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It happens that, for any $m\geq 1$, $$\sum_{n=0}^{+\infty}\frac{(-1)^n}{(2n+1)^{2m+1}}=\frac{E_{2m}}{2\cdot(2m)!}\left(\frac{\pi}{2}\right)^{2m+1}\tag{1}$$ where $E_{2m}$ is an integer number.

My proof works through the following lines: the LHS is: $$\frac{1}{(2m)!}\int_{0}^{1}\frac{(\log x)^{2m}}{1+x^2}dx = \frac{1}{2\cdot(2m)!}\int_{0}^{+\infty}\frac{(\log x)^{2m}}{1+x^2}dx,$$ so we just need to compute: $$\left.\frac{d^{2m}}{dk^{2m}}\int_{0}^{+\infty}\frac{x^k}{1+x^2}\right|_{k=0},\tag{2}$$ but: $$ \int_{0}^{+\infty}\frac{x^{1/r}}{1+x^2}\,dx = r\int_{0}^{+\infty}\frac{y^r}{1+y^{2r}}\,dy = \frac{\pi/2}{\cos(\pi/(2r))}$$ by the residue theorem, so $E_{2m}$ is just the absolute value of an Euler number, that belongs to $\mathbb{N}$.

Can someone provide a real-analytic proof of $(1)$?

Jack D'Aurizio
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  • You could express the partial fractions expansion of $\sec x$ as an iterated sum, switch the order of summation, and then compare it to the Taylor series of $\sec x$ at $x=0$. – Random Variable Jun 28 '14 at 15:04
  • @RandomVariable: please post your argument as an answer, I'll be glad to upvote and accept it. – Jack D'Aurizio Jun 28 '14 at 15:07
  • I actually already posted it in another thread. I just don't like to link to my own answers. It's the second answer in my post. Robjohn also posted an answer. http://math.stackexchange.com/questions/762813/sum-n-0-infty-frac-1n2n12m1-frac-1m-e-2m-pi2m14 – Random Variable Jun 28 '14 at 15:15
  • All right, I just posted my solution in the other thread, thanks for pointing that out. – Jack D'Aurizio Jun 28 '14 at 15:24
  • @JackD'Aurizio : I don't get why $\zeta(2m+1)$ and $\eta(2m+1)$ are complicated (in contrary to $\zeta(2m),\eta(2m)$), but not $\beta(2m+1)$, do you have an explanation for this ? – reuns Jun 19 '16 at 03:32
  • @user1952009: that can be seen as a consequence of logarithmic differentiation of Weierstrass products. By applying $\frac{d}{dz}\log(\cdot)$ to the sine and cosine function we easily get $\zeta(2m)$ and $\eta(2m+1)$ as Taylor coefficients of $\tan$ and $x \cot$, but the periodicity of the zeroes of $\sin$ and $\cos$ is in fact an obstruction for obtaining $\zeta(2m+1)$ in the same way. – Jack D'Aurizio Jun 19 '16 at 09:47
  • I think a reason is also that the trivial zeros of $\beta(s)$ are at the odd negative integers its functional equation being $\beta(s) = \ldots \sin(\pi \frac{s-1}{2}) \Gamma(1-s) \beta(1-s)$ while $\zeta(s)$ has its trivial zeros at the even negative integers $\zeta(s) = \ldots \sin(\pi \frac{s}{2}) \Gamma(1-s) \zeta(1-s)$ – reuns Jun 19 '16 at 10:23
  • Do you seek a proof that the numerator is specifically Euler's numbers, or would a proof that it's an integer suffice? – Simply Beautiful Art Aug 04 '18 at 20:58
  • @SimplyBeautifulArt: once we have the connection with the Laurent series of $\sec$ everything is pretty clear, my original question (which has been answered in the comments) was about a real-analytic proof of (1). – Jack D'Aurizio Aug 04 '18 at 21:00
  • Okay... was just asking for clarification as to whether you wanted to prove (1) in the form "where $E_{2k}$ is an integer number" or "$E_{2m}$ is an Euler number". – Simply Beautiful Art Aug 04 '18 at 21:04

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