Integral representation of the Euler-Mascheroni constant involving $\pi$ 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!
 A: 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
$\gamma = \frac 12 + \int_0^{\frac 1{\pi}} \arctan(\cot(\frac 1u)) \,du$
A: 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
$$\begin{align}
\gamma
&=\frac12+\int_0^{1/\pi}\pi\left(\frac1\pi\arctan{\left(\cot{\left(\frac\pi{\pi x}\right)}\right)}\right)\mathrm{d}x\\
&=\frac12+\int_0^{1/\pi}\pi\left(\frac12-\left\{\frac1{\pi x}\right\}\right)\mathrm{d}x\\
&=\frac12+\int_0^1\left(\frac12-\left\{\frac1u\right\}\right)\mathrm{d}u\\
&=1-\int_0^1\left\{\frac1u\right\}\mathrm{d}u\\
&=1-(1-\gamma)\\
&=\gamma\\
\end{align}$$
A: Hint:
$\arctan(\cot(\frac1t))$ 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.
