Formula for $\zeta(3)$ -verification

By simple manipulating with some series I have found the following formula for $\zeta(3)$: $$\zeta(3)=\frac27\sum_{k=0}^{\infty}(-1)^kB_{2k}\frac{\pi^{2k+2}}{(2k+2)!},$$ where $b_k$ are Bernoulli numbers, defined from the equations: $$ B_0=1,\quad B_k=-\frac{1}{k+1}\sum_{i=0}^{k-1}\binom{k+1}{i} B_i,\quad k=1,2,3,\dots $$ Questions: Is this formula for $\zeta(3)$ correct? Can somebody check it numerically? If it's correct, is this well known identity or not? Thanks for your help.

Added. By usimg my identities another interesting formulas follows: $$\ln 2=\sum_{k=0}^{\infty}(-1)^kB_{2k}\frac{\pi^{2k}}{(2k+1)!}$$ $$\zeta(3)=\frac45\sum_{k=0}^{\infty}(-1)^kB_{2k}\frac{\pi^{2k+2}}{(2k+3)!}$$

Your formula is a variant of a formula from Euler ($1772$) (exposed in Ayoub's paper Euler and the Zeta Function (1974) p. $1085$) rediscovered by Ramaswami ($1934$) and more recently by Ewell ($1990$).
See too Adamchik's paper and Boros and Moll's nice book 'Irresisitible Integrals' $(11.4.6)$ :

\begin{align} \zeta(3)&=\frac{\pi^2}7\left(1-4\sum_{n=1}^\infty \frac{\zeta(2\,n)}{(2n+2)(2n+1)\,2^{2n}}\right)\qquad\text{or since}\;\zeta(0)=-\frac12\\ \tag{1}\zeta(3)&=-\frac{4\pi^2}7\sum_{n=0}^\infty \frac{\zeta(2\,n)}{(2n+2)(2n+1)\,2^{2n}}\\ \end{align} This becomes indeed, after substitution of $\,\displaystyle \frac{\zeta(2n)}{2^{2n}}=(-1)^{n+1}\frac{B_{2n}\,\pi^{2n}}{2(2n)!}$ : \begin{align} \zeta(3)&=\frac{2\,\pi^2}7\sum_{n=1}^\infty (-1)^n\frac{B_{2n}\,\pi^{2n}}{(2n+2)(2n+1)(2n)!}\\ \tag{2}\zeta(3)&=\frac27\sum_{n=0}^{\infty}(-1)^nB_{2n}\frac{\pi^{2n+2}}{(2n+2)!}\\ \end{align}

Let's reproduce here Boros and Moll's proof of $(1)$ and start with the generating function for central binomial coefficients : $$\tag{3}\frac 1{\sqrt{1-4t}}=\sum_{n=0}^\infty \binom{2n}{n}t^n$$ Setting $\,t:=\left(\dfrac x2\right)^2$ and integrating we get : $$\tag{4}\arcsin(x)=\sum_{n=0}^\infty \frac{\binom{2n}{n}}{2^{2n}}\frac{x^{2n+1}}{2n+1}$$ (btw Euler found too this neat expression for the square : $\displaystyle \arcsin(x)^2=\sum_{n=1}^\infty \frac{2^{2n}}{\binom{2n}{n}}\frac{x^{2n}}{2\,n^2}\;$)

Dividing $(4)$ by $x$ and integrating we get : $$\tag{5}\int\frac{\arcsin(x)}xdx=\sum_{n=0}^\infty \frac{\binom{2n}{n}}{2^{2n}}\frac{x^{2n+1}}{(2n+1)^2}$$ For $x:=\sin(t)$ this becomes : $$\tag{6}\int\frac{t\cos(t)}{\sin(t)}dt=\sum_{n=0}^\infty \frac{\binom{2n}{n}}{2^{2n}}\frac{(\sin(t))^{2n+1}}{(2n+1)^2}$$

At this point we may use the well known generating function for $\zeta(2n)$ : $$\pi\;x\;\cot(\pi\;x)=1-2\sum_{n=1}^\infty \zeta(2n)\;x^{2n}$$ So that setting $t:=\pi x\,$ and integrating we get using $(6)$ : $$\tag{7}\int t\,\cot(t)\,dt=t-2\sum_{n=1}^\infty \frac{\zeta(2n)\;t^{2n+1}}{\pi^{2n}(2n+1)}=\sum_{n=0}^\infty \frac{\binom{2n}{n}}{2^{2n}}\frac{(\sin(t))^{2n+1}}{(2n+1)^2}$$ Integrating this again from $0$ to $\frac {\pi}2$ returns : $$\frac {(\pi/2)^2}2-2\sum_{n=1}^\infty \frac{\zeta(2n)\;(\pi/2)^{2n+2}}{\pi^{2n}(2n+2)(2n+1)}=\sum_{n=0}^\infty \frac{\binom{2n}{n}}{2^{2n}}\frac{1}{(2n+1)^2}\int_0^{\pi/2}\sin^{2n+1}(t)\,dt$$ The integral at the right may be evaluated with the Wallis formula : $$\frac{\binom{2n}{n}}{2^{2n}}\int_0^{\pi/2}\sin^{2n+1}(t)\,dt=\frac1{2n+1}$$ and we get $$\tag{8}\frac {\pi^2}8-\frac{\pi^2}2\sum_{n=1}^\infty \frac{\zeta(2n)}{2^{2n}(2n+2)(2n+1)}=\sum_{n=0}^\infty\frac{1}{(2n+1)^3}=\frac 78\zeta(3)$$ (separating the even and odd terms of zeta it is easy to prove the more general identity $\;\displaystyle\sum_{n=0}^\infty\frac{1}{(2n+1)^m}=\left(1-\frac 1{2^m}\right)\zeta(m)\;$)

From $(8)$ we deduce $(1)$ and thus your formula $(2)$.

GENERALIZATION

Let's define $$f(m):=\sum_{n=0}^{\infty}(-1)^nB_{2n}\frac{\pi^{2n+m-1}}{(2n+m-1)!}=-\pi^{m-1}\,\sum_{n=0}^{\infty}\frac{(2n)!\;\zeta(2n)}{(2n+m-1)!\,2^{2n-1}}$$

then using the method nicely exposed in your answer (computing further integrals of $\dfrac{x}{e^x-1}$ that is evaluating $\;\displaystyle \int \cdots \int \frac{x}{e^x-1} dx\cdots\,dx=\int_0^x \frac{t}{e^t-1}\frac{(x-t)^{n-1}}{(n-1)!} dt\,$ for $x=-\pi\,i$)

we may obtain : \begin{align} f(1)&=0\\ f(2)&=\pi\,\log(2)\\ f(3)&=\frac 72\zeta(3)\\ f(4)&=\frac 54\pi\zeta(3)\\ \tag{9}f(5)&=-\frac {31}4\zeta(5)+\pi^2\zeta(3)\\ f(6)&= \frac{\pi^3}3\zeta(3) -\frac{49}{16}\pi\zeta(5)\\ f(7)&=\frac{381}{32}\zeta(7)-2\pi^2\zeta(5)+\frac{\pi^4}{12}\zeta(3)\\ f(8)&= \frac{\pi^5}{60}\zeta(3)-\frac 23\pi^3\zeta(5)+\frac{321}{64}\pi\zeta(7)\\ f(9)&=-\frac{511}{32}\zeta(9)+3\pi^2\zeta(7)-\frac{\pi^4}6\zeta(5)+\frac{\pi^6}{‌​360}\zeta(3)\\ \end{align} and so on and thus conjecture that for $\;n>1$ we have : $\qquad\qquad\qquad\qquad\qquad\qquad\qquad(10)$ \begin{align} f(2n+1)&=(-1)^{n+1}\left[n\left(4-2^{1-2n}\right)\zeta(2n+1)+2\sum_{k=1}^{n-1} \frac{(-1)^k(n-k)}{(2k)!}\pi^{2k}\ \zeta(2(n-k)+1)\right]\\ f(2n+2)&=(-1)^{n+1}\pi\left[\left(2n-1+2^{-2n}\right)\zeta(2n+1)+2\sum_{k=1}^{n‌​-1} \frac{(-1)^k(n-k)}{(2k+1)!}\pi^{2k}\,\zeta(2(n-k)+1)\right]\\ \end{align}

For a proof see the Theorem A of Cvijovic and Klinowski ($1997$) "New rapidly convergent series representations for $\zeta(2n+1)$" with the result (similar to the first equation $(10)$) : $$\tag{11}\zeta(2n+1)=\frac{(-1)^n\,(2\pi)^{2n}}{n(2^{2n+1}-1)}\left[\sum_{k=1}^{n-1}\frac{(-1)^{k-1}\,k\,\zeta(2k+1)}{(2n-2k)!\,\pi^{2k}}+\sum_{k=0}^\infty\frac{(2k)!\,\zeta(2k)}{(2n+2k)!\,2^{2k}}\right]$$

This allows to find whole families of formulae for $\zeta(2n+1)$ with some neat instances like these :
(cf the comments for $\zeta(5)$) \begin{align} \zeta(5)&=\frac{16}{23}\sum_{n=0}^{\infty}(-1)^nB_{2n}\frac{(2n+2)\ \pi^{2n+4}}{(2n+5)!}\\ \zeta(5)&=-\frac{2^5\,\pi^4}{23}\sum_{n=0}^\infty \frac{\zeta(2\,n)}{(2n+5)(2n+4)(2n+3)(2n+1)\,2^{2n}}\\ \zeta(7)&=-\frac{2^7\,\pi^6}{4719}\sum_{n=0}^\infty \frac{(128\,n+211)\quad\zeta(2\,n)}{(2n+7)(2n+6)(2n+5)(2n+4)(2n+3)(2n+1)\,2^{2n}}\\ \zeta(9)&=-\frac{2^{10}\,\pi^8}{77259}\sum_{n=0}^\infty \frac{(449\,n+780)\quad\quad\zeta(2\,n)}{(2n+9)(2n+8)(2n+7)(2n+6)(2n+5)(2n+3)(2n+1)\,2^{2n}}\\ \zeta(11)&=-\frac{2^{12}\,\pi^{10}}{\small{395681475}}\sum_{n=0}^\infty\frac{(2497024\,n^2+10676923\,n+11093808)\quad\zeta(2n)}{(2n+11)(2n+10)\cdots(2n+6)(2n+5)(2n+3)(2n+1)\,2^{2n}}\\ \zeta(13)&=-\frac{2^{13}\,\pi^{12}}{\small{75159854595}}\sum_{n=0}^\infty\frac{\small{(619308920\,n^2+2659184244\,n+2771839831)\quad}\zeta(2n)}{\small{(2n+13)(2n+12)\cdots(2n+8)(2n+7)(2n+5)(2n+3)(2n+1)\,2^{2n}}}\\ \zeta(15)&=-\frac{2^{15}\,\pi^{14}}{\small{1697182926260535}}\sum_{n=0}^\infty\frac{\small{(21253850808320\,n^3+165886464354888\,n^2+415352534250460\,n+332739769444737)}\zeta(2n)}{\small{(2n+15)(2n+14)\cdots(2n+8)(2n+7)(2n+5)(2n+3)(2n+1)\,2^{2n}}}\\ \end{align} The coefficients for these formulae may possibly be obtained using recurrences (by rewriting $(11)$ or using alternative relations) but I obtained them numerically with high precision (i.e. these identities are only conjectured).

We may compare them with the Euler formula $(1)$ and its variants : \begin{align} \zeta(3)&=-\frac{4\pi^2}7\sum_{n=0}^\infty \frac{\zeta(2\,n)}{(2n+2)(2n+1)\,2^{2n}}\\ \zeta(3)&=-\frac{8\pi^2}5\sum_{n=0}^\infty \frac{\zeta(2\,n)}{(2n+3)(2n+2)(2n+1)\,2^{2n}}\\ \zeta(3)&=\frac{2\pi^2}7\left(\log(2)+\sum_{n=0}^\infty \frac{\zeta(2\,n)}{(n+1)\,2^{2n}}\right)\\ \end{align} These two last formulae were obtained by Chen and Srivastava in "Some Families of Series Representations for the Riemann $\zeta(3)$" and reproduced in Srivastava and Choi's book "Zeta and q-Zeta Functions and Associated Series and Integrals" (see p.$405$ "$4.2$ Rapidly Convergent Series for $\zeta(2n + 1)$") that contains many other series for $\zeta$ and general formulae like this one : \begin{align} \zeta(2n+1)&=\frac{(-1)^{n-1}(2\pi)^{2n}}{2^{2n+1}-1}\left[\frac{H_{2n}-\log(\pi)}{(2n)!}+\sum_{k=1}^{n-1}\frac{(-1)^k\,\zeta(2k+1)}{(2n-2k)!\,\pi^{2k}}+2\sum_{k=1}^\infty\frac{(2k-1)!\,\zeta(2k)}{(2n+2k)!\,2^{2k}}\right]\\ \end{align} A review is proposed in Katsurada's paper "Rapidly convergent series representations for $\zeta(2n+1)$ and their $\chi$-analogue".

OTHER GENERALIZATIONS

The OP asked what would happen with the upper bound of the integral $x=-\pi\,i\,$ replaced by $\,x=-\frac{\pi}2i\,$ so let's investigate this by defining $$g(m):=\sum_{n=0}^{\infty}(-1)^nB_{2n}\frac{\left(\frac{\pi}2\right)^{2n+m-1}}{(2n+m-1)!}=-\left(\frac{\pi}2\right)^{m-1}\,\sum_{n=0}^{\infty}\frac{2\,(2n)!\;\zeta(2n)}{(2n+m-1)!\;2^{4n}}$$

then evaluating $\;\displaystyle \int \cdots \int \frac{x}{e^x-1} dx\cdots\,dx=\int_0^x \frac{t}{e^t-1}\frac{(x-t)^{n-1}}{(n-1)!} dt\,$ for $x=-\frac{\pi}2\,i\;$ returns if $\;\displaystyle \beta(s):=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^s}$ is the Dirichlet beta function : \begin{align} g(1)&=\frac{\pi}4\\ g(2)&=\frac{\pi}4\log(2)+\beta(2)\\ g(3)&=\frac{35}{16}\zeta(3) -\frac 12\pi\,\beta(2)\\ \tag{12}g(4)&=\frac{61}{64}\pi\,\zeta(3)-3\,\beta(4)\\ g(5)&=\frac 14\pi^2\,\zeta(3)-\frac{527}{128}\zeta(5)+\frac 12\pi\,\beta(4)\\ g(6)&=\frac 1{24}\pi^3\,\zeta(3)-\frac{2033}{1024}\pi\,\zeta(5)+5\,\beta(6)\\ g(7)&=\frac 1{192}\pi^4\,\zeta(3)-\frac 12\pi^2\,\zeta(5)+\frac{24765}{4096}\zeta(7)-\frac 12\pi\,\beta(6)\\ \end{align}

and so on so that we may conjecture that for $\;n>1$ : $\qquad\qquad\qquad\qquad\qquad\qquad\qquad(13)$ \begin{align} g(2n+1)&=(-1)^n\left[\frac {\pi}2\beta(2n)-n\frac{2^{4n+1}+2^{2n}-1}{2^{4n}}\zeta(2n+1)-\\2\sum_{k=1}^{n-1}(-1)^k\frac{n-k}{(2k)!}\left(\frac{\pi}2\right)^{2k}\zeta(2(n-k)+1))\right]\\ g(2n+2)&=(-1)^n\left[(2n+1)\beta(2n+2)-\left(n-\frac{2^{2n}-1}{2^{4n+2}}\right)\pi\,\zeta(2n+1)-\\2\sum_{k=1}^{n-1}(-1)^k\,\frac{n-k}{(2k+1)!}\left(\frac{\pi}2\right)^{2k+1}\zeta(2(n-k)+1)\right]\\ \end{align}

Returning some expressions of interest for the Catalan constant $G$ and other $\beta(2n)$ constants :

\begin{align} G=\beta(2)&=-\frac{\pi}4\log(2)-\pi\,\sum_{n=0}^{\infty}\frac{\zeta(2n)}{(2n+1)\;2^{4n}}\\ \beta(4)&=-\frac {\pi^3}{840}\left[\frac{61}4\log(2)+\sum_{n=0}^\infty \frac{(244(n+2)(n+1)-9)\,\zeta(2n)}{(2n+3)(2n+2)(2n+1)\,2^{4n}}\right]\\ \beta(6)&=-\frac{\pi^5}{3541440}\left[\frac{50345}8\log(2)+\\\sum_{n=0}^\infty\frac{(402760n^4 + 3020700n^3 + 8300346n^2 + 9777801n + 4077233)\zeta(2n)}{(2n+5)(2n+4)(2n+3)(2n+2)(2n+1)\,2^{4n}}\right]\\ \end{align} as well as fast series for $\zeta(\text{odd})$ : \begin{align} \zeta(3)&=-\frac{2\;\pi^2}{35}\left[\log(2)+4\sum_{n=0}^\infty\frac{(2n+3)\quad\zeta(2n)}{(2n+2)(2n+1)\;2^{4n}}\right]\\ \zeta(5)&=-\frac{2\;\pi^4}{55335}\left[157\,\log(2)+8\sum_{n=0}^\infty\frac{(628\,n^3+3140\,n^2+5111\,n+2581)\;\zeta(2n)}{(2n+4)(2n+3)(2n+2)(2n+1)\;2^{4n}}\right]\\ \end{align}

We could get slower converging results with the upper bound of the integral replaced by $\,x:=-2\pi\,i$. For this let's define $$h(m):=\sum_{n=0}^{\infty}(-1)^nB_{2n}\frac{(2\,\pi)^{2n+m-1}}{(2n+m-1)!}=-(2\,\pi)^{m-1}\,\sum_{n=0}^{\infty}\frac{2\,(2n)!\;\zeta(2n)}{(2n+m-1)!}$$

then evaluating $\;\displaystyle \int \cdots \int \frac{x}{e^x-1} dx\cdots\,dx=\int_0^x \frac{t}{e^t-1}\frac{(x-t)^{n-1}}{(n-1)!} dt\,$ for $x=-2\,\pi\,i$)

we may obtain : \begin{align} h(3)&=0\\ h(4)&=6\,\pi\,\zeta(3)\\ h(5)&=4\,\pi^2\,\zeta(3)\\ \tag{14}h(6)&=\frac 83\pi^3\,\zeta(3)-10\,\pi\,\zeta(5)\\ h(7)&=\frac 43\pi^4\,\zeta(3)-8\,\pi^2\,\zeta(5)\\ h(8)&= \frac 8{15}\pi^5\,\zeta(3)-\frac{16}3\pi^3\,\zeta(5)+14\,\pi\,\zeta(7)\\ h(9)&= \frac 8{45}\pi^6\,\zeta(3)-\frac 83\,\pi^4\,\zeta(5)+12\,\pi^2\,\zeta(7)\\ h(10)&=\frac{16}{315}\pi^7\,\zeta(3)-\frac{16}{15}\pi^5\,\zeta(5)+8\,\pi^3\,\zeta(7)-18\,\pi\,\zeta(9)\\ \end{align} And find further expressions for $\zeta(\text{odd})$ (using these expressions or combining them with the earlier results) but since the subject appears endless I'll stop here.

So,I put here my derivation, but I'm not sure, if all my steps are fully correct. The key idea is very simple. We just put $x=-i\pi$ in this formula: \begin{equation} 2\zeta(3)+\zeta(2)x+\frac{x^3}{12}+\sum_{k=0}^{\infty}B_{2k}\frac{x^{2k+2}}{(2k+2)!}=2\sum_{k=1}^{\infty}\frac{e^{kx}}{k^3}-x\sum_{k=1}^{\infty}\frac{e^{kx}}{k^2},\quad x\in(-2\pi,0\,\rangle\end{equation} plus some simplifacations. That's all.

(Formula can be easily obtaided from two expansions: $$\frac{x}{e^x-1}=-\frac{x}2+\sum_{k=0}^{\infty}B_{2k}\frac{x^{2k}}{(2k)!},\quad x\in(-2\pi,2\pi),$$ $$\frac{x}{e^x-1}=-x\frac{1}{1-e^x}=-x\sum_{k=0}^{\infty}e^{kx},\quad x\in(-\infty,0),$$ so formal double integration gives: $$\int_0^x\!\!\int_0^x\frac{x}{e^x-1} dx dx=-\frac{x^3}{12}+\sum_{k=0}^{\infty}B_{2k}\frac{x^{2k+2}}{(2k+2)!}$$ $$\int_0^x\!\!\int_0^x\frac{x}{e^x-1} dx dx=2\sum_{k=1}^{\infty}\frac{e^{kx}}{k^3}-x\sum_{k=1}^{\infty}\frac{e^{kx}}{k^2}-\frac{x^{3}}{6}-\zeta(2)x-2\zeta(3)$$ which implies the desired identity.)

Now back to my derivation of formula for $\zeta(3)$. Putting $x=-i\pi$ we have: $$ 2\zeta(3)-i\pi\zeta(2)+\frac{(-i\pi)^3}{12}+\sum_{k=0}^{\infty}B_{2k} \frac{(-i\pi)^{2k+2}}{(2k+2)!}=2\sum_{k=1}^{\infty}\frac{e^{(-i\pi)k}}{k^3}+i\pi\sum_{k=1}^{\infty}\frac{e^{(-i\pi)k}}{k^2} $$ and because $$e^{-i\pi}=\cos(-\pi)+i\sin(-\pi)=-1$$ $$\sum_{k=1}^{\infty}\frac{(-1)^k}{k^2}=-\frac12\zeta(2)$$ $$\sum_{k=1}^{\infty}\frac{(-1)^k}{k^3}=-\frac34\zeta(3),$$ we have step by step: \begin{align} 2\zeta(3)-i\pi\zeta(2)+\frac{i\pi^3}{12}-\sum_{k=0}^{\infty}B_{2k}\frac{(-1)^{k}\pi^{2k+2}}{(2k+2)!}&=2\sum_{k=1}^{\infty}\frac{(-1)^k}{k^3}+i\pi\sum_{k=1}^{\infty}\frac{(-1)^k}{k^2}\\ 2\zeta(3)-i\pi\zeta(2)+\frac{i\pi^3}{12}-\sum_{k=0}^{\infty}B_{2k}\frac{(-1)^{k}\pi^{2k+2}}{(2k+2)!}&=-\frac32\zeta(3)-i\pi\frac12\zeta(2)\\ \frac72\zeta(3)-\frac{i\pi}2\zeta(2)&=\sum_{k=0}^{\infty}B_{2k}\frac{(-1)^{k}\pi^{2k+2}}{(2k+2)!}-\frac{i\pi^3}{12}, \end{align} so we immediately obtain even two formulas: $$\zeta(2)=\frac{\pi^2}6$$ $$\zeta(3)=\frac{2}7\sum_{k=0}^{\infty}(-1)^{k}B_{2k}\frac{\pi^{2k+2}}{(2k+2)!}.$$