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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.

  • 1
    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 $x$.2012-01-20
  • 2
    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 $-\frac{1}{2} \log^2(2) = -.2402265070$ approximately2012-01-20
  • 2
    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.2012-01-20
  • 0
    This is really annoying. I removed the !. I am very ver sorry for this error.2012-01-21

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