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How related are $G$ (Catalan's constant) and $\pi$?

I seem to encounter $G$ a lot when computing definite integrals involving logarithms and trig functions.

Example:

It is well known that $$G=\int_0^{\pi/4}\log\cot x\,\mathrm{d}x$$ So we see that $$G=\int_0^{\pi/4}\log\sin(x+\pi/2)\,\mathrm{d}x-\int_0^{\pi/4}\log\sin x\,\mathrm{d}x$$ So we set out on the evaluation of $$L(\phi)=\int_0^\phi\log\sin x\,\mathrm{d}x,\qquad \phi\in(0,\pi)$$ we recall that $$\sin x=x\prod_{n\geq1}\frac{\pi^2n^2-x^2}{\pi^2n^2}$$ Applying $\log$ on both sides, $$\log\sin x=\log x+\sum_{n\geq1}\log\frac{\pi^2n^2-x^2}{\pi^2n^2}$$ integrating both sides from $0$ to $\phi$, $$L(\phi)=\phi(\log\phi-3)+\sum_{n\geq1}\phi\log\frac{\pi^2n^2-\phi^2}{\pi^2n^2}+\pi n\log\frac{\pi n+\phi}{\pi n-\phi}$$ With the substitution $u=x+\pi/2$, $$ \begin{align} \int_0^\phi \log\cos x\,\mathrm{d}x=&\int_0^{\phi}\log\sin(x+\pi/2)\,\mathrm{d}x\\ =&\int_{\pi/2}^{\phi+\pi/2}\log\sin x\,\mathrm{d}x\\ =&\int_{0}^{\phi+\pi/2}\log\sin x\,\mathrm{d}x-\int_{0}^{\pi/2}\log\sin x\,\mathrm{d}x\\ =&L(\phi+\pi/2)+\frac\pi2\log2 \end{align} $$ So $$G=L\bigg(\frac{3\pi}4\bigg)-L\bigg(\frac\pi4\bigg)+\frac\pi2\log2$$ And after a lot of algebra, $$G=\frac\pi4\bigg(\log\frac{27\pi^2}{16}+2\log2-6\bigg)+\pi\sum_{n\geq1}\bigg[\frac14\log\frac{(16n^2-9)^3}{256n^4(16n^2-1)}+n\log\frac{(4n+3)(4n-1)}{(4n-3)(4n+1)}\bigg]$$

So yeah I guess I found a series for $G$ in terms of $\pi$, but are there any other sort of these representations of $G$ in terms of $\pi$?

really important edit

As it turns out, the series $$\frac\pi4\bigg(\log\frac{27\pi^2}{16}+2\log2-6\bigg)+\pi\sum_{n\geq1}\bigg[\frac14\log\frac{(16n^2-9)^3}{256n^4(16n^2-1)}+n\log\frac{(4n+3)(4n-1)}{(4n-3)(4n+1)}\bigg]$$ does not converge, however it is a simple fix, and the series $$G=\frac\pi4\bigg(\log\frac{3\pi\sqrt{3}}2-1\bigg)+\pi\sum_{n\geq1}\bigg[\frac14\log\frac{(16n^2-9)^3}{256n^4(16n^2-1)}+n\log\frac{(4n+3)(4n-1)}{(4n-3)(4n+1)}-1\bigg]$$ does converge to $G$.

Quite amazingly, we can use this to find a really neat infinite product identity. Here's how.

Using the rules of exponents and logarithms, we may see that $$\frac{G}\pi+\frac12-\log\bigg(3^{3/4}\sqrt{\frac\pi2}\bigg)=\sum_{n\geq1}\log\bigg[\frac1{4en}\bigg(\frac{(16n^2-9)^3}{16n^2-1}\bigg)^{1/4}\bigg(\frac{(4n+3)(4n-1)}{(4n-3)(4n+1)}\bigg)^n\bigg]$$ Then using the fact that $$\log\prod_{i}a_i=\sum_{i}\log a_i$$ We have $$\frac{G}\pi+\frac12-\log\bigg(3^{3/4}\sqrt{\frac\pi2}\bigg)=\log\bigg[\prod_{n\geq1}\frac1{4en}\bigg(\frac{(16n^2-9)^3}{16n^2-1}\bigg)^{1/4}\bigg(\frac{(4n+3)(4n-1)}{(4n-3)(4n+1)}\bigg)^n\bigg]$$ Then taking $\exp$ on both sides, $$\prod_{n\geq1}\frac1{4en}\bigg(\frac{(16n^2-9)^3}{16n^2-1}\bigg)^{1/4}\bigg(\frac{(4n+3)(4n-1)}{(4n-3)(4n+1)}\bigg)^n=\sqrt{\frac{2e}{3\pi\sqrt{3}}}e^{G/\pi}$$ Or perhaps more aesthetically, $$\prod_{n\geq1}\frac1{4en}\bigg(\frac{(16n^2-9)^3}{16n^2-1}\bigg)^{1/4}\bigg(\frac{(4n+3)(4n-1)}{(4n-3)(4n+1)}\bigg)^n=\sqrt{\frac{2}{3\pi\sqrt{3}}}\exp\bigg(\frac{G}{\pi}+\frac12\bigg)$$

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    $\begingroup$ Hello. I hope this and this will help you. $\endgroup$ – Rohan Dec 27 '18 at 8:50
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    $\begingroup$ The $\pi G$ constant appears many times in the explicit evaluation of hypergeometric series at $\pm 1$, see for instance arxiv.org/abs/1710.03221 $\endgroup$ – Jack D'Aurizio Dec 27 '18 at 9:03
  • $\begingroup$ Are you sure that your series is convergent? $\endgroup$ – FDP Dec 27 '18 at 9:11
  • $\begingroup$ @FDP I may have made some simplification errors, but I am almost certain that $$G=L(3\pi/4)-L(\pi/4)+\pi\log\sqrt2$$ See here: desmos.com/calculator/hndn0teed7 $\endgroup$ – clathratus Dec 27 '18 at 9:14
  • $\begingroup$ @JackD'Aurizio Thank you for that link, it's a fascinating paper! $\endgroup$ – clathratus Dec 27 '18 at 9:32
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\begin{align}\sum_{n=0}^\infty \frac{\binom{2n}{n}^2}{4^{2n+1}(2n+1)}=\frac{\text{G}}{\pi}\tag1\end{align}

(see p81, Deriving Forsyth-Glaisher type series for $\frac{1}{\pi}$ and Catalan's constant by an elementary method. )

From the same source,

\begin{align}\sum_{n=0}^\infty \frac{\binom{2n}{n}^2}{16^n(2n+3)}=\frac{\text{G}}{\pi}+\frac{1}{2\pi}\tag2\end{align}

ADDENDUM:

Proof for (1),

It is well known that for $n\geq 0$ integer,

\begin{align}\int_0^{\frac{\pi}{2}}\cos^{2n} x\,dx=\frac{\pi}{2}\cdot\frac{\binom{2n}{n}}{4^n}\end{align}

(Wallis formula)

Therefore for $n\geq 0$ integer,

\begin{align}\frac{\binom{2n}{n}^2\pi^2}{4^{2n+1}(2n+1)}=\int_0^1 \left(\int_0^\infty \int_0^\infty t^{2n}\cos^{2n}x \cos^{2n}y \,dx\,dy \right)\,dt\end{align}

therefore,

\begin{align}\pi^2\sum_{n=0}^{\infty}\frac{\binom{2n}{n}^2}{4^{2n+1}(2n+1)}&=\sum_{n=0}^{\infty}\left(\int_0^1 \left(\int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}} t^{2n}\cos^{2n}x \cos^{2n}y \,dx\,dy \right)\,dt\right)\\ &=\int_0^1 \left(\int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}} \left(\sum_{n=0}^{\infty}t^{2n}\cos^{2n}x \cos^{2n}y\right) \,dx\,dy \right)\,dt\\ &=\int_0^1 \left(\int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}} \frac{1}{1-t^2\cos^2 x\cos^2 y}\,dx\,dy \right)\,dt\\ \end{align}

Perform the change of variable $u=\tan x$,$v=\tan y$,

\begin{align}\pi^2\sum_{n=0}^{\infty}\frac{\binom{2n}{n}^2}{4^{2n+1}(2n+1)}&= \int_0^1 \left(\int_0^{\infty} \int_0^{\infty}\frac{1}{(1+u^2)(1+v^2)-t^2}\,du\,dv \right)\,dt\\ &=\int_0^1 \left(\int_0^\infty \frac{1}{\sqrt{1+v^2}}\left[\frac{\arctan\left(\frac{u\sqrt{1+v^2}}{\sqrt{1+v^2-t^2}}\right)}{\sqrt{1+v^2-t^2}}\right]_{u=0}^{u=\infty}\,dv\right)\,dt\\ &=\frac{\pi}{2}\int_0^1 \left(\int_0^\infty \frac{1}{\sqrt{1+v^2}\sqrt{1+v^2-t^2}}\,dv\right)\,dt\\ &=\frac{\pi}{2}\int_0^\infty \frac{1}{\sqrt{1+v^2}}\left[\arctan\left(\frac{t}{\sqrt{1+v^2-t^2}}\right)\right]_{t=0}^{t=1}\,dv\\ &=\frac{\pi}{2}\int_0^\infty \frac{\arctan\left(\frac{1}{v}\right)}{\sqrt{1+v^2}}\,dv\\ \end{align}

Perform the change of variable $y=\dfrac{1}{x}$,

\begin{align}\pi^2\sum_{n=0}^{\infty}\frac{\binom{2n}{n}^2}{4^{2n+1}(2n+1)}&=\frac{\pi}{2}\int_0^\infty \frac{\arctan x}{x\sqrt{1+x^2}}\,dx\\ \end{align}

Perform the change of variable $y=\arctan x$,

\begin{align}\pi^2\sum_{n=0}^{\infty}\frac{\binom{2n}{n}^2}{4^{2n+1}(2n+1)}&=\frac{\pi}{2}\int_0^{\frac{\pi}{2}}\frac{x}{\sin x} \,dx\\ &=\frac{\pi}{2}\Big[x\ln\left(\tan\left(\frac{x}{2}\right)\right)\Big]_0^{\frac{\pi}{2}}-\frac{\pi}{2}\int_0^{\frac{\pi}{2}}\ln\left(\tan\left(\frac{x}{2}\right)\right)\,dx\\ &=-\frac{\pi}{2}\int_0^{\frac{\pi}{2}}\ln\left(\tan\left(\frac{x}{2}\right)\right)\,dx\\ \end{align}

Perform the change of variable $y=\frac{x}{2}$,

\begin{align}\pi^2\sum_{n=0}^{\infty}\frac{\binom{2n}{n}^2}{4^{2n+1}(2n+1)}&= -\pi\int_0^{\frac{\pi}{4}}\ln(\tan x)\,dx\\ &=\pi\times \text{G}\\ \end{align}

Therefore,

\begin{align}\boxed{\sum_{n=0}^\infty \frac{\binom{2n}{n}^2}{4^{2n+1}(2n+1)}=\frac{\text{G}}{\pi}}\end{align}

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  • $\begingroup$ I just looked at this again and I was struck by it's elegance. Really nice work. $\endgroup$ – clathratus Jan 30 at 22:23
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As is detailed here, there are many representations of Catalan's constant, even in terms of alternating infinite sums of polynomial reciprocals - see equations $(20)$ through $(32)$. Equation $(9)$ provides a very nice form including $\pi$, $$G=\frac{\pi^2}8-2\sum_{k\ge 0}\frac1{(4k+3)^2}$$ but it is derived from $\zeta(2)$. Therefore it shouldn't be surprising as values of $\zeta(2s)$ for a positive integer $s$ are fractions of $\pi^2$. Another one from Wikipedia gives $$8G=\pi\log(2+\sqrt3)+\sum_{k\ge0}\frac3{(2k+1)^2\binom{2k}k}.$$

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  • $\begingroup$ These are really nice (+1). Would you happen to know if any of them are a simplified version of the series I found? $\endgroup$ – clathratus Dec 27 '18 at 9:00
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    $\begingroup$ Obviously all of these expressions and yours should be equivalent. Perhaps yours is a combination of the two, as I see factors of $4k\pm3$, $4k\pm1$ and logarithms in various places in the expression. $\endgroup$ – TheSimpliFire Dec 27 '18 at 9:03
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    $\begingroup$ Minor nitpick: the last central binomial coefficient should be $\binom{2k}{k}$, I believe. The last identity can be proved through the Beta function and the dilogarithms machinery. $\endgroup$ – Jack D'Aurizio Dec 27 '18 at 9:05
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Let us give a self-contained proof of Ramanujan's identity $$\sum_{n\geq 0}\left[\frac{1}{4^n}\binom{2n}{n}\right]^2\frac{1}{2n+1}=\frac{4G}{\pi}.\tag{1}$$ We may recall the Maclaurin series of the complete elliptic integral of the first kind (in the following, the argument of $K$ is the elliptic modulus) $$ K(x)=\frac{\pi}{2}\sum_{n\geq 0}\left[\frac{1}{4^n}\binom{2n}{n}\right]^2 x^n \tag{2}$$ such that the LHS of $(1)$ blatantly is $\frac{2}{\pi}\int_{0}^{1}K(x^2)\,dx$ or $$ \frac{1}{\pi}\int_{0}^{1}\frac{K(x)}{\sqrt{x}}\,dx.\tag{3}$$ Due to the generating function for Legendre polynomials, both $K(x)$ and $\frac{1}{\sqrt{x}}$ have very simple FL (Fourier-Legendre) expansions, namely $$ K(x)=\sum_{m\geq 0}\frac{2}{2m+1}P_m(2x-1),\qquad \frac{1}{\sqrt{x}}=\sum_{m\geq 0}2(-1)^m P_m(2x-1) \tag{4} $$ hence by the orthogonality relation $\int_{0}^{1}P_n(2x-1)P_m(2x-1)\,dx=\frac{\delta(m,n)}{2n+1}$ we get $$ \sum_{n\geq 0}\left[\frac{1}{4^n}\binom{2n}{n}\right]^2\frac{1}{2n+1} = \frac{4}{\pi}\sum_{m\geq 0}\frac{(-1)^m}{(2m+1)^2}=\frac{4G}{\pi}\tag{5}$$ QED.

This approach is powerful enough to let you compute much worse.

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  • $\begingroup$ An elementary proof is possible using Wallis formula. (see my answer) $\endgroup$ – FDP Dec 27 '18 at 18:58
  • $\begingroup$ What is the $P_m(\cdots)$ function here? $\endgroup$ – clathratus Jan 1 at 21:05
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    $\begingroup$ @clathratus: $P_m$ is the $m$-th Legendre polynomial. $\endgroup$ – Jack D'Aurizio Jan 1 at 21:07
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For some integrals: $$\color{blue}{\int_0^1 \ln\left(\frac{1-x}{1+x}\right)\ln\left(\frac{1-x^2}{1+x^2}\right)\frac{dx}{x}=\pi G}$$ $$\color{red}{\int_0^\frac{\pi}{2} x\ln\left(\cot\left(\frac{x}{2}\right)\left(\frac{\sec x}{2}\right)^4\right)dx=\pi G}$$

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    $\begingroup$ In the left hand of the first formula perform the change of variable $y=\dfrac{1-x}{1+x}$ and you are ready to apply my favorite method ;) $\endgroup$ – FDP Dec 27 '18 at 11:26
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    $\begingroup$ Since \begin{align}\pi^2 \times \frac{\text{G}}{\pi}=\pi\text{G}\end{align} your formula can be probably used to prove the one i have written below (the first one) $\endgroup$ – FDP Dec 27 '18 at 11:33
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Here is a selection of formulas stated in section 1.7 Catalan's Constant, $G$ of Mathematical constants by Steven R. Finch

A nice coincidence:

\begin{align*} \frac{\pi^2}{12\ln(2)}&=\left(1-\frac{1}{2^2}+\frac{1}{3^2}-\frac{1}{4^2}+-\cdots\right)\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+-\cdots\right)^{-1}\\ \frac{4G}{\pi}&=\left(1-\frac{1}{3^2}+\frac{1}{5^2}-\frac{1}{7^2}+-\cdots\right)\left(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+-\cdots\right)^{-1}\\ \end{align*} and the variation \begin{align*} \frac{8G}{\pi^2}&=\left(1-\frac{1}{3^2}+\frac{1}{5^2}-\frac{1}{7^2}+-\cdots\right)\left(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots\right)^{-1}\\ \end{align*}


Series:

\begin{align*} \sum_{k=0}^\infty \frac{1}{(2k+1)^2\binom{2k}{k}}&=\frac{8}{3}G-\frac{\pi}{3}\ln(2+\sqrt{3})\\ \sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^2}\sum_{k=1}^n\frac{1}{k+n}&=\pi G-\frac{33}{16}\zeta(3) \end{align*}

A series obtained by Ramanujan:

\begin{align*} G=\frac{5}{48}\pi^2-2\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)^2\left(e^{\pi (2k+1)}-1\right)}-\frac{1}{4}\sum_{k=1}^\infty\frac{\mathrm{sech} (\pi k)}{k^2} \end{align*}


Integrals:

\begin{align*} 4\int_{0}^1\frac{\arctan(x)^2}{x}\,dx=\int_0^{\frac{\pi}{2}}\frac{x^2}{\sin (x)}\,dx=2\pi G-\frac{7}{2}\zeta(3) \end{align*}

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  • $\begingroup$ These are really nice (+1). Thanks for your answer $\endgroup$ – clathratus Dec 27 '18 at 19:11
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    $\begingroup$ @clathratus: You're welcome. $\endgroup$ – Markus Scheuer Dec 27 '18 at 19:13

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