This morning I wrote with the help of a CAS, and integral representation for the Apéry's constant $\zeta(3)$ and some standard formulas two formulas involving this constant. I would like to know if these were in the litetature (my purpose is to know if if it possible to justify these and/or exploit to get more nice statements than mine, or some formula more interesting than mine using the way that I've used).


The integral representation wich I've combined is a formula that appears on [1], or see the first formula from the post in this site [MathOverflow](https://mathoverflow.net/questions/270286/a-closed-form-for-an-integral-expressed-as-a-finite-series-of-zeta2k1-p). I've combined with standard formulas that you can see in Google Books (currently this page is free view on Google Books) or your library as page 91 of [2] *Section 6.3 The Gamma Function*. 



> 
> **Claim.** It seems that the following formulas
>  $$\zeta(3)=\frac{1}{7\pi}\int\_0^{\pi}x(\pi-x)\left(\psi\left(\frac{x+\pi}{2\pi}\right)-\psi\left(\frac{x}{2\pi}\right)\right)dx,\tag{1}$$
>  where $\psi(s)$ denotes the digamma function, and 
>  $$\zeta(3)=\frac{\pi^2}{7}\int\_0^{\pi/4}\frac{\cos2x+\log\tan x}{\log^3\tan x}\cdot\frac{dx}{(\sin x)(\cos x)}\tag{2}$$
>  are right.
> 
> 
> 


I found the formula $(1)$ in an attempt to combine Zurab's integral (the cited formula in MO's post) and Exercise 6.3.1 from [2] with the help of Wolfram Alpha online calculator; while than $(2)$ is a combination of Exercise 6.3.4 (from [2]) and Zurab's integral, combined with Exercise 6.3.2 (from [2]) and the help of mentioned online calculator.



> 
> **Questions.** 
> 
> 
> 1) Are known these formulas from the literature? Or are these a particular case of theorems from the literature? Then, please answer this Questions as a reference request and I try to search the articles and study those formulas. 
> 
> 
> 2) In other case and with the purpose to improve my results...
> 
> 
> ... my attempt was to write explicitly the difference of the terms involving the digamma function using a series expansion, with the purpose to swap the integral sign and the infinite series that I've evoked, but my calculations are failed by some undeterminate term. I would like to know if it is feasible to expand such terms involving the digamma function and interchange the integral sign and the series, or well a different approach is better to rewrite $(1)$ as a formula for the Apéry's constant with a good mathematical meaning.
> 
> 
> ... my attempt was using the help of the mentioned online calculator (standard time of computation, my code and scarce knowledges of programming with this CAS) get the closed-form for $(2)$. But currently for me $(2)$ is a conjecture. Can you justify than $(2)$ is right?
>  **Many thanks.**
> 
> 
> 


Of course, if these formulas were not in the literature feel free to provide me general feedback about these formulas adding a comment. 


References:
-----------


[1] Zurab Silagadze, *Sums of Generalized Harmonic Series. For Kids from Five to Fifteen*, RESONANCE, September 2015.


[2] Ram Murty, *Problems in Analytic Number Theory*, Second Edition, Sringer (2008).



Two integral representations for $\zeta(3)$ from Zurab's integral and standard formulas for the gamma function
For the second integral, with the substitutions $t=\tan x$, then $t=\exp u$ and taking advantage of the symmetry, it comes
\begin{align}
I\_2&=\frac{\pi^2}{7}\int\_0^{\pi/4}\frac{\cos2x+\log\tan x}{\log^3\tan x}\cdot\frac{dx}{(\sin x)(\cos x)}\\
&=\frac{\pi^2}{7}\int\_0^{1}\frac{\tfrac{1-t^2}{1+t^2}+\ln t}{\ln^3 t}\frac{dt}{t}\\
&=\frac{\pi^2}{7}\int\_{-\infty}^0 \frac{u-\tanh u}{u^3}\,du\\
&=\frac{\pi^2}{14}\int\_{-\infty}^\infty \frac{u-\tanh u}{u^3}\,du
\end{align} 
This last integral can be calculated using the residue theorem, closing the contour by the upper half-circle. Poles are situated at $u=(2n+1)i\pi/2$ with $n\geq 0$:
\begin{align}
I\_2&=2i\pi\frac{\pi^2}{14}\sum\_{n\geq 0}\frac{-1}{\left( (2n+1)i\pi/2 \right)^3}\\
&=\frac{8}{7}\sum\_{n\geq 0}\frac{1}{ (2n+1)^3}\\
&=\zeta(3)
\end{align} 


For the first, integral, changing $x\to \pi -x$, one obtains:
\begin{align}
I\_1&=\frac{1}{7\pi}\int\_0^{\pi}x(\pi-x)\left(\psi\left(\frac{x+\pi}{2\pi}\right)-\psi\left(\frac{x}{2\pi}\right)\right)dx\\
&=\frac{1}{7\pi}\int\_0^{\pi}x(\pi-x)\left(\psi\left(1-\frac{x}{2\pi}\right)-\psi\left(\frac{1}{2}-\frac{x}{2\pi}\right)\right)dx
\end{align}
Summing both expressions,
\begin{equation}
2I\_1=\frac{1}{7\pi}\int\_0^{\pi}x(\pi-x)\left[\psi\left(\frac{1}{2}+\frac{x}{2\pi}\right)-\psi\left(\frac{x}{2\pi}\right)+\psi\left(1-\frac{x}{2\pi}\right)-\psi\left(\frac{1}{2}-\frac{x}{2\pi}\right)\right]dx
\end{equation} 
Now, with the [reflection formula for the digamma function](http://dlmf.nist.gov/5.5.E4)
\begin{equation}
\psi\left(z\right)-\psi\left(1-z\right)=-\pi/\tan\left(\pi z\right)
\end{equation} 
it can be written as
\begin{align}
I\_1&=\frac{1}{14}\int\_0^{\pi}x(\pi-x)\left[\tan\left( \frac{x}{2\pi} \right)+\frac{1}{\tan\left( \frac{x}{2\pi} \right)}\right]\,dx\\
&=\frac{1}{7}\int\_0^{\pi}\frac{x(\pi-x)}{\sin\left( \frac{x}{\pi} \right)}dx\\
&=\frac{\pi^3}{7}\int\_0^{1}\frac{t(1-t)}{\sin\left( t \right)}dt\\
&=\zeta(3)
\end{align} 
where we have used the representation given [here](https://mathoverflow.net/questions/271001/are-these-two-new-ways-of-representing-odd-zeta-values-as-integrals-known/282087#282087).


