Justify $\zeta(3)=2\int_0^1 \left(Li_2(e^{-2\pi i x})+Li_2(e^{2\pi i x}\right))\log \Gamma(x)dx$

Question

I don't know if this approach to get a formula involving the Apéry constant was in the literature. This idea was a simple idea few minutes ago, when I was studying the answers in this site Math Stack Exchange for the question Integral $\int_0^1 \log \Gamma(x)\cos (2\pi n x)\, dx=\frac{1}{4n}$.

One has that since Wolfram Alpha said that $$\sum_{k=1}^\infty\frac{\cos (2 \pi k x)}{k^2}=\frac{Li_2(e^{-2\pi i x})+Li_2(e^{2\pi i x})}{2},$$ where $Li_s(z)$ is the polylogarithm function. Then using the dominated convergence theorem we should have then $$\zeta(3)=2\int_0^1 \left(Li_2(e^{-2\pi i x})+Li_2(e^{2\pi i x}\right))\log \Gamma(x)dx .$$

Question. Please can you justify all these claims to provide us this nice exercise for this site Mathematics Stack Exchange? I say justify the closed-form for the series involving the cosine function (if you find a reference in this site, only is required add it) and after jusfity how one uses the dominated convergence theorem. Thanks in advance.

With respect the use of the theorem, I know that I can bound the cosine function, but how can one bounds $|\log \Gamma(x) |$ for $0<x<1$?

You can see the calculation from the online calculator with this code

int_0^1 log (Gamma(x))2(Li_2(e^(-2 i pi x))+Li_2(e^(2 i pi x)))dx.

Notice that the imaginary part is $\approx 0$.

Answer 1

So we have to prove that $$ \frac{\zeta(3)}{4} = \int_{0}^{1}\sum_{k\geq 1}\frac{\cos(2\pi k x)}{k^2}\log\Gamma(x)\,dx $$ but over the iterval $(0,1)$ the graph of the Fourier cosine series $\sum_{k\geq 1}\frac{\cos(2\pi k x)}{k^2}$ is just the graph of a parabola, since $\sum_{k\geq 1}\frac{\sin(2\pi k x)}{k}$ is a sawtooth wave. Moreover, such a parabola is symmetric with respect to the point $x=\frac{1}{2}$. The claim boils down to: $$ \frac{\zeta(3)}{4} = \int_{0}^{1/2}\left(\frac{\pi^2}{6}-\pi^2(x-x^2)\right)\log\left(\Gamma(x)\Gamma(1-x)\right)\,dx $$ or, through the reflection formula for the $\Gamma$ function and the substitution $x\to\frac{1}{2}-x$: $$ \frac{\zeta(3)}{4}=\frac{\pi^2}{12}\int_{0}^{1/2}(12x^2-1)\log\left(\frac{\pi}{\cos(\pi x)}\right)\,dx$$ that is equivalent to: $$ 3\zeta(3)=\pi^2\int_{0}^{1/2}(1-12x^2)\log(\cos(\pi x))\,dx \tag{1a}$$ or to: $$ \frac{3\zeta(3)}{\pi^3} = \int_{0}^{1/2}\tan(\pi x)(x-4x^3)\,dx,\tag{1b}$$ $$ \frac{\zeta(3)}{\zeta(2)} = \int_{0}^{1}\frac{\pi}{2}\cot(\pi x/2)x(x-1)(x-2)\,dx.\tag{1c}$$ Since the Weierstrass product of the cosine function gives $$ \cos(\pi x)=\prod_{n\geq 0}\left(1-\frac{4 x^2}{(2n+1)^2}\right)\tag{2}$$ and $$ \int_{0}^{1/2}(1-12x^2)\log\left(1-\frac{4 x^2}{(2n+1)^2}\right)\,dx \\= \frac{1}{3}+4 n (1+n)+2 n (1+n) (1+2 n) (\log(n)-\log(n+1)) \tag{3}$$ the whole question boils down to an exercise in summation by parts involving the derivative of the $\zeta$ function (in particular, $\zeta'(-2)$, that is related with $\zeta(3)$ via the reflection formula for the $\zeta$ function). As an alternative, $(1)$ can be checked by expanding both $12x^2-1$ and $\log\cos(\pi x)$ as Fourier series. On the other hand, Chen Wang's solution to the linked question just relies on Kummer's Fourier series for $\log\Gamma$. The Weierstrass product also provides the identities $$ \frac{\pi}{2}\cot\left(\frac{\pi x}{2}\right)=\frac{1}{x}+\sum_{n\geq 1}\left(\frac{1}{x-2n}+\frac{1}{x+2n}\right) \tag{4}$$ $$ \tan\left(\frac{\pi x}{2}\right)=\frac{4}{\pi}\sum_{n\geq 0}\frac{x}{(2n+1)^2-x^2} \tag{5} $$ that can be used to directly tackle $(1b)$ or $(1c)$.

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Now we have a situation that often occurs in mathematics: when proving something, we go through some lemma that is crucial for other situations. In this case we have that the RHS of $(3)$ is summable over $n\geq 1$, so the "smallness" of such term gives us that $$ \frac{4n(n+1)+\frac{1}{3}}{2n(2n+1)(n+1)} $$ is an excellent approximation of $\log\left(1+\frac{1}{n}\right)$ for any $n\geq 1$, and we may even estimate the error by approximating the associated integral with the Cauchy-Schwarz inequality, integration by parts or other techniques. In our case we get that the approximation error is less than $1.3\cdot 10^{-3}$ for any $n\geq 1$, and $$ \log(2)\approx\left(\frac{5}{6}\right)^2. $$

Answer 2

Comment : from here http://functions.wolfram.com/ZetaFuncti ... owAll.html We know that $$\displaystyle{L{i_2}\left( z \right) + L{i_2}\left( {\frac{1}{z}} \right) = - \frac{1}{2}{\log ^2}\left( { - z} \right) - \frac{{{\pi ^2}}}{6}} ,$$

Consequently $$\displaystyle{L{i_2}\left( {{e^{ - 2i\pi x}}} \right) + L{i_2}\left( {{e^{2i\pi x}}} \right) = - \frac{1}{2}{\log ^2}\left( { - {e^{2i\pi x}}} \right) - \frac{{{\pi ^2}}}{6}}$$

$$\displaystyle{S = 2\int\limits_0^1 {\left( {L{i_2}\left( {{e^{ - 2i\pi x}}} \right) + L{i_2}\left( {{e^{2i\pi x}}} \right)} \right)\log \left( {\Gamma \left( x \right)} \right)dx} = }$$ $$\displaystyle{2\int\limits_0^1 {\left( { - \frac{1}{2}{{\log }^2}\left( { - {e^{2i\pi x}}} \right) - \frac{{{\pi ^2}}}{6}} \right)\log \left( {\Gamma \left( x \right)} \right)dx} = }$$

$$\displaystyle{ = \int\limits_0^1 {\left( { - {{\log }^2}\left( {{e^{i\pi \left( {2x - 1} \right)}}} \right) - \frac{{{\pi ^2}}}{3}} \right)\log \left( {\Gamma \left( x \right)} \right)dx} = \int\limits_0^1 {\left( { - {{\left( {i\pi \left( {2x - 1} \right)} \right)}^2} - \frac{{{\pi ^2}}}{3}} \right)\log \left( {\Gamma \left( x \right)} \right)dx} = }$$

$$\displaystyle{ = \int\limits_0^1 {\left( {{\pi ^2}{{\left( {2x - 1} \right)}^2} - \frac{{{\pi ^2}}}{3}} \right)\log \left( {\Gamma \left( x \right)} \right)dx} = {\pi ^2}\int\limits_0^1 {\left( {{{\left( {2x - 1} \right)}^2} - \frac{1}{3}} \right)\log \left( {\Gamma \left( x \right)} \right)dx} \mathop { = = = = }\limits^{x \to 1 - x} }$$

$$\displaystyle{ = {\pi ^2}\int\limits_0^1 {\left( {{{\left( {2x - 1} \right)}^2} - \frac{1}{3}} \right)\log \left( {\Gamma \left( {1 - x} \right)} \right)dx} \Rightarrow 2S = }$$ $$\displaystyle{{\pi ^2}\int\limits_0^1 {\left( {{{\left( {2x - 1} \right)}^2} - \frac{1}{3}} \right)\left( {\log \left( {\Gamma \left( {1 - x} \right)} \right) + \log \left( {\Gamma \left( x \right)} \right)} \right)dx} = }$$

$$\displaystyle{ = {\pi ^2}\int\limits_0^1 {\left( {{{\left( {2x - 1} \right)}^2} - \frac{1}{3}} \right)\log \left( {\Gamma \left( x \right)\Gamma \left( {1 - x} \right)} \right)dx} = {\pi ^2}\int\limits_0^1 {\left( {{{\left( {2x - 1} \right)}^2} - \frac{1}{3}} \right)\log \left( {\frac{\pi }{{\sin \pi x}}} \right)dx} = }$$

$$\displaystyle{ = {\pi ^2}\log \left( \pi \right)\underbrace {\int\limits_0^1 {\left( {{{\left( {2x - 1} \right)}^2} - \frac{1}{3}} \right)dx} }_{ = 0} - {\pi ^2}\int\limits_0^1 {\left( {{{\left( {2x - 1} \right)}^2} - \frac{1}{3}} \right)\log \left( {\sin \pi x} \right)dx} }$$ $>\displaystyle{ = - {\pi ^2}\int\limits_0^1 {\left( {{{\left( {2x - 1} \right)}^2} - \frac{1}{3}} \right)\log \left( {\sin \pi x} \right)dx} }$$

Then $$\displaystyle{S = - \frac{{{\pi ^2}}}{2}\int\limits_0^1 {\left( {{{\left( {2x - 1} \right)}^2} - \frac{1}{3}} \right)\log \left( {\sin \pi x} \right)dx} }.$$ But $$\displaystyle{\sum\limits_{n = 1}^\infty {\frac{{\cos nx}}{n}} = - \log 2 - \log \left( {\sin \frac{x}{2}} \right)}$$ (it turned out multiple times)

so $$\displaystyle{\log \left( {\sin \pi x} \right) = - \log 2 - \sum\limits_{n = 1}^\infty {\frac{{\cos 2n\pi x}}{n}} }.$$ Consequently $$\displaystyle{S = \frac{{{\pi ^2}}}{2}\int\limits_0^1 {\left( {{{\left( {2x - 1} \right)}^2} - \frac{1}{3}} \right)\left( {\log 2 + \sum\limits_{n = 1}^\infty {\frac{{\cos 2n\pi x}}{n}} } \right)dx} = }$$

$$\displaystyle{ = \frac{{{\pi ^2}}}{2}\log 2\int\limits_0^1 {\left( {{{\left( {2x - 1} \right)}^2} - \frac{1}{3}} \right)dx} + \frac{{{\pi ^2}}}{2}\sum\limits_{n = 1}^\infty {\frac{1}{n}\int\limits_0^1 {\left( {{{\left( {2x - 1} \right)}^2} - \frac{1}{3}} \right)\cos 2n\pi xdx} } }$$ $$\displaystyle{ = \frac{{{\pi ^2}}}{2}\sum\limits_{n = 1}^\infty {\frac{1}{n}\int\limits_0^1 {{{\left( {2x - 1} \right)}^2}\cos 2n\pi x\,dx} } }$$

But $$\displaystyle{\int\limits_0^1 {{{\left( {2x - 1} \right)}^2}\cos 2n\pi x\,dx} = \frac{2}{{{n^2}{\pi ^2}}}} (elementary) and finally \displaystyle{S = \frac{{{\pi ^2}}}{2}\sum\limits_{n = 1}^\infty {\frac{2}{{{n^3}{\pi ^2}}}} = \zeta \left( 3 \right)}$$ :) :)

Answer 3

With completely Fourier ( somewhat similar ) we get the following:

One can easily verify $(\dagger)$ identity $$\displaystyle{\sum_{n=1}^\infty\frac{\cos (2 \pi n x)}{n^2}=\frac{{\rm Li}_2(e^{-2\pi i x})+{\rm Li}_2(e^{2\pi i x})}{2}}$$ Then we have sequentially: $$\displaystyle{\begin{aligned} \int_0^1 \bigg({\rm Li}_2 \left(e^{-2\pi i x} \right)+{\rm Li}_2 \left(e^{2\pi i x} \right) \bigg)\log \Gamma(x) \; {\rm d}x &= 2\int_{0}^{1} \log \Gamma(x) \sum_{n=1}^{\infty} \frac{\cos 2 \pi n x}{n^2} \, {\rm d}x\\ &= 2\sum_{n=1}^{\infty} \frac{1}{n^2} \int_{0}^{1}\cos 2 n \pi x \log \Gamma(x) \, {\rm d}x\\ &\overset{(*)}{=} 2 \sum_{n=1}^{\infty} \frac{1}{4n^3} \\ &= \frac{\zeta(3)}{2} \end{aligned}}$$ $\dagger)$ It is left as an exercise to the reader. Of course, the sum of polylogarithms falls into something elementary, since for example it is true $$\displaystyle{\sum_{n=1}^{\infty} \frac{\sin nx}{n} = \frac{\pi-x}{2} \quad , \quad x \in (0, 2\pi)}$$ where with proper manipulation and integration we get the formula which does not contain polylogarithms.

(*) These are the Fourier a_n coefficients of the spread $\log \Gamma.$ Namely $$\displaystyle{\int_{0}^{1} \cos 2 n \pi x \log \Gamma(x) \, {\rm d}x = \frac{1}{4n}}$$