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).


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 :)

  • 1
    $\begingroup$ Of course, you want a nontrivial solution that doesn't mention $G$, so e.g. $\int_0^\infty e^{-Gx/\pi}dx$ is banned. $\endgroup$
    – J.G.
    Commented Apr 12, 2021 at 18:49
  • $\begingroup$ @J.G. yes of course $\endgroup$
    – clathratus
    Commented Apr 12, 2021 at 19:00
  • $\begingroup$ Sorry, I've just noticed your original edit mentions that anyway. $\endgroup$
    – J.G.
    Commented Apr 12, 2021 at 19:03
  • 2
    $\begingroup$ Integrating $f(z)=\sum_{n\ge 1} \mu(n)e^{-nz}$ over the right interval(s) gives every $1/L(s,\chi)$. But... $\endgroup$
    – reuns
    Commented Apr 12, 2021 at 19:04

1 Answer 1


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 $n$.

  • 1
    $\begingroup$ What, explicitly, are $\chi_4$ and $\phi$ here? $\endgroup$
    – clathratus
    Commented Apr 12, 2021 at 22:47
  • 1
    $\begingroup$ @clathratus $\chi$ is a Dirichlet character and $\phi(n)$ is Euler's Totient function $\endgroup$
    – Graviton
    Commented Apr 13, 2021 at 0:15

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