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An interesting identity involving powers of $\pi$ and values of $\eta(s)$ [duplicate]

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 ...
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What is $f(2s+1)$ when $f(s)=\sum_{n=0}^\infty {\frac{(-1)^n}{(2n+1)^s}}=1-\frac{1}{3^s}+\frac{1}{5^s}-\frac{1}{7^s}+\dots$? [duplicate]

Is there an exact form of $$f(s)=\sum_{n=0}^\infty {\frac{(-1)^n}{(2n+1)^s}}=1-\frac{1}{3^s}+\frac{1}{5^s}-\frac{1}{7^s}+\dots$$ when $s$ is odd? Discussion I have been exploring infinite series ...
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Proving that $\frac{\pi^{3}}{32}=1-\sum_{k=1}^{\infty}\frac{2k(2k+1)\zeta(2k+2)}{4^{2k+2}}$

After numerical analysis it seems that $$\frac{\pi^{3}}{32}=1-\sum_{k=1}^{\infty}\frac{2k(2k+1)\zeta(2k+2)}{4^{2k+2}}$$ Could someone prove the validity of such identity?
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Sum : $\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^3}$

Prove that : $$\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2k+1)^3}=\frac{\pi^3}{32}.$$ I think this is known (see here), I appreciate any hint or link for the solution (or the full solution).
Hi I am trying to prove the relation $$I:=\int_0^\pi \log^2\left(\tan\frac{ x}{4}\right)dx=\frac{\pi^3}{4}.$$ I tried expanding the log argument by using $\sin x/ \cos x=\tan x,$ and than used $\log(... 3answers 1k views Recursion relation for Euler numbers I am trying to solve the following: The Euler numbers$E_n$are defined by the power series expansion $$\frac{1}{\cos z}=\sum_{n=0}^\infty \frac{E_n}{n!}z^n\text{ for }|z|<\pi/2$$ (a) ... 1answer 1k views Calculating residues of multiple poles? How would I calculate $$\mathrm{Res}\left(\frac{\pi}{\sin(\pi z)(2z+1)^3}\right)?$$ I understand it has singularities at$z=n$and$z=-1/2$, I'm interested in the residue when$z=-1/2$. I know that ... 0answers 276 views Odd values for Dirichlet beta function Hello there I want to find a proof for the generating formula for odd values of Dirichlet beta function given by wikipedia: link I searched MSE and didnt find something similar. My try was to start ... 1answer 68 views How can I expand this How can I expand$\dfrac{\pi \csc(z\pi)}{(2z+1)^3}\$? so then I can find the residue ? thanks
I need hints on the proof of: $$\int_0^\infty\dfrac{\ln(x)^2}{1+x^2}{\rm{d}x}=\dfrac{\pi^3}{8}$$ and then: $$\sum\limits_{n=0}^\infty\left((-1)^n\dfrac{1}{(2n+1)^3}\right)=\dfrac{\pi^3}{32}$$ ...