Cfrgtkky

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
$$G=int_0^{pi/4}logcot x,mathrm{d}x$$
So we see that
$$G=int_0^{pi/4}logsin(x+pi/2),mathrm{d}x-int_0^{pi/4}logsin x,mathrm{d}x$$
So we set out on the evaluation of
$$L(phi)=int_0^philogsin x,mathrm{d}x,qquad phiin(0,pi)$$
we recall that
$$sin x=xprod_{ngeq1}frac{pi^2n^2-x^2}{pi^2n^2}$$
Applying $log$ on both sides,
$$logsin x=log x+sum_{ngeq1}logfrac{pi^2n^2-x^2}{pi^2n^2}$$
integrating both sides from $0$ to $phi$,
$$L(phi)=phi(logphi-3)+sum_{ngeq1}philogfrac{pi^2n^2-phi^2}{pi^2n^2}+pi nlogfrac{pi n+phi}{pi n-phi}$$
With the substitution $u=x+pi/2$,
$$
begin{align}
int_0^phi logcos x,mathrm{d}x=&int_0^{phi}logsin(x+pi/2),mathrm{d}x\
=&int_{pi/2}^{phi+pi/2}logsin x,mathrm{d}x\
=&int_{0}^{phi+pi/2}logsin x,mathrm{d}x-int_{0}^{pi/2}logsin x,mathrm{d}x\
=&L(phi+pi/2)+fracpi2log2
end{align}
$$
So
$$G=Lbigg(frac{3pi}4bigg)-Lbigg(fracpi4bigg)+fracpi2log2$$
And after a lot of algebra,
$$G=fracpi4bigg(logfrac{27pi^2}{16}+2log2-6bigg)+pisum_{ngeq1}bigg[frac14logfrac{(16n^2-9)^3}{256n^4(16n^2-1)}+nlogfrac{(4n+3)(4n-1)}{(4n-3)(4n+1)}bigg]$$

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
$$fracpi4bigg(logfrac{27pi^2}{16}+2log2-6bigg)+pisum_{ngeq1}bigg[frac14logfrac{(16n^2-9)^3}{256n^4(16n^2-1)}+nlogfrac{(4n+3)(4n-1)}{(4n-3)(4n+1)}bigg]$$
does not converge, however it is a simple fix, and the series
$$G=fracpi4bigg(logfrac{3pisqrt{3}}2-1bigg)+pisum_{ngeq1}bigg[frac14logfrac{(16n^2-9)^3}{256n^4(16n^2-1)}+nlogfrac{(4n+3)(4n-1)}{(4n-3)(4n+1)}-1bigg]$$
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
$$frac{G}pi+frac12-logbigg(3^{3/4}sqrt{fracpi2}bigg)=sum_{ngeq1}logbigg[frac1{4en}bigg(frac{(16n^2-9)^3}{16n^2-1}bigg)^{1/4}bigg(frac{(4n+3)(4n-1)}{(4n-3)(4n+1)}bigg)^nbigg]$$
Then using the fact that
$$logprod_{i}a_i=sum_{i}log a_i$$
We have
$$frac{G}pi+frac12-logbigg(3^{3/4}sqrt{fracpi2}bigg)=logbigg[prod_{ngeq1}frac1{4en}bigg(frac{(16n^2-9)^3}{16n^2-1}bigg)^{1/4}bigg(frac{(4n+3)(4n-1)}{(4n-3)(4n+1)}bigg)^nbigg]$$
Then taking $exp$ on both sides,
$$prod_{ngeq1}frac1{4en}bigg(frac{(16n^2-9)^3}{16n^2-1}bigg)^{1/4}bigg(frac{(4n+3)(4n-1)}{(4n-3)(4n+1)}bigg)^n=sqrt{frac{2e}{3pisqrt{3}}}e^{G/pi}$$
Or perhaps more aesthetically,
$$prod_{ngeq1}frac1{4en}bigg(frac{(16n^2-9)^3}{16n^2-1}bigg)^{1/4}bigg(frac{(4n+3)(4n-1)}{(4n-3)(4n+1)}bigg)^n=sqrt{frac{2}{3pisqrt{3}}}expbigg(frac{G}{pi}+frac12bigg)$$

begin{align}sum_{n=0}^infty frac{binom{2n}{n}^2}{4^{2n+1}(2n+1)}=frac{text{G}}{pi}tag1end{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}{2pi}tag2end{align}

ADDENDUM:

Proof for (1),

It is well known that for $ngeq 0$ integer,

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

(Wallis formula)

Therefore for $ngeq 0$ integer,

begin{align}frac{binom{2n}{n}^2pi^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),dtend{align}

therefore,

begin{align}pi^2sum_{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),dtright)\
&=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}yright) ,dx,dy right),dt\
&=int_0^1 left(int_0^{frac{pi}{2}} int_0^{frac{pi}{2}} frac{1}{1-t^2cos^2 xcos^2 y},dx,dy right),dt\
end{align}

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

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

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

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

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

begin{align}pi^2sum_{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[xlnleft(tanleft(frac{x}{2}right)right)Big]_0^{frac{pi}{2}}-frac{pi}{2}int_0^{frac{pi}{2}}lnleft(tanleft(frac{x}{2}right)right),dx\
&=-frac{pi}{2}int_0^{frac{pi}{2}}lnleft(tanleft(frac{x}{2}right)right),dx\
end{align}

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

begin{align}pi^2sum_{n=0}^{infty}frac{binom{2n}{n}^2}{4^{2n+1}(2n+1)}&=
-piint_0^{frac{pi}{4}}ln(tan x),dx\
&=pitimes 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}

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 $pi$, $$G=frac{pi^2}8-2sum_{kge 0}frac1{(4k+3)^2}$$ 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 $$8G=pilog(2+sqrt3)+sum_{kge0}frac3{(2k+1)^2binom{2k}k}.$$

Let us give a self-contained proof of Ramanujan's identity
$$sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^2frac{1}{2n+1}=frac{4G}{pi}.tag{1}$$
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)
$$ K(x)=frac{pi}{2}sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^2 x^n tag{2}$$
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
$$ K(x)=sum_{mgeq 0}frac{2}{2m+1}P_m(2x-1),qquad frac{1}{sqrt{x}}=sum_{mgeq 0}2(-1)^m P_m(2x-1) tag{4} $$
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
$$ sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^2frac{1}{2n+1} = frac{4}{pi}sum_{mgeq 0}frac{(-1)^m}{(2m+1)^2}=frac{4G}{pi}tag{5}$$
QED.

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

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