euler-mascheroni-constant
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!
From the continuous fraction expansion, the seventh convergent is
$$\gamma \approx \frac{15}{26}$$
From the limit definition
$$\begin{align} \gamma &= \lim_{n \to \infty} {\left(2H_n-\frac{1}{6}H_{n^2+n-1}-\frac{5}{6}H_{n^2+n}\right)} \\ &= \frac{7}{12}+\sum_{n=1}^{\infty}\left(\frac{2}{n+1}-\frac{1}{3n(n+1)(n+2)}-\sum_{k=n(n+1)+1}^{(n+1)(n+2)}\frac{1}{k}\right) \\ &=\frac{7}{12}-\frac{1}{180}+\sum_{n=2}^{\infty}\left(\frac{2}{n+1}-\frac{1}{3n(n+1)(n+2)}-\sum_{k=n(n+1)+1}^{(n+1)(n+2)}\frac{1}{k}\right) \\ &=\frac{26}{45}+\sum_{n=2}^{\infty}\left(\frac{2}{n+1}-\frac{1}{3n(n+1)(n+2)}-\sum_{k=n(n+1)+1}^{(n+1)(n+2)}\frac{1}{k}\right) \\ \end{align} $$
so $$\gamma \approx \frac{26}{45}$$
Multiplying both approximations,
$$\gamma^2 \approx \frac{1}{3}$$
The origin of this limit definition is improving the convergence of Macys formula from $o(n^{-2})$ to $o(n^{-4})$ downweighting the last term, without additional fractions (https://math.stackexchange.com/a/129808/134791)
In fact, the convergent approximation is not necessary.
Given $$\gamma \approx \frac{26}{45}$$
we have $$3\gamma^2\approx3\left(\frac{26}{45}\right)^2=3\frac{676}{2025}=3\frac{676}{3\cdot675}=\frac{676}{675}\approx 1$$
which also yields $$\gamma^2 \approx \frac{1}{3}$$
Towards proving that $\gamma < \frac{1}{ \sqrt {3} }$, we may take one more term out of the summation: $$\gamma \approx \frac{7}{12}-\frac{1}{180}-\frac{1}{2310}=\frac{4001}{6930}<\frac{1}{\sqrt{3}}$$
Pictorial proof.
In the picture, take $n=11$. The red graph is $1/x$, so the area under the red graph from $1$ to $n$ is
$$
\int_1^n\frac{dx}{x} = \ln n.
$$ The area of the white rectangles under the graph is
$$
\frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}
$$
The difference is shown in green,
$$
\text{area}(\text{green}_n) = \ln(n) - \left(\frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}\right)
$$
Take the green triangles, translate them to the left, so they are in the strip between $x=0$ and $x=1$. They are disjoint. So their areas add to at most the area of the whole strip $(0,1) \times (0,1)$; that is $1$...
$$
\text{area}(\text{green}_n) < 1
$$
Now as $n$ increases, we add more green regions, so this value increases with $n$, and it bounded above by $1$, so it converges and has limit $\le 1$:
$$
0 < \lim_{n\to\infty}\left[\ln(n) - \left(\frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}\right)\right] \le 1
$$
The value you want is obtained by subtracting all of this from $1$:
$$
0 \le \lim_{n\to\infty}\left[ \left(1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}\right)- \ln(n)\right] < 1
$$
If we like, we can think of this as the sum of the gray areas.
These little gray and green regions are approximately triangles... if they were exactly triangles, then our result would be exactly $1/2$. As it is, $\gamma$ is approximately $1/2$.
Best Answer
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
$\gamma = \frac 12 + \int_0^{\frac 1{\pi}} \arctan(\cot(\frac 1u)) \,du$