1}^{\infty} \frac{1}{n^{2j+2}} x^{2j+1}}_{\textrm{Rearrangement}} \\ =&\frac{1}{\pi x}- \frac{2}{\pi}\sum_{j=0}^{\infty}\zeta(2j+2) x^{2j+1} \\ =&\frac{1}{\pi x}- \frac{2}{\pi}\sum_{k=1}^{\infty}\zeta(2k) x^{2k-1} \\ \end{align*}$$ The rearrangement of the sums is possible due to the fact that the series $$\sum_{j=0}^{\infty} \left(\frac{x}{n}\right)^{2j}$$ is absolutely convergent and the series $$\zeta(v) = \sum_{n=1}^{\infty} \frac{1}{n^v}$$ is convergent. In a very similar way we can show the proof of $\coth(x)$ Therefore $$\boxed{ \cot(\pi x) =\frac{1}{\pi x}- \frac{2}{\pi}\sum_{k=1}^{\infty}\zeta(2k) x^{2k-1} \quad |x|\lt 1 } \tag{4}$$ $$\boxed{ \coth(\pi x) = \frac{1}{\pi x}- \frac{2}{\pi}\sum_{k=1}^{\infty}(-1)^k\zeta(2k) x^{2k-1} \quad |x|\lt 1} \tag{5}$$
4. We will apply the residue theorem


It is easy to show that if $\displaystyle f(z) = \frac{\pi \operatorname{coth}(\pi z) }{z^{4m-1}}$ $$\lim_{N \to \infty} \oint_{C_{N}} \pi\cot(\pi z)f(z) dz = 0$$ where $C_{N}$ be a square with vertices $\left(N+\frac{1}{2}\right)(1+i), \left(N+\frac{1}{2}\right)(-1+i), \left(N+\frac{1}{2}\right)(-1-i), \left(N+\frac{1}{2}\right)(1-i)$

Therefore, we can apply the following result : $$\sum_{\substack{n=-\infty \\ n\neq 0}}^{\infty}\frac{\pi\coth(\pi n)}{n^{4m-1}} = -\sum \left\{\textrm{Residues of } f(z)\pi\cot(\pi z) \textrm{ at the sungularities of } f \right\}$$ Given that the function $f(z)$ has a singularity at $z=0$ and a countable number of singularities at $\displaystyle z=in \quad n\in\mathbb{N}\setminus\left\{0\right\}$

We have: $$\sum_{\substack{n=-\infty \\ n\neq 0}}^{\infty}\frac{\pi\coth(\pi n)}{n^{4m-1}} = -\sum_{n=1}^{\infty}\operatorname{Res}\left(\pi\cot(\pi z),in\right) -\sum_{n=1}^{\infty}\operatorname{Res}\left(f(z)\pi\cot(\pi z),-in\right) -\operatorname{Res}\left(f(z)\pi \cot(\pi z),0\right)$$ With the L'Hopital rule can be proven that $$\lim_{z \to \pm in} (z\mp in)\cot(\pi z) \coth (\pi z) = \mp \frac{i\coth(n \pi)}{\pi} \quad \forall n>0$$ $$\therefore \operatorname{Res}\left(f(z)\pi\cot(\pi z),\mp in\right) = \lim_{z \to \pm in} \frac{(z\mp in)\pi^2 \cot(\pi z)\coth(\pi z)}{ z^{4m-1}} = \frac{\pi \coth(n \pi)}{ n^{4m-1}}$$ Therefore $$\sum_{\substack{n=-\infty \\ n\neq 0}}^{\infty}\frac{\pi\coth(\pi n)}{n^{4m-1}} = -\sum_{n=1}^{\infty}\frac{\pi\coth(\pi n)}{n^{4m-1}} -\sum_{n=1}^{\infty}\frac{\pi\coth(\pi n)}{n^{4m-1}} -\operatorname{Res}\left(f(z)\pi \cot(\pi z),0\right)$$ $$2\sum_{\substack{n=-\infty \\ n\neq 0}}^{\infty}\frac{\pi\coth(\pi n)}{n^{4m-1}} = -\operatorname{Res}\left(f(z)\pi \cot(\pi z),0\right)$$
We just have to find the residue at $z = 0$.

For that end, we will make use of the expansion in zeta numbers of $\cot$ and $\coth$

Note $$\begin{align*} \coth(\pi z)\cot(\pi z) =& \left[ \frac{1}{\pi z}- \frac{2}{\pi}\sum_{k=1}^{\infty}(-1)^k\zeta(2k) z^{2k-1}\right]\left[\frac{1}{\pi z}- \frac{2}{\pi}\sum_{k=1}^{\infty}\zeta(2k) z^{2k-1} \right] \quad (\textrm{from (4) and (5)})\\ &= \frac{1}{\pi^2 z^2} -\frac{2}{\pi^2} \sum_{k=1}^{\infty}\zeta(2k) z^{2k-2} - \frac{2}{\pi^2} \sum_{k=1}^{\infty}(-1)^k\zeta(2k) z^{2k-2} + \frac{4}{\pi^2} \underbrace{\sum_{k=1}^{\infty}\sum_{j=1}^{k} (-1)^{j}\zeta(2j)\zeta(2k-2j+2)z^{2k}}_{\textrm{Cauchy product}}\\ &= \frac{1}{\pi^2 z^2} -\frac{4}{\pi^2} \sum_{k=1}^{\infty}\zeta(4k) z^{4k-2} - \frac{4}{\pi^2} \sum_{k=1}^{\infty}\sum_{j=1}^{k} (-1)^{j}\zeta(2j)\zeta(2k-2j+2)z^{2k}\\ \end{align*}$$ Hence $$\frac{\pi^2 \cot(\pi z)\cot(\pi z)}{z^{4m-1}} = z^{-4m-1-2}- 4\sum_{k=1}^{\infty}\zeta(4k) z^{4k-4m-1} + 4 \sum_{k=1}^{\infty}\sum_{j=1}^{k} (-1)^{j}\zeta(2j)\zeta(2k-2j+2)z^{2k-4m+1}$$ The residue at $z=0$ implies that $$z^{4k-4m-1}= z^{-1} \Longrightarrow 4k-4m-1 = -1 \Longrightarrow k=m$$ $$z^{2k-4m+1} = z^{-1} \Longrightarrow 2k-4m+1 = -1 \Longrightarrow k=2m-1$$ Therefore $$\operatorname{Res}\left(\frac{\pi^2\cot(\pi z)\coth(\pi z)}{z^{4m-1}}, 0\right) = -4\zeta(4m) + 4\sum_{j=1}^{2m-1} (-1)^{j}\zeta(2j)\zeta(4m-2j)$$ Therefore $$2\sum_{\substack{n=-\infty \\ n\neq 0}}^{\infty} \frac{\pi \coth(\pi n)}{n^{4m-1}} = 4\zeta(4m) + 4\sum_{j=1}^{2m-1} (-1)^{j-1}\zeta(2j)\zeta(4m-2j)$$ Given that the function $\displaystyle f(z) = \frac{\pi \coth(\pi z)}{z^{4m-1}}$ is even we have $$4\sum_{n=1}^{\infty} \frac{\pi \coth(\pi n)}{n^{4m-1}} = 4\zeta(4m) + 4\sum_{j=1}^{2m-1} (-1)^{j-1}\zeta(2j)\zeta(4m-2j)$$ Hence, we can conclude $$\boxed{\sum_{n=1}^{\infty} \frac{\pi \coth(\pi n)}{n^{4m-1}} = \zeta(4m) + \sum_{j=1}^{2m-1} (-1)^{j-1}\zeta(2j)\zeta(4m-2j) }$$

Integral of the day VI

Integral involving the Apéry's constant $\zeta(3)$

Today we show the solution to this integral posted by @sounansya_29 on Twitter $$\int_{0}^{\infty} \frac{x\ln(\sinh x)}{\sinh x \cosh x} dx = -\frac{7}{16}\zeta(3)$$
The solution involves the integral representation of the Dirichlet lambda function as well as some known results of Euler sums.

Proof

$$\begin{align*} I=\int_{0}^{\infty} \frac{x\ln(\sinh x)}{\sinh x \cosh x} dx =& 2\int_{0}^{\infty} \frac{x\ln(\sinh(x))}{\sinh(2x)} \quad \left( \sinh x \cosh x = \frac{1}{2}\sinh(2x)\right)\\ =& 2\int_{0}^{\infty} \frac{x\ln\left(\frac{1}{2}(e^{x}-e^{-x})\right)}{\sinh(2x)}\\ =& -2\ln(2)\int_{0}^{\infty} \frac{x}{\sinh(2x)}dx + 2\int_{0}^{\infty} \frac{x\ln(e^{x}(1-e^{-2x}))}{\sinh(2x)}dx\\ =& -2\ln(2)\underbrace{\int_{0}^{\infty} \frac{x}{\sinh(2x)}dx}_{I_{1}} +2\underbrace{\int_{0}^{\infty} \frac{x^2}{\sinh(2x)}dx}_{I_{2}}+ 2\underbrace{\int_{0}^{\infty} \frac{x\ln(1-e^{-2x})}{\sinh(2x)}dx}_{I_{3}}\\ \end{align*}$$ Recall the beautiful integral representation of the Dirichlet lambda function: $$\lambda(v) = \frac{1}{2\Gamma(v)} = \int_{0}^{\infty} \frac{t^{v-1}}{\sinh(t)}dt$$ Therefore, for $I_{1}, I_{2}$ we have: $$I_{1} = \int_{0}^{\infty} \frac{x}{\sinh(2x)}dx = \frac{1}{4}\int_{0}^{\infty} \frac{w}{\sinh(w)} dw = \frac{1}{2}\Gamma(2) \lambda(2) = \frac{\pi^2}{16} \quad (w \mapsto 2x)$$ $$I_{2} = \int_{0}^{\infty} \frac{x^2}{\sinh(2x)}dx = \frac{1}{8}\int_{0}^{\infty} \frac{w^2}{\sinh(w)} dw = \frac{1}{4}\Gamma(3) \lambda(3) = \frac{7\zeta(3)}{16} \quad (w \mapsto 2x)$$
For $\displaystyle I_{3}$ we can split the integral in two:
$$\begin{align*} I_{3} = \int_{0}^{\infty} \frac{x\ln(1-e^{-2x})}{\sinh(2x)}dx =& 2\int_{0}^{\infty} \frac{x\ln(1-e^{-2x})}{e^{2x}-e^{-2x}}dx\\ =& -\frac{1}{2}\int_{0}^{1} \frac{\ln(w) \ln(1-w)}{(1-w)(1+w)}dw \quad (w \mapsto e^{-2x})\\ =& -\frac{1}{4}\underbrace{\int_{0}^{1} \frac{\ln(w) \ln(1-w)}{(1+w)}dw}_{J_{1}} - \frac{1}{4}\underbrace{\int_{0}^{1} \frac{\ln(w) \ln(1-w)}{(1-w)}dw}_{J_{2}} \end{align*}$$ $$J_{1} = \int_{0}^{1} \frac{\ln(w) \ln(1-w)}{(1+w)}dw \stackrel{IBP}{=} \underbrace{\ln(w)\ln(w+1)\ln(1-w)\Big|_{0}^{1}}_{=0} - \underbrace{\int_{0}^{1} \frac{\ln(w+1)\ln(1-w)}{w} dw}_{K_{1}} + \underbrace{\int_{0}^{1}\frac{\ln(w)\ln(w+1)}{1-w}dw}_{K_{2}}$$ $$\begin{align*} K_{1}=\int_{0}^{1}\frac{\ln(1-x)\ln(1+x)}{x} dx= & \int_{0}^{1}\sum_{n=1}^{\infty} \frac{(-1)^{n+1}x^n}{n}\frac{\ln(1-x)}{x} dx\\ =&\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\int_{0}^{1} x^{n-1}\ln(1-x) dx\\ =&\sum \frac{(-1)^{n+1}}{n}\int_{0}^{1}\sum_{n=1}^{\infty} x^{n-1} \left[\frac{d}{dt}\Big|_{t=0+} (1-x)^t\right] dx\\ =& \frac{d}{dt}\Big|_{t=0+} \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\int_{0}^{1}\sum x^{n-1}(1-x)^t dx\\ =&\frac{d}{dt}\Big|_{t=0+} \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}B(n,t+1)\\ =& \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n}\left(\psi^{(0)}(n+1)+\gamma\right)B(n,1)\\ =& \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^2}H_{n}\\ \end{align*}$$ $$\begin{align*} K_{2} = \int_{0}^{1}\frac{\ln(w)\ln(w+1)}{1-w}dw = &\int_{0}^{1} \frac{\ln(1-s)\ln(2-s)}{s}ds \\ =& \int_{0}^{1} \frac{\ln(1-s)\ln\left(2\left(1-\frac{s}{2}\right)\right)}{s}ds\\ =&\ln(2) \int_{0}^{1} \frac{\ln(1-s)}{s}ds + \int_{0}^{1} \frac{\ln(1-s)\ln\left(1-\frac{s}{2}\right)}{s}ds \\ =& -\ln(2) \int_{1}^{0} \frac{\ln(1-s)}{s}ds - \int_{0}^{1} \frac{\ln(1-s)}{s} \sum_{n=1}^{\infty} \frac{s^{n}}{2^nn}ds \\ =& -\ln(2)\operatorname{Li}_{2}(1) - \sum_{n=1}^{\infty} \frac{1}{2^nn}\int_{0}^{1} s^{n-1}\ln(1-s)ds \\ =& -\frac{\ln(2)\pi^2}{6} - \sum_{n=1}^{\infty} \frac{1}{2^nn}\int_{0}^{1} s^{n-1} \left(\frac{d}{dt}\Big|_{t=0+} (1-s)^t\right) ds\\ =& -\frac{\ln(2)\pi^2}{6} - \frac{d}{dt}\Big|_{t=0+} \sum_{n=1}^{\infty} \frac{1}{2^nn}\int_{0}^{1} s^{n-1} (1-s)^t ds\\ =& -\frac{\ln(2)\pi^2}{6} - \frac{d}{dt}\Big|_{t=0+} \sum_{n=1}^{\infty} \frac{1}{2^nn}B(n,t+1)\\ =& -\frac{\ln(2)\pi^2}{6} + \sum_{n=1}^{\infty} \frac{H_{n}}{2^nn^2}\\ \end{align*}$$ $$\begin{align*} J_{2} = \int_{0}^{1} \frac{\ln(1-w)\ln(w)}{1-w}dw =& \underbrace{-\frac{1}{2} \ln(w)\ln^2(1-w)\Big|_{0}^{1}}_{=0} +\frac{1}{2}\int_{0}^{1} \frac{\ln^2(1-w)}{w}dw \\ =& \frac{1}{2}\int_{0}^{1} \frac{\ln^2(1-w)}{w}dw \\ =& -\frac{1}{2} \int_{0}^{1} \sum_{n=1}^{\infty} \frac{w^n}{n} \frac{\ln(1-w)}{w} dw \\ =& - \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n} \int_{0}^{1} w^{n-1} \left(\frac{d}{dt}\Big|_{t=0+} (1-w)^t \right) dw \\ =&\frac{d}{dt}\Big|_{t=0+} - \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n} \int_{0}^{1} w^{n-1}(1-w)^t dw\\ =& \frac{d}{dt}\Big|_{t=0+} - \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n}B(n,t+1)\\ =& \frac{1}{2} \sum_{n=1}^{\infty}\frac{H_{n}}{n^2} \end{align*}$$ From the generating function (I will add the proof of the generating functions to the appendix later) $$\sum_{j=1}^{\infty} \frac{H_{j}(1-z)^j}{j^2} = \frac{1}{2}\ln^2(z)\ln(1-z)+\operatorname{Li}_{2}(z)\ln(z) - \operatorname{Li}_{3}(z) + \operatorname{Li}_{3}(1-z)$$ If we put $z = \frac{1}{2}$ we have: $$\sum_{n=1}^{\infty} \frac{H_{n}}{2^nn^2} = \zeta(3) - \frac{\pi^2\ln(2)}{12}$$ If $z \to 0$ $$\sum_{n=1}^{\infty} \frac{H_{n}}{n^2} = \zeta(3)$$ while from the generating function $$\sum_{j=1}^{\infty} \frac{H_{j}(z)^j}{j^2} = \frac{1}{3}\ln^3(1-z) +\ln(1-z)\operatorname{Li}_{2}\left(\frac{z}{z-1}\right) + \operatorname{Li}_{3}\left(\frac{z}{z-1}\right) +2 \operatorname{Li}_{3}(z) \quad z\in\left[-1,\frac{1}{2}\right]$$ If we put $\displaystyle z = -1$ we have $$\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^2}H_{n} = -\frac{5}{8}\zeta(3)$$ Therefore $$J_{1} = -K_{1} + K_{2} = -\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^2}H_{n} + -\frac{\ln(2)\pi^2}{6} + \sum_{n=1}^{\infty} \frac{H_{n}}{2^nn^2} = \frac{13}{8}\zeta(3) - \frac{\pi^2\ln(2)}{4}$$ $$J_{2} = \frac{1}{2} \sum_{n=1}^{\infty}\frac{H_{n}}{n^2} = \zeta(3)$$ $$I_{3} = -\frac{1}{4}J_{1} -\frac{1}{4}J_{2} = -\frac{21}{32}\zeta(3)+\frac{\pi^2\ln(2)}{16}$$ $$I = -2\ln(2)I_{1} + 2I_{2} +2I_{3} = -2\ln(2)\frac{\pi^2}{16} + 2\frac{7\zeta(3)}{16} -\frac{42}{32}\zeta(3)+\frac{\pi^2\ln(2)}{8} = -\frac{7}{16}\zeta(3)$$ Hence, we can conclude $$\boxed{ \int_{0}^{\infty} \frac{x\ln(\sinh x)}{\sinh x \cosh x} dx = -\frac{7}{16}\zeta(3)}$$