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. 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.


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.


[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).


1 Answer 1


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 \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.


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