TITLE: A new formula for Apery's constant and other zeta(s)?
QUESTION [31 upvotes]: I recently found these Plouffe-like formulas using Mathematica's LatticeReduce. Has anybody seen/can prove these are indeed true? 
$$\begin{aligned}\frac{3}{2}\,\zeta(3) &= \frac{\pi^3}{24}\sqrt{2}-2\sum_{k=1}^\infty \frac{1}{k^3(e^{\pi k\sqrt{2}}-1)}-\sum_{k=1}^\infty\frac{1}{k^3(e^{2\pi k\sqrt{2}}-1)}\\
\frac{3}{2}\,\zeta(5) &= \frac{\pi^5}{270}\sqrt{2}-4\sum_{k=1}^\infty \frac{1}{k^5(e^{\pi k\sqrt{2}}-1)}+\sum_{k=1}^\infty \frac{1}{k^5(e^{2\pi k\sqrt{2}}-1)}\\
\frac{9}{2}\,\zeta(7) &= \frac{41\pi^7}{37800}\sqrt{2}-8\sum_{k=1}^\infty\frac{1}{k^7(e^{\pi k\sqrt{2}}-1)}-\sum_{k=1}^\infty\frac{1}{k^7(e^{2\pi k\sqrt{2}}-1)} \end{aligned}$$
And so on for other $\zeta(2n+1)$.  The background for these are in my blog.

REPLY [17 votes]: It seems  to have escaped attention  that these sums  may be evaluated
using harmonic summation techniques.
Introduce the sum
$$S(x; \alpha, p) = \sum_{n\ge 1} \frac{1}{n^p(e^{\alpha n x}-1)}$$
with $p$ a positive odd integer and $\alpha>1$, so that we seek e.g.
$2 S(1; \pi\sqrt{2}, 3)+S(1; 2\pi\sqrt{2}, 3).$
The sum term is harmonic and  may be evaluated by inverting its Mellin
transform.
Recall the harmonic sum identity
$$\mathfrak{M}\left(\sum_{k\ge 1} \lambda_k g(\mu_k x);s\right) =
\left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} \right) g^*(s)$$
where $g^*(s)$ is the Mellin transform of $g(x).$
In the present case we have
$$\lambda_k = \frac{1}{k^p}, \quad \mu_k = k 
\quad \text{and} \quad
g(x) = \frac{1}{e^{\alpha x}-1}.$$
We need the Mellin transform $g^*(s)$ of $g(x)$ which is
$$\int_0^\infty \frac{1}{e^{\alpha x}-1} x^{s-1} dx
= \int_0^\infty \frac{e^{-\alpha x}}{1-e^{-\alpha x}} x^{s-1} dx
\\ = \int_0^\infty \sum_{q\ge 1} e^{-\alpha q x} x^{s-1} dx
=  \sum_{q\ge 1} \int_0^\infty e^{-\alpha q x} x^{s-1} dx 
\\= \Gamma(s) \sum_{q\ge 1} \frac{1}{(\alpha q)^s}
= \frac{1}{\alpha^s} \Gamma(s) \zeta(s).$$
It follows that the Mellin transform $Q(s)$ of the harmonic sum 
$S(x;\alpha,p)$ is given by
$$Q(s) = \frac{1}{\alpha^s} \Gamma(s) \zeta(s) \zeta(s+p)
\quad\text{because}\quad
\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} = 
\sum_{k\ge 1} \frac{1}{k^p} \frac{1}{k^s}
= \zeta(s+p)$$
for $\Re(s) > 1-p.$
The Mellin inversion integral here is
$$\frac{1}{2\pi i} \int_{3/2-i\infty}^{3/2+i\infty} Q(s)/x^s ds$$
which we evaluate  by shifting it to the left  for an expansion about
zero.
First formula.
We take
$$Q(s) = \frac{1}{\pi^s\sqrt{2}^s} 
\left(2 + \frac{1}{2^s}\right)
\Gamma(s) \zeta(s) \zeta(s+3).$$
We shift the Mellin inversion integral to the line $s=-1$, integrating
right through  the pole at  $s=-1$ picking up the  following residues:
$$\mathrm{Res}(Q(s)/x^s; s=1) =
\frac{\pi^3\sqrt{2}}{72 x}
\quad\text{and}\quad
\mathrm{Res}(Q(s)/x^s; s=0) =
-\frac{3}{2}\zeta(3)$$
and
$$\frac{1}{2}\mathrm{Res}(Q(s)/x^s; s=-1) =
\frac{\pi^3\sqrt{2}x}{36}.$$
This almost  concludes the proof of  the first formula if  we can show
that the  integral on the line  $\Re(s) = -1$ vanishes  when $x=1.$ To
accomplish this we must show that the integrand is odd on this line.
Put $s = -1 - it$ in the integrand to get
$$\pi^{1+it} \sqrt{2}^{1+it} 
\left(2 + 2^{1+it}\right)
\Gamma(-1-it) \zeta(-1-it) \zeta(2-it).$$
Now use  the functional equation of  the Riemann Zeta  function in the
following form:
$$\zeta(1-s) = \frac{2}{2^s\pi^s} 
\cos\left(\frac{\pi s}{2}\right) \Gamma(s) \zeta(s)$$
to obtain (with $s=-1-it$)
$$\pi^{1+it} \sqrt{2}^{1+it} 
\left(2 + 2^{1+it}\right)
\zeta(2+it) 2^{-1-it} \pi^{-1-it} 
\frac{1}{2\cos\left(\frac{\pi (-1-it)}{2}\right)}
\zeta(2-it)$$
which is
$$ \sqrt{2}^{1+it} 
\left(2^{-it} + 1\right)
\zeta(2+it)
\frac{1}{2\cos\left(\frac{\pi (1+it)}{2}\right)}
\zeta(2-it)$$
and finally yields
$$-\frac{1}{\sin(\pi i t/2)}
\left(\sqrt{2}^{-1-it}+\sqrt{2}^{-1+it}\right)
\zeta(2+it)\zeta(2-it).$$
It is now possible to  conclude by inspection: the zeta function terms
and the powers of the square root are even in $t$ and the sine term is
odd, so the whole term is odd and the integral vanishes. (We get exponential decay from the sine term.)
Second formula.
We take
$$Q(s) = \frac{1}{\pi^s\sqrt{2}^s} 
\left(4 - \frac{1}{2^s}\right)
\Gamma(s) \zeta(s) \zeta(s+5).$$
We shift the Mellin inversion integral  to the line $s=-2$ (no pole on
the line this time) picking up the following residues:
$$\mathrm{Res}(Q(s)/x^s; s=1) =
\frac{\pi^5\sqrt{2}}{540 x}
\quad\text{and}\quad
\mathrm{Res}(Q(s)/x^s; s=0) =
-\frac{3}{2}\zeta(5)$$
and
$$\mathrm{Res}(Q(s)/x^s; s=-1) =
\frac{\pi^5\sqrt{2}x}{540}.$$
It remains to verify that the integrand on the line $\Re(s)=-2$ is odd
when $x=1$. Put $s=-2-it$ in the integrand to get
$$\pi^{2+it} \sqrt{2}^{2+it} 
\left(4 - 2^{2+it}\right)
\Gamma(-2-it) \zeta(-2-it) \zeta(3-it).$$
Applying the functional equation once again with $s=-2-it$ we obtain
$$\pi^{2+it} \sqrt{2}^{2+it} 
\left(4 - 2^{2+it}\right)
\zeta(3+it) 2^{-2-it} \pi^{-2-it} 
\frac{1}{2\cos\left(\frac{\pi(-2-it)}{2}\right)}
\zeta(3-it)$$
which is
$$\sqrt{2}^{2+it} 
\left(2^{-it} - 1\right)
\zeta(3+it)
\frac{1}{2\cos\left(\frac{\pi(-2-it)}{2}\right)}
\zeta(3-it)$$
which is in turn
$$\left(\sqrt{2}^{2-it} - \sqrt{2}^{2+it}  \right)
\frac{1}{2\cos\left(\frac{\pi(2+it)}{2}\right)}
\zeta(3+it)
\zeta(3-it)$$
which finally yields
$$-\left(\sqrt{2}^{2-it} - \sqrt{2}^{2+it}  \right)
\frac{1}{2\cos(\pi i t /2)}
\zeta(3+it)
\zeta(3-it)$$
The  product of  the zeta  function terms  is even,  as is  the cosine
term. The term in front is odd, so the integrand is odd as claimed.
(We get exponential decay from the cosine term.)
Third formula.
We take
$$Q(s) = \frac{1}{\pi^s\sqrt{2}^s} 
\left(8 + \frac{1}{2^s}\right)
\Gamma(s) \zeta(s) \zeta(s+7).$$
We shift the Mellin inversion integral to the line $s=-3$, integrating
right through  the pole at  $s=-3$ picking up the  following residues:
$$\mathrm{Res}(Q(s)/x^s; s=1) =
\frac{17 \pi^7\sqrt{2}}{37800 x}
\quad\text{and}\quad
\mathrm{Res}(Q(s)/x^s; s=0) =
-\frac{9}{2}\zeta(7)$$
and
$$\mathrm{Res}(Q(s)/x^s; s=-1) =
\frac{\pi^7\sqrt{2}x}{1134}
\quad\text{and}\quad
\frac{1}{2} \mathrm{Res}(Q(s)/x^s; s=-3) =
-\frac{\pi^7\sqrt{2}x^3}{4050}.$$
This almost concludes  the proof of this third formula  if we can show
that the  integral on the line  $\Re(s) = -3$ vanishes  when $x=1.$ To
accomplish this  we must show once  more that the integrand  is odd on
this line.
Put $s = -3 - it$ in the integrand to get
$$\pi^{3+it} \sqrt{2}^{3+it} 
\left(8 + 2^{3+it}\right)
\Gamma(-3-it) \zeta(-3-it) \zeta(4-it).$$
By the functional equation we obtain with $s = -3-it$
$$\pi^{3+it} \sqrt{2}^{3+it} 
\left(8 + 2^{3+it}\right)
\zeta(4+it)
2^{-3-it} \pi^{-3-it} 
\frac{1}{2\cos\left(\pi(-3-it)/2\right)}
\zeta(4-it)$$
which is
$$\sqrt{2}^{3+it} 
\left(2^{-it} + 1\right)
\zeta(4+it)
\frac{1}{2\cos\left(\pi(3+it)/2\right)}
\zeta(4-it)$$
which finally yields
$$\frac{1}{2\sin(\pi it/2)}
\left(\sqrt{2}^{3-it} + \sqrt{2}^{3+it}\right)
\zeta(4+it)
\zeta(4-it).$$
This concludes it since the  two zeta function terms together are even
as  is the  square root  term while  the sine  term is  odd,  so their
product is odd.
A similar yet not quite the same computation can be found at this MSE link.
Another computation in the same spirit is at this MSE link II.