Is there an integral for $\frac{\pi}{\mathrm{G}}$? 
I would like to find an integral of the form
$$\int_a^bf(x)dx=\frac{\pi}{\mathrm G},$$
or at least an infinite series
$$\sum_{n\ge k}a_n=\frac{\pi}{\mathrm G},$$ where $\mathrm G$ is Catalan's constant. These identities should be nontrivial (that is, $\int_0^1\pi/\mathrm G dx=\pi/\mathrm G$ and anything sufficiently similar does not count).

Context:
A while ago, I posted this question (Relationship between Catalan's Constant and $\pi$), and in the answers there were many wonderful integrals and infinite series involving $\pi\mathrm G$ and $\mathrm G/\pi$, but none for $\pi/\mathrm G$. I got the idea to search for an integral for $\pi/\mathrm G$, because it, along with the integrals
$$\color{blue}{\int_0^1 \ln\left(\frac{1-x}{1+x}\right)\ln\left(\frac{1-x^2}{1+x^2}\right)\frac{dx}{x}=\pi G}$$
$$\color{red}{\int_0^\frac{\pi}{2} x\ln\left(\cot\left(\frac{x}{2}\right)\left(\frac{\sec x}{2}\right)^4\right)dx=\pi G}$$
(from @Zacky) and
$$\frac {G}\pi=\int_0^1\frac{dx}{4\text{sech}^{-1}x}$$
(from @Quanto), would give a complete description of the multiplicative relationship between $\pi$ and $\mathrm G$.
Really, I guess we would only need an integral of the form $$\int_a^bg(x)dx=\frac1{\mathrm G},$$
because there are a plethora of sufficient integrals for $\pi$ (see here) such as $$\int_{-1}^{1}\frac{dx}{\sqrt{1-x^2}}=\pi,$$
and we could then write
$$\int_a^b\int_{-1}^1\frac{g(y)}{\sqrt{1-x^2}}dxdy=\frac{\pi}{\mathrm G},$$
which would probably be pretty easy to convert from a double integral to a single integral.
Of course there is the natural question "is this even possible?" I don't know. I have (clearly) never seen an integral for $1/\mathrm G$.
Do you have any ideas? Thanks :)
 A: An infinite series whose sum is $\pi/4G$.
Applying the idea of @reuns in the comment.
Let $\beta(s)=L(s,\chi_4)$ be Dirichlet beta function. Then we have
$$
\sum_{n=1}^{\infty} \frac{\chi_4(n)\phi(n)}{n^{s+1}} = \frac{\beta(s)}{\beta(s+1)}.  \ \ \ (1)
$$
Put $s=1$, we obtain the infinite series
$$
\sum_{n=1}^{\infty} \frac{\chi_4(n)\phi(n)}{n^2}=\frac{\beta(1)}{\beta(2)}=\frac{\pi}{4G}.
$$
By definition of $\chi_4$ below, we may also write
$$
\sum_{k=0}^{\infty} \frac{(-1)^k\phi(2k+1)}{(2k+1)^2}=\frac{\pi}{4G}.
$$
Derivation of (1):
$$
\begin{align}\frac{L(s,\chi_4)}{L(s+1,\chi_4)}&= \prod_p \frac{ 1-\frac{\chi_4(p)}{p^{s+1}}}{1-\frac{\chi_4(p)}{p^s}}\\
&=\prod_p \left(1-\frac{\chi_4(p)}{p^{s+1}}\right)\left(1+\frac{\chi_4(p)}{p^s} + \frac{\chi_4^2(p)}{p^{2s}} + \cdots \right)\\
&=\prod_p \left(1+\frac{\chi_4(p)}{p^s}\left(1-\frac1p\right) + \frac{\chi_4^2(p)}{p^{2s}}\left(1-\frac1p\right)+\cdots\right)\\
&=\sum_{n=1}^{\infty} \frac{\chi_4(n) \frac{\phi(n)}n}{n^s} \ \textrm{ if } \ \Re s >1.
\end{align}
$$
This derivation requires the change of order of summation, which is valid under absolute convergence. So, this is okay if $\Re s >1$.
The convergence of (1) at $s=1$.
For $s=1$, the convergence is a bit more subtle. We proceed as follows.
$$
\begin{align}
\sum_{n\leq x} \frac{\chi_4(n)\phi(n)}{n^2} &= \sum_{n\leq x} \frac{\chi_4(n)}n \sum_{d|n} \frac{\mu(d)}d \ \left(\textrm{We have} \ \frac{\phi(n)}n=\sum_{d|n}\frac{\mu(d)}d\right)\\
&=\sum_{d\leq x} \sum_{k\leq \frac xd} \frac{\mu(d)\chi_4(d)\chi_4(k)}{d^2k} \ (\textrm{Here, we use} \  n=dk )\\
&=\sum_{d\leq x} \frac{\mu(d)\chi_4(d)}{d^2} \sum_{k\leq \frac xd} \frac{\chi_4(k)}k\\
&=\sum_{d\leq x}\frac{\mu(d)\chi_4(d)}{d^2} \left( L(1,\chi_4) + O\left(\frac dx\right)\right)\\
&=\frac{L(1,\chi_4)}{L(2,\chi_4)} +O\left(\frac1x\right) + O\left(\sum_{d\leq x} \frac1{dx}\right)\\
&=\frac{L(1,\chi_4)}{L(2,\chi_4)} + O\left(\frac{\log x}x\right) \\
&=\frac{\pi}{4G}+ O\left(\frac{\log x}x\right).
\end{align}
$$
Definitions of $\chi_4, \phi, \mu$.
Here $\chi_4(n)$ (Dirichlet character modulo $4$) and $\phi(n)$ (Euler Totient function), $\mu(n)$ (Mobius function) are defined by
$$
\chi_4(n)=\begin{cases} 0 &\mbox{ if } n \mbox{ is even}\\
(-1)^k &\mbox{ if } n=2k+1\end{cases}
$$
$$
\phi(n)=n \prod_{p|n} \left(1-\frac 1p\right) \mbox{ is the number of $1\leq k\leq n$ coprime to $n$. }
$$
$$
\mu(n)=\begin{cases} 0 &\mbox{ if } n \ \mbox{ is not square-free}\\
(-1)^{\omega(n)} &\mbox{ if } n \ \mbox{is square-free, }\end{cases}
$$
where $\omega(n)$ is the number of distinct prime factors of $n$.
