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While trying to prove $\int_0^{\infty } \frac{\log (x)}{e^x+1} \, dx = -\frac{1}{2} \log ^2(2)$ How to show? in an alternative way, I came to this solution:

$$\sum_{k=0}^{+\infty}(-1)^{k+1}\frac{\log (k+1)+\gamma }{(k+1)}.$$

As both solutions have to be the same, the following equality should be valid:

$$\sum_{k=0}^{+\infty}(-1)^{k+1}\frac{\log (k+1)+\gamma }{(k+1)}=- \frac{1}{2}{{\log }^2(2)}. $$

Can anyone give me some advice on how to prove this equality.

p.s. You can be sure that the equality is correct, as I checked it numerically.

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wnvl
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    So, obviously the hard part is $\displaystyle \sum_k \frac{(-1)^k\log(k)}{k!}$. I think a possible route would be to see if you can find $\displaystyle \sum_k \frac{(-1)^kk^x}{k!}$ for a general $x$ (it's easy for integral $x$) and then you can differentiate both sides and let $x=0$. I have to run, but that should work, assuming that sum isn't actually too crazy for non-integral $x$. – Alex Youcis Jan 20 '12 at 23:26
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    Are you sure you wrote this right? I get $\sum_{k=0}^\infty (-1)^{k+1} \frac{\log(k+1)+\gamma}{(k+1)!} = -.1548995048$ and $-\frac{1}{2} \log^2(2) = -.2402265070$ approximately – Robert Israel Jan 20 '12 at 23:39
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    The ! should be removed from the denominator: $\sum_{k=0}^{\infty} (-1)^{k+1} \frac{\log(k+1)+\gamma}{k+1}=-\frac{1}{2}(\log 2)^2.$ This is obtained from the integral by expanding $(1+e^x)^{-1}=e^{-x}-e^{-2x}+{\dots}$ and integrating term-by-term. – David Moews Jan 20 '12 at 23:52
  • This is really annoying. I removed the !. I am very ver sorry for this error. – wnvl Jan 21 '12 at 00:18

3 Answers3

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Note that $$ \begin{align} \sum_{k=1}^\infty(-1)^k\frac{\log(k)+\gamma}{k} &=\lim_{n\to\infty}\left(2\sum_{k=1}^{n}\frac{\log(2k)+\gamma}{2k}-\sum_{k=1}^{2n}\frac{\log(k)+\gamma}{k}\right)\\ &=\lim_{n\to\infty}\left(\sum_{k=1}^{n}\frac{\log(2)+\log(k)+\gamma}{k}-\sum_{k=1}^{2n}\frac{\log(k)+\gamma}{k}\right)\tag{1} \end{align} $$ Using the Euler-Maclaurin Sum Formula, we get that $$ \sum_{k=1}^n\frac{1}{k}=\log(n)+\gamma+\frac{1}{2n}-\frac{1}{12n^2}+O\left(\frac{1}{n^4}\right)\tag{2} $$ and that $$ \sum_{k=1}^n\frac{\log(k)}{k}=\frac12\log(n)^2+C+\frac{\log(n)}{2n}-\frac{\log(n)-1}{12n^2}+O\left(\frac{\log(n)}{n^4}\right)\tag{3} $$ Applying $(2)$ and $(3)$ to $(1)$, leaving out the terms which vanish, we get $$ \begin{align} &\sum_{k=1}^\infty(-1)^k\frac{\log(k)+\gamma}{k}\\ &=\small{\lim_{n\to\infty}\left(\log(2)(\log(n){+}\gamma)+\left(\frac12\log(n)^2+C+\gamma(\log(n){+}\gamma)\right)-\left(\frac12\log(2n)^2+C+\gamma(\log(2n){+}\gamma)\right)\right)}\\ &=\lim_{n\to\infty}\left(\log(2)(\log(n)+\gamma)-\log(2)\log(n)-\frac12\log(2)^2-\gamma\log(2)\right)\\ &=-\frac12\log(2)^2\tag{4} \end{align} $$

robjohn
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This can be solved similarly to the original problem. The Dirichlet eta function is defined by $$ \eta(s):=\sum_{n\ge 1} \frac{(-1)^{n-1}}{n^s}. $$ The given sum can be rewritten as $$ \sum_{n\ge 1} (-1)^n \frac{\log n}{n}+\sum_{n\ge 1} (-1)^n \frac{\gamma}{n}= \eta'(1)-\gamma \log 2.\qquad (*)$$ We have $$ \eta(s)=\sum_{n\ge 1} \frac{1}{n^s}-2\sum_{n\ge 1} \frac{1}{(2n)^s}=(1-2^{1-s})\zeta(s) $$ so, using the expansions $$\zeta(s)=\frac{1}{s-1}+\gamma+O(s-1),$$ $$ 2^{1-s}=e^{(1-s)\log 2}=1-(s-1)\log 2+\frac{1}{2}(\log 2)^2 (s-1)^2+O((s-1)^3), $$ we get $\eta'(1)=\gamma \log 2 -\frac{1}{2}(\log 2)^2$, so (*) equals $-\frac{1}{2}(\log 2)^2$.

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David Moews
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  • In fact, I use the result from my answer to derive the expansion for $\zeta(s)$ you cite above. See Riemann zeta near 1 – robjohn Jan 28 '12 at 14:16
  • Great answer but something's off by a sign and I can't figure out what. The term $-\frac{1}{2}(\log 2)^2$ comes from the term $\left ( \zeta'=-\frac{1}{(s-1)^2}\right )* \color{red}{-}(2^{1-s}_{\text{third term}})=\color{red}{+}\frac{1}{(s-1)^2}*\frac{1}{2}(\log 2)^2 (s-1)^2=\frac{1}{2}(\log 2)^2$ – user5389726598465 Jun 05 '17 at 05:23
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The problem is equivalent to showing

$$ \sum_{n\ge1}{(-1)^n\log n\over n}=\gamma\log2+\frac12\log^22 $$

Now, let's first consider the finite case:

$$ \begin{aligned} \sum_{n\le N}{(-1)^n\log n\over n} &=2\sum_{n\le N/2}{\log(2n)\over2n}-\sum_{n\le N}{\log n\over n}+\mathcal O\left(\log N\over N\right) \\ &=\sum_{n\le N/2}{\log2+\log n\over n}-\sum_{n\le N}{\log n\over n}+\mathcal O\left(\log N\over N\right) \\ &=\log2\sum_{n\le N/2}\frac1n-\sum_{N/2<n\le N}{\log n\over n}+\mathcal O\left(\log N\over N\right) \\ \end{aligned} $$

In fact, using Riemann-Stieltjes integral, we can show

$$ \begin{aligned} \sum_{N/2<n\le N}{\log n\over n} &=\int_{N/2}^N{\log x\over x}\mathrm d\lfloor x\rfloor \\ &={N\log(N)-N\log(N/2)\over N}-\int_{N/2}^N[x-\{x\}]\mathrm d\left(\log x\over x\right)+\mathcal O\left(\frac1n\right) \\ &=\log2-\int_{N/2}^N\left({1-\log x\over x}\right)\mathrm dx+\mathcal O\left(\log N\over N\right) \\ &=\int_{N/2}^N{\log x\over x}\mathrm dx+\mathcal O\left(\log N\over N\right) \\ &=\frac12[\log^2N-\log^2(N/2)]+\mathcal O\left(\log N\over N\right) \\ &=\frac12[\log N+\log(N/2)][\log N-\log(N/2)]+\mathcal O\left(\log N\over N\right) \\ &=\frac12\log2[2\log N-\log2]+\mathcal O\left(\log N\over N\right) \\ &=\log2\log N-\frac12\log^22+\mathcal O\left(\log N\over N\right) \end{aligned} $$

Now, employing this obtained identity and the asymptotic formula for harmonic series yields:

$$ \begin{aligned} \sum_{n\le N}{(-1)^n\log n\over n} &=\log2(\log N+\gamma)-\log2\log N+\frac12\log^22+\mathcal O\left(\log N\over N\right) \\ &=\gamma\log2+\frac12\log^22+\mathcal O\left(\log N\over N\right) \end{aligned} $$

Now, take the limit $N\to\infty$ on both side gives

$$ \sum_{n\ge1}{(-1)^n\log n\over n}=\gamma\log2+\frac12\log^22 $$

TravorLZH
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