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 Post subject: Series & IntegralPosted: Sun Jul 10, 2016 9:50 am

Joined: Sat Nov 07, 2015 6:12 pm
Posts: 841
Location: Larisa
Show that: $\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^n\ln n}{n}=\gamma \ln2 -\frac{\ln^2 2}{2}$

hence prove that: $\displaystyle \int_{0}^{\infty}\frac{\ln x}{e^x+1}\,dx=\frac{-\ln^2 2}{2}$.

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 Post subject: Re: Series & IntegralPosted: Sun Jul 10, 2016 1:07 pm

Joined: Sat Nov 14, 2015 6:32 am
Posts: 159
Location: Melbourne, Australia
We recall that the eta function is defined as

$$\eta(s)=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^s} = \left(1- 2^{1-s} \right) \zeta(s)$$

Hence differentiating once we have that:

\begin{align*}
\eta'(s) &= \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \log n}{n^s} \\
&= \sum_{n=2}^{\infty} \frac{(-1)^{n-1} \log n}{n^s}\\
&= 2^{1-s} \zeta(s )\log 2 + \left ( 1-2^{1-s} \right ) \zeta'(s)
\end{align*}

Thus taking limit as $s \rightarrow 1$ we have that:

\begin{align*}
\eta'(1) &=\sum_{n=2}^{\infty} \frac{(-1)^{n-1} \log n}{n} \\
&= \lim_{s \rightarrow 1} \left [ 2^{1-s} \zeta (s) \log 2 + \left ( 1-2^{1-s} \right ) \zeta'(s) \right ]\\
&=\lim_{s \rightarrow 1} 2^{1-s} \zeta(s) \log 2 + \lim_{s\rightarrow 1} \left ( 1-2^{1-s} \right ) \zeta'(s) \\
&= \gamma \ln 2 -\frac{\ln^2 2}{2}
\end{align*}

since $\zeta(s)$ behaves very well near $1$ as its Cauchy Values exists because the limit $\displaystyle \lim_{\epsilon \rightarrow 0} \frac{\zeta(1+\epsilon) + \zeta(1-\epsilon)}{2}= \gamma$. Also using the facts that:

$$\mathbf{(1)} \quad \lim_{s \rightarrow 1} \zeta(s) (s-1) =1 \quad \mathbf{(2)} \quad \lim_{s \rightarrow 1} \left ( \zeta(s) +(s-1) \zeta'(s) \right )=\gamma$$

we get that the second limit is $\displaystyle \lim_{s\rightarrow 1} \left ( 1-2^{1-s} \right ) \zeta'(s)=-\frac{\ln^2 2}{2}$. Of course DLH will be used.

Now for the integral we have successively:

\begin{align*}
\int_{0}^{\infty}\frac{\ln x}{e^x+1} \, {\rm d}x &= \lim_{s \rightarrow 1}\int_{0}^{\infty} \frac{\partial }{\partial s} \frac{x^{s-1}}{e^x+1} \, {\rm d}x \\
&= \lim_{s \rightarrow 1}\frac{\mathrm{d} }{\mathrm{d} s} \int_{0}^{\infty} \frac{x^{s-1}}{e^x+1} \, {\rm d}x\\
&= \lim_{s \rightarrow 1}\frac{\mathrm{d} }{\mathrm{d} s} \mathcal{M} \left \{ \frac{1}{e^x+1} \right \}\\
&= \lim_{s \rightarrow 1}\frac{\mathrm{d} }{\mathrm{d} s} \left ( \Gamma(s) \zeta(s) \right ) \\
&= \lim_{s \rightarrow 1} \left [\Gamma'(s) \zeta(s) + \Gamma (s) \zeta'(s) \right ] \\
&= \lim_{s \rightarrow 1}\left [\Gamma (s)\psi(s) \zeta(s) + \Gamma (s) \zeta'(s) \right ] \\
&=\lim_{s \rightarrow 1} \Gamma(s) \left [ \psi(s) \zeta(s) + \zeta'(s) \right ] \\
&= - \frac{\ln^2 2}{2}
\end{align*}

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$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$

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 Post subject: Re: Series & IntegralPosted: Sun Jul 10, 2016 1:09 pm

Joined: Sat Nov 14, 2015 6:32 am
Posts: 159
Location: Melbourne, Australia
In more general if $k \in \mathbb{N}$ then:

$$\sum_{n=1}^{\infty}\frac{(-1)^{n}\log^{k}n}{n}=\frac{(-1)^{n}\left(\log2\right)^{k+1}}{k+1}+(-1)^{k-1}\sum_{j=0}^{k-1}\gamma_{j}\binom{k}{j}\log^{k-j}2$$

where $\gamma_i$ are the Stieljes constants.

_________________
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$

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 Post subject: Re: Series & IntegralPosted: Sat Jun 03, 2017 8:18 pm

Joined: Mon Oct 26, 2015 12:27 pm
Posts: 40
Two more solutions for the first series can be found here.

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