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Is there a known expansion of $\pi$ as a function of the Euler-Mascheroni $\gamma$? As in, $$f(\gamma)=\pi h(\pi,\gamma)$$ where $\gamma$ appears alone with only rational arguments like $f(1/2,\gamma)$, and $h(\pi,\gamma)$ is a single, or finite closed form expression of the two constants, like $\pi \zeta(\pi\gamma)$, not considering (trivial) change of variables in integrals as in here. This page seems to have only $$h(\pi,\gamma)\gamma=f(\pi)$$ the inverse of what I ask.

nmasanta
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user687639
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  • I don't know which formula at mathworld you are referring to. $\int_0^{\infty}e^{-x^2}\log x,dx=-(1/4)\sqrt{\pi}(\gamma+2\log2)$ has both $\pi$ and $\gamma$, but the integral maybe doesn't count as "closed form". Similarly for $\int_0^{\infty}e^{-x}(\log x)^2,dx=\gamma^2+(1/6)\pi^2$. $\gamma=\log(4/\pi)-\sum_1^{\infty}(-2)^{-n}(n+1)^{-1}\zeta(n+1)$ has an infinite series. There are also a couple of formulas with infinite products. I don't see any other formula on that page with both $\pi$ and $\gamma$. – Gerry Myerson Jul 18 '19 at 04:35
  • Right, the integral doesn't have $\gamma$ in it, so it doesn't apply. The other expressions expand $\gamma$ in terms of $\pi$, like $\zeta(n+1)$ in the last formula that you wrote. I'm asking about the converse, where $\pi$ is expanded as a function, say, series of $\gamma$. – user687639 Jul 18 '19 at 04:43
  • So what's to stop you from solving for $\pi$? $\pi=4e^{-\gamma-\sum(-2)^{-n}(n+1)^{-1}\zeta(n+1)}$. – Gerry Myerson Jul 18 '19 at 04:45
  • like $f(\gamma,1)a_{1}+f(\gamma,2)a_{2}....=\pi g(\pi,\gamma)$, where g is some function. – user687639 Jul 18 '19 at 04:45
  • What you wrote is expanding $\gamma$ in terms of $\pi$. There are many such, but I ask the opposite which I couldn't find. – user687639 Jul 18 '19 at 04:48
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    What I wrote has $\pi=$ on the left, and an expression in $\gamma$ on the right, which would seem to be exactly what you want. How is $\gamma=\log(4/\pi)-\sum\cdots$ expanding $\gamma$ in terms of $\pi$, but $\pi=4e^{-\gamma-\sum\cdots}$ not expanding $\pi$ in terms of $\gamma$? – Gerry Myerson Jul 18 '19 at 04:58
  • Haha, alright then. – user687639 Jul 18 '19 at 05:00

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