A month ago, I came up with a proof that

$\gamma = \frac12 + \int_0^{\frac1\pi} \arctan(\cot(\frac1x)) \,dx$

where $\gamma$ is the Euler-Mascheroni constant and $\arctan$ is the inverse $\tan$ function.

My proof is based on the idea, that $\lfloor x\rfloor = \frac {\arctan(\cot(x\pi))}\pi - \frac 12 + x$

Is that well known? Did somebody came up with it before? If so, where can i find some references? I want to know if there are any other ways to prove that.

Thank you in advance!

  • 1
    $\begingroup$ You are right, my bad. $\endgroup$ – Yves Daoust Jul 16 '19 at 12:09
  • $\begingroup$ "Any other ways" than what ? A proof by decomposition of the domain seems easy. $\endgroup$ – Yves Daoust Jul 16 '19 at 12:10
  • $\begingroup$ Could you provide the proof you came up with? I've never seen thus before $\endgroup$ – miraunpajaro Jul 16 '19 at 12:15

It's well known that $$\int_0^1 \left\{\frac1x\right\}\mathrm{d}x=1-\gamma$$ where $\{x\}=x-\lfloor x\rfloor$ is the fractional part of $x$ (see a proof here and also here) and we have that $$\frac1\pi\arctan{\left(\cot{\left(\frac\pi{x}\right)}\right)}=\frac12-\left\{\frac1x\right\}$$ for all $x\in\mathbb{R}$. Hence your identity is $$\begin{align} \gamma &=\frac12+\int_0^{1/\pi}\pi\left(\frac1\pi\arctan{\left(\cot{\left(\frac\pi{\pi x}\right)}\right)}\right)\mathrm{d}x\\ &=\frac12+\int_0^{1/\pi}\pi\left(\frac12-\left\{\frac1{\pi x}\right\}\right)\mathrm{d}x\\ &=\frac12+\int_0^1\left(\frac12-\left\{\frac1u\right\}\right)\mathrm{d}u\\ &=1-\int_0^1\left\{\frac1u\right\}\mathrm{d}u\\ &=1-(1-\gamma)\\ &=\gamma\\ \end{align}$$


So here is my proof:

Two important things that i wont proof here are

$\lfloor x\rfloor = \frac {\arctan(\cot(x\pi))}\pi - \frac 12 + x$ and

$\sum_{n=1}^x \frac 1n = \frac {\lfloor x\rfloor}x + \int_1^x \frac {\lfloor t\rfloor}{t^2} \,dt$

But those are well known facts.

From this it follows that

$\sum_{n=1}^x \frac 1n = \frac {\arctan(\cot(x\pi))}{\pi x} - \frac 1{2x} + 1 + \int_1^x \frac {\arctan(\cot(t\pi))}{\pi t^2} - \frac 1{2t^2} + \frac 1t\,dt$

(simply substitute the first into the second)

$\sum_{n=1}^x \frac 1n = \frac {\arctan(\cot(x\pi))}{\pi x} - \frac 1{2x} + 1 + \frac 1{2x} - \frac 12 + \ln(x) + \int_1^x \frac {\arctan(\cot(t\pi))}{\pi t^2} \,dt$

$\sum_{n=1}^x \frac 1n - \ln(x) = \frac 12 + \int_1^x \frac {\arctan(\cot(t\pi))}{\pi t^2} \,dt + \frac {\arctan(cot(x\pi))}{\pi x}$

Now take the limit as x goes to $\infty$

$\gamma = \frac 12 + \int_1^\infty \frac {\arctan(\cot(t\pi))}{\pi t^2} \,dt$

substituting in $u = \frac 1{\pi t}$ gives

$\gamma = \frac 12 + \int_0^{\frac 1{\pi}} \arctan(\cot(\frac 1u)) \,du$



$\arctan(\cot(\frac1t))$ amounts to a "modulo $\pi$" operation. Hence $\frac1t-\arctan(\cot(\frac1t))$ is a piecewise hyperbolic function with pieces proportional to $\frac1t$, and the summation generates an harmonic series. This is how $\gamma$ appears.


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