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PostPosted: Thu Jul 07, 2016 12:38 pm 

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


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PostPosted: Thu Jul 07, 2016 12:41 pm 
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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 \(-\gamma\).

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Grigorios Kostakos


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PostPosted: Thu Jul 07, 2016 12:42 pm 

Joined: Wed Nov 11, 2015 12:47 pm
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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)}\\
&=e^{x\left(\log(p)-1-\frac{1}{2}-\frac{1}{3}-\ldots-\frac{1}{p}\right)}\cdot e^{x\left(1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{p}\right)}\quad\Rightarrow\\
&\qquad\qquad\boxed{p^{x}=e^{x\left(\log(p)-\mathcal{H}_{p}\right)}e^{x\mathcal{H}_{p}}}
\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)}\,.\\
\displaystyle\lim_{p\to \infty}\Gamma_{p}(x)\,&=\lim_{p\to\infty}\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)}\quad\Rightarrow\quad \Gamma(x)=\frac{e^{-\gamma x}\prod_{k=1}^{\infty}e^{\frac{x}{k}}}{x\prod_{k=1}^{\infty}\left(1+\frac{x}{k}\right)}=\frac{1}{xe^{\gamma x }}\prod_{k=1}^{\infty}\frac{e^{\frac{x}{k}}}{\left(1+\frac{x}{k}\right)}\quad\Rightarrow\\
& \qquad\qquad \qquad\qquad\boxed{\displaystyle\frac{1}{\Gamma(x)}=xe^{\gamma x}\prod_{k=1}^{\infty}\left(1+\frac{x}{k}\right)e^{-\frac{x}{k}}}
\end{alignat*}
Taking the last relation for \(\displaystyle k\in\mathbb{R}-\{0,\mathbb{Z}_{\small -}\}\) \begin{align*}
&\boxed{\displaystyle\frac{1}{\Gamma(x)}=xe^{\gamma x }\prod_{k=1}^{\infty}\left(1+\frac{x}{k}\right)e^{-\frac{x}{k}}}\quad\Rightarrow\quad \log\left(\frac{1}{\Gamma(x)}\right)=\log\left(xe^{\gamma x }\prod_{k=1}^{\infty}\left(1+\frac{x}{k}\right)e^{-\frac{x}{k}}\right)\quad\Rightarrow\\
&\displaystyle-\log\left(\Gamma(x)\right)=\log(x)+\gamma x +\sum_{k=1}^{\infty}\left(\log\left(1+\frac{x}{k}\right)-\frac{x}{k}\right)\quad\Rightarrow\\
& - \frac{\Gamma'(x)}{\Gamma(x)}=\frac{1}{x}+\gamma +\sum_{k=1}^{\infty}\left(\frac{1}{k}-\frac{1}{x-k}\right)\quad\Rightarrow\quad\underbrace{\frac{\Gamma'(x)}{\Gamma(x)}}_{\Psi(x)}=-\gamma+\sum_{k=1}^{\infty}\left(\frac{1}{k}-\frac{1}{x+k-1}\right)\quad\Rightarrow \\
&\Psi(x)=-\gamma +\sum_{k=1}^{\infty}\left(\frac{x-1}{k(x+k-1)}\right)\quad\Rightarrow\\
& \Psi(1)=\frac{\Gamma'(1)}{\underset{\underset{=1}{\downarrow}}{\Gamma(1)}}=\Gamma'(1)=-\gamma +\sum_{k=1}^{\infty}\frac{1-1}{k(1+k-1)}=-\gamma\quad\Rightarrow\quad \Gamma'(1)=-\gamma\,.
\end{align*}
Now since \begin{align*}
&\displaystyle\Gamma(z):=\int_{0}^{\infty}e^{-t}t^{z-1}\;dz\quad\Rightarrow\quad \Gamma'(z)=\int_{0}^{\infty}e^{-t}\left(\frac{d}{dz}t^{z-1}\right)\;dt=\int_{0}^{\infty}e^{-t}\log(t)t^{z-1}\;dt\quad\Rightarrow\\
& \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*}
&\displaystyle\int_{0}^{\infty}e^{-t}\log(t)\;dt=-\gamma\quad\Rightarrow\quad \gamma=-\int_{0}^{\infty}e^{-t}\log(t)\;dt\quad\Rightarrow\\ &\displaystyle\lim_{n\to\infty}\mathcal{H}_{n}-\log(n)=-\int_{0}^{\infty}e^{-t}\log(t)\;dt\,.
\end{align*}


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