Articles of eulers constant

Euler-Mascheroni constant: understanding why $\lim_{m\rightarrow \infty} \sum_{n=1}^{m} (\ln (1 + \frac{1}{n})-\frac{1}{n+1})= 1 – \gamma$

I am trying to understand why the Euler-Mascheroni constant $\displaystyle \gamma = \lim_{n \rightarrow \infty} \left ( \sum_{k=1}^n \frac{1}{k} – \ln n \right )$ is equal to $1 – \displaystyle \int_{1}^{\infty} \frac{t-[t]}{t^2}\; \mathrm{d}t$ where $[t]$ is the floor function. There has been an answered question here already: Integral form for the euler-mascheroni gamma constant using […]

Define integral for $\gamma,\zeta(i) i\in\mathbb{N}$ and Stirling numbers of the first kind

Consider the integral $$\int\limits_0^{\infty}e^{-x}x^k\ln(x)^n\dfrac{dx}x$$ For $n=3$ we have $$(-\gamma^2-2\zeta(3)-3\zeta(2)\gamma)\genfrac{[}{]}{0pt}{}{k}{1}+3(\gamma^2+\zeta(2))\genfrac{[}{]}{0pt}{}{k}{2}-6\gamma\genfrac{[}{]}{0pt}{}{k}{3}+6\genfrac{[}{]}{0pt}{}{k}{4}$$ where $\genfrac{[}{]}{0pt}{}{n}{k}$ is Stirling number of the first kind. For different n there are simmilar formulas. In the general case $$\int\limits_0^{\infty}e^{-x}x^k\ln(x)^n\dfrac{dx}x\in\bigoplus_{j=1}^{n+1}\genfrac{[}{]}{0pt}{}{k}{j}\mathbb{Z}[\gamma,\zeta(i)]_{n+1-j}$$ where $$\mathbb{Z}[\gamma,\zeta(i)]=\bigoplus_{j=0}^{\infty}\mathbb{Z}[\gamma,\zeta(i)]_{j}$$ is graded ring. Does anyone know a simple proof? Are there any ideas on how to generalize on multiple zeta values?

How does one show that $\int_{0}^{\pi/4}\sin(2x)\ln{(\ln^2{\cot{x}})}\mathrm dx=\ln{\pi\over4}-\gamma?$

Vardi’s type integral Consider $(1)$ $$\int_{0}^{\pi/4}\sin(2x)\ln{(\ln^2{\cot{x}})}\mathrm dx=\ln{\pi\over4}-\gamma\tag1$$ An attempt : maybe? $u=\ln{(\ln{\cot{x}})}$ then $du=-{\cos{x}\over \sin^3{x}}dx$

Why is $-\gamma = \int_0^1 \frac{e^{-z}-1}{z}dz+\int_1^\infty \frac{e^{-z}}{z}dz$

It seems like the sum of the two RHS integrals is “well known”$^\dagger$ to be Euler’s constant: $$\gamma \equiv \int_1^\infty \frac{1}{\lfloor z\rfloor} – \frac{1}{z}dz \quad\stackrel{?}{=}\quad -\int_0^1 \frac{e^{-z}-1}{z}dz-\int_1^\infty \frac{e^{-z}}{z}dz$$ How can I prove this is so? Edit: I can prove this converges to a constant by showing that this is equivalent to: $$\int_0^1 \frac{e^{-1/z}+e^{-z}-1}{z}dz$$ And that […]

A fractional part integral giving $\frac{F_{n-1}}{F_n}-\frac{(-1)^n}{F_n^2}\ln\left(\!\frac{F_{n+2}-F_n\gamma}{F_{n+1}-F_n\gamma}\right)$

I’ve been asked to elaborate on the following evaluation: $$ \begin{align}\\ \displaystyle {\large\int_0^{1}} \!\cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {\ddots + \cfrac 1 { 1 + \psi (\left\{1/x\right\}+1)}}}} \:\mathrm{d}x & = \dfrac{F_{n-1}}{F_{n}} – \dfrac{(-1)^{n}}{F_{n}^2} \ln \!\left(\!\dfrac{F_{n+2}-F_{n}\gamma}{F_{n+1}-F_{n}\gamma} \right)\\\\ \end{align} $$ where $\left\{x\right\}=x-\lfloor x\rfloor$ denotes the fractional part of $x$, $\gamma$ is the […]

A closed form of the series $\sum_{n=1}^{\infty} \frac{H_n^2-(\gamma + \ln n)^2}{n}$

I have found a closed form of the following new series involving non-linear harmonic numbers. Proposition. $$\sum_{n=1}^{\infty} \dfrac{H_n^2-(\gamma + \ln n)^2}{n} = \dfrac{5}{3}\zeta(3)-\dfrac{2}{3}\gamma^3-2\gamma \gamma_{1}-\gamma_{2} $$ where \begin{align} & H_{n}: =\sum_{k=1}^{n}\frac{1}{k} \\ &\gamma: =\lim_{n\to\infty} \left(\sum_{k=1}^n \frac{1}{k}-\ln n\right) \\ & \gamma_{1}:=\lim_{n\to\infty} \left(\sum_{k=1}^n \frac{\ln k}{k}-\frac{1}{2}\ln^2 n\right)\\& \gamma_{2}: =\lim_{n\to\infty} \left(\sum_{k=1}^n \frac{\ln^2 k}{k}-\frac{1}{3}\ln^3 n\right), \end{align} $\gamma_1, \gamma_2$ being Stieltjes constants. […]

Please help me to prove the convergence of $\gamma_{n} = 1+\frac{1}{2}+\frac{1}{3}+…+\frac{1}{n}-\ln n$

Please help me to prove tha $$\gamma_{n} = 1+\frac{1}{2}+\frac{1}{3}+….+\frac{1}{n}-\ln n$$ converges and its limit is $\gamma \in [0,1]$. Then, using $\gamma_{n}$ find the sum: $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n}$.

Need to show that $\lim_{x\to\infty}\left(\sum_{n\le x}^{}\frac{1}{n}-\ln x \right)$ exist and is less than $1$

This question already has an answer here: Showing that $\lim_{n\to\infty}\sum^n_{k=1}\frac{1}{k}-\ln(n)=0.5772\ldots$ 2 answers

Showing $\gamma < \sqrt{1/3}$ without a computer

In 1735 Euler gave the value of $\gamma$ as $0.577218.$ The constant is generally defined as the limit of the difference between the harmonic series and $\log n:~\gamma= \lim_{n\to\infty}\sum_{k=1}^{n}\frac{1}{k}-\log n.$ Euler apparently relished this sort of calculation and must have taken quite a few terms to get such a good approximation. My question is whether […]

Euler-Mascheroni constant expression, further simplification

The Euler-Mascheroni constant gamma is defined as: $$\gamma=\lim\limits_{n \rightarrow \infty}\left(\sum\limits_{m=1}^{n} \frac{1}{m} – \log(n)\right)$$ From this previous question Do these series converge to logarithms? $\log(n)$ can be written: $$\log(n)=\sum\limits_{k=1}^\infty \sum\limits_{a=1}^{n-1}\frac{1}{kn-a}-\sum\limits_{k=1}^\infty\frac{n-1}{kn}$$ $$\gamma=\lim\limits_{n \rightarrow \infty}\left(\sum\limits_{m=1}^{n} \frac{1}{m} – \left(\sum\limits_{k=1}^\infty \sum\limits_{a=1}^{n-1}\frac{1}{kn-a}-\sum\limits_{k=1}^\infty\frac{n-1}{kn}\right)\right)$$ $$\gamma=\lim\limits_{n \rightarrow \infty}\left(\sum\limits_{m=1}^{n} \frac{1}{m} – \sum\limits_{k=1}^\infty \sum\limits_{a=1}^{n-1}\frac{1}{kn-a} + \sum\limits_{k=1}^\infty\frac{n-1}{kn}\right)$$ Can this last expression be further simplified?