# 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 :)

• Of course, you want a nontrivial solution that doesn't mention $G$, so e.g. $\int_0^\infty e^{-Gx/\pi}dx$ is banned.
– J.G.
Commented Apr 12, 2021 at 18:49
• @J.G. yes of course Commented Apr 12, 2021 at 19:00
• Sorry, I've just noticed your original edit mentions that anyway.
– J.G.
Commented Apr 12, 2021 at 19:03
• Integrating $f(z)=\sum_{n\ge 1} \mu(n)e^{-nz}$ over the right interval(s) gives every $1/L(s,\chi)$. But... Commented Apr 12, 2021 at 19:04

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 $$n$$.

• What, explicitly, are $\chi_4$ and $\phi$ here? Commented Apr 12, 2021 at 22:47
• @clathratus $\chi$ is a Dirichlet character and $\phi(n)$ is Euler's Totient function Commented Apr 13, 2021 at 0:15