Relationship between Catalan's constant and $\pi$

Question

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, $$\prod_{n\geq1}\frac1{4en}\bigg(\frac{(16n^2-9)^3}{16n^2-1}\bigg)^{1/4}\bigg(\frac{(4n+3)(4n-1)}{(4n-3)(4n+1)}\bigg)^n=\sqrt{\frac{2}{3\pi\sqrt{3}}}\exp\bigg(\frac{G}{\pi}+\frac12\bigg)$$

Answer 1

\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^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}} 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}

Answer 2

For some integrals: $$\color{blue}{\int_0^1 \ln\left(\frac{1-x}{1+x}\right)\ln\left(\frac{1-x^2}{1+x^2}\right)\frac{dx}{x}=\pi G}$$ $$\color{red}{\int_0^\frac{\pi}{2} x\ln\left(\cot\left(\frac{x}{2}\right)\left(\frac{\sec x}{2}\right)^4\right)dx=\pi G}$$

Answer 3

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.

Answer 4

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 $$8G=\pi\log(2+\sqrt3)+\sum_{k\ge0}\frac3{(2k+1)^2\binom{2k}k}.$$

Answer 5

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*}

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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*}

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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*}

Answer 6

Two neat examples from (Almost) Impossible Integrals, Sums, and Series are

$$i) \ \int_0^1\frac{\log(x)\operatorname{Li}_2(-x)}{1+x^2}\textrm{d}x=\frac{1}{48}\pi^2 G;$$

$$ii) \ \int_0^1 \arctan(x)\log(x) \operatorname{Li}_2(-x) \textrm{d}x$$ $$=\frac{1}{2} G^2+\frac{\pi}{4} G+\frac{\log(2)}{2}G-\frac{13}{2560}\pi^4+\frac{\pi^3}{192} -\frac{9}{32}\zeta(3)-\frac{7}{96}\log(2)\pi^2-\frac{3}{4}\log(2)\pi$$ $$+\frac{3}{4}\pi-\frac{ 3}{2}\log(2)+ \frac{1}{4}\log^2(2).$$

I would also find interesting the cases involving a product between $\log(2)$ and $G$ (like at the point $ii)$).

I also add a magical example from Romanian Mathematical Magazine, proposed by the author of the book previously mentioned (for $\sim2$ years the magazine has received no solution).

$$ iii) \ \int _0^{\pi/4}\int _0^{\pi/4}\frac{\left(\tan ^2(x)+\tan ^2(y)\right) \log (\tan (x)) \log (\tan (y)) \operatorname{Ti_2(\tan(x)\tan(y))}}{\tan (x) \tan (y)}\textrm{d}x \textrm{d}y$$ $$=\frac{1}{368640}\psi ^{(5)}\left(\frac{1}{4}\right)-\frac{\pi^6}{1440}-\frac{2 }{3}G^3.$$

I noticed in comments the OP enjoyed the closed-form with $G^2$. Thus, I also added one with $G^3$.

Answer 7

They are related via the integral

$$\frac {G}\pi=\int_0^1\frac{dx}{4\text{sech}^{-1}x} $$

Answer 8

If $\mu(k)=((2^k-1)/2^k)\,\lambda(k)=((2^k-1)/2^k)^2\,\zeta(k)$ then

$$\sum_{k=1}^\infty \frac{\mu(2k)}{k(2k+1)}=\sum_{k=1}^\infty \left(\frac{2^{2k}-1}{2^{2k}}\right)\frac{\lambda(2k)}{k(2k+1)}=\sum_{k=1}^\infty \left(\frac{2^{2k}-1}{2^{2k}}\right)^2\frac{\zeta(2k)}{k(2k+1)}=\frac{2G}{\pi}\tag{1}$$

which arises from the series expansion $$\ln\left( \frac{\cos(x/2)}{\cos(x)}\right)=\sum_{k=1}^\infty \frac{2^{2k}\mu(2k)}{k\,\pi^{2k}}x^{2k}\tag{2}$$

When either side of this identity is integrated between the limits $0$ and $\pi/2$, $G$ results.

It is similar to another slowly converging series for $\ln2$

$$\sum_{k=1}^\infty \frac{\lambda(2k)}{k(2k+1)}=\ln2\tag{3} $$

Another similar series for $G/\pi$ is

$$\frac{G}{\pi}=\sum_{k=1}^\infty \frac{\lambda(2k)}{2^{2k}k}\left( 1-\frac{1}{2(2k+1)}\right)\tag{4}$$

which is obtained by combining the identities

$$\sum_{k=1}^\infty \frac{\lambda(2k)}{2^{2k}k}=\frac{\ln2}{2}$$

and

$$\sum_{k=1}^\infty \frac{\lambda(2k)}{2^{2k}k(2k+1)}=\ln2-\frac{2G}\pi{}$$

Answer 9

A relation between $\mathrm G$ and even $\zeta$ values.

Use the infinite product for the sine $$\sin t= t\prod_{k\ge1}\left(1-\frac{t^2}{\pi^2k^2}\right)$$ and $$\ln(1-z)=-\sum_{n\ge1}\frac{z^n}{n}\qquad z\in \{q\in\Bbb C\setminus \{1\}: |q|\le1\}$$ to get $$\begin{align} \ln\sin t &=\ln t+\sum_{k\ge1}\ln\left(1-\frac{t^2}{\pi^2k^2}\right)\\ &=\ln t+-\sum_{k\ge1}\sum_{n\ge1}\frac1n\left(\frac{t^2}{\pi^2k^2}\right)^n\\ &=\ln t-\sum_{n\ge1}\frac{t^{2n}}{\pi^{2n}n}\sum_{k\ge1}\frac1{k^{2n}}\\ &=\ln t-\sum_{n\ge1}\frac{\zeta(2n)}{\pi^{2n}n}t^{2n}. \end{align}$$ Thus $$\mathrm{Cl}_2(x)=-\int_0^x\ln\left(2\sin\tfrac{t}{2}\right)dt=-\int_0^x \ln tdt+\sum_{n\ge1}\frac{\zeta(2n)x^{2n+1}}{(2\pi)^{2n}(2n^2+n)}.\qquad 0<x<2\pi$$ That is, $$\sum_{n\ge1}\frac{\zeta(2n)x^{2n+1}}{2^{2n}(2n^2+n)}=x\ln\frac{e}{\pi x}+\frac1\pi\mathrm{Cl}_2(\pi x)\tag 1$$ which gives

$$\sum_{n\ge1}\frac{\zeta(2n)}{16^n (2n^2+n)}=1+\ln\frac2\pi-\frac{2\mathrm{G}}{\pi}.$$