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 Post subject: Identity regarding the euler gamma costantPosted: Thu Jul 07, 2016 12:38 pm

Joined: Wed Nov 11, 2015 12:47 pm
Posts: 13
Prove that $\displaystyle \lim_{n\to \infty} \mathcal{H}_{n}-\log(n)=-\int_{0}^{\infty}e^{-t}\log(t)\;dt$.

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 Post subject: Re: Identity regarding the euler gamma costantPosted: Thu Jul 07, 2016 12:41 pm
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Joined: Mon Nov 09, 2015 1:36 am
Posts: 443
Location: Ioannina, Greece
The Euler-Mascheroni constant denoted by the letter $\gamma$ is defined as $\gamma=\displaystyle\lim_{n \rightarrow \infty } \biggl( \sum_{k=1}^n \frac{1}{k} - \ln(n) \biggr)= \lim_{n\to \infty} ({\mathcal{H}_{n}-\log(n)})\,.$ So we must prove that $\gamma=-\displaystyle\int_{0}^{\infty}e^{-t}\log(t)\;dt\,.$
The following (beautiful) proof was given by Sangchul Lee here

Proof: It is easy to prove that the sequence of functions
$f_n(x) = \begin{cases} \left( 1 - \frac{x}{n}\right)^n & 0 \leqslant x \leqslant n \\ 0 & x > n \end{cases}$ satisfies $0 \leq f_n(x) \uparrow e^{-x}$. Thus by dominated convergence theorem, we have
$\int_{0}^{\infty} e^{-x} \log x \; dx = \lim_{n\to\infty} \int_{0}^{n} \left( 1 - \frac{x}{n}\right)^n \log x \; dx.$ Now by the substitution $x=nu$, we have
\begin{align*}
\int_{0}^{n} \Bigl({1-\frac{x}{n}}\Bigr)^n\log x \; dx
&= n\int_{0}^{1} ({1 - u})^n (\log n + \log u) \; du \\
&= \frac{n}{n+1}\log n + n\int_{0}^{1} (1-u)^n \log u \; du \\
&= \frac{n}{n+1}\log n + n\int_{0}^{1} v^n \log (1-v) \; dv \\
&= \frac{n}{n+1}\log n - n\int_{0}^{1} v^n \biggl( \sum_{k=1}^{\infty} \frac{v^k}{k} \biggr) \; dv \\
&= \frac{n}{n+1}\log n - n \sum_{k=1}^{\infty} \frac{1}{k(n+k+1)} \\
&= \frac{n}{n+1}\log n - \frac{n}{n+1} \sum_{k=1}^{\infty} \Bigl( \frac{1}{k} - \frac{1}{n+k+1}\Bigr) \\
&= \frac{n}{n+1} \biggl( \log n - \sum_{k=1}^{n+1} \frac{1}{k} \biggr)\,.
\end{align*} Therefore taking $n\to+\infty$ yields $-\gamma$.

_________________
Grigorios Kostakos

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 Post subject: Re: Identity regarding the euler gamma costantPosted: Thu Jul 07, 2016 12:42 pm

Joined: Wed Nov 11, 2015 12:47 pm
Posts: 13
By definition $\displaystyle \Gamma_{p}(x):=\frac{p^{x}p!}{x(x+1)(x+2)\cdot\ldots\cdot (x+p)}=\frac{p^{x}}{x\left(1+\frac{x}{1}\right)\left(1+\frac{x}{2}\right)\cdot\ldots\cdot\left(1+\frac{x}{p}\right)}$ and $\Gamma(x):=\displaystyle\lim_{p\to \infty}\Gamma_{p}(x)\,.$
\begin{align*}
p^{x}&=e^{x\log(p)}=e^{x\left(\log(p)-1-\frac{1}{2}-\frac{1}{3}-\ldots-\frac{1}{p}\right)+x\left(1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{p}\right)}\\
\end{align*}
\begin{alignat*}{2}
\displaystyle\Gamma_{p}(x)\,&=\frac{1}{x}\cdot \frac{1}{x+1}\cdot \frac{2}{x+2}\cdot\ldots\cdot \frac{p}{x+p}\cdot p^{x}=\frac{1}{x}\cdot\frac{1}{1+\frac{x}{1}}\cdot\frac{1}{1+\frac{x}{2}}\cdot\ldots\cdot\frac{1}{1+\frac{x}{p}}\cdot e^{x\left(\log(p)-\mathcal{H}_{p}\right)}e^{x\mathcal{H}_{p}}\\
&=\frac{e^{x\left(\log(p)-\mathcal{H}_{p}\right)}e^{x\mathcal{H}_{p}}}{x\left(1+x\right)\left(1+\frac{x}{2}\right)\ldots\left(1+\frac{x}{p}\right)}=\frac{e^{x\left(\log(p)-\mathcal{H}_{p}\right)}e^{x\mathcal{H}_{p}}}{x\prod_{k=1}^{p}\left(1+\frac{x}{k}\right)}\,.\\
\end{alignat*}
Taking the last relation for $\displaystyle k\in\mathbb{R}-\{0,\mathbb{Z}_{\small -}\}$ \begin{align*}
\end{align*}
Now since \begin{align*}
& \Gamma'(1)=\int_{0}^{\infty}e^{-t}\log(t)t^{0}\;dt=\int_{0}^{\infty}e^{-t}\log(t)\;dt\,.
\end{align*}
And so \begin{align*}
\end{align*}

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