TITLE: Relationship between Catalan's constant and $\pi$
QUESTION [21 upvotes]: 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,
$$\prod_{n\geq1}\frac1{4en}\bigg(\frac{(16n^2-9)^3}{16n^2-1}\bigg)^{1/4}\bigg(\frac{(4n+3)(4n-1)}{(4n-3)(4n+1)}\bigg)^n=\sqrt{\frac{2}{3\pi\sqrt{3}}}\exp\bigg(\frac{G}{\pi}+\frac12\bigg)$$

REPLY [2 votes]: They are related via the integral
$$\frac {G}\pi=\int_0^1\frac{dx}{4\text{sech}^{-1}x}
$$