Infinite series with inverse trigonometric functions Consider the infinite series
$$
F(s)=\sum_{k=1}^\infty \frac{1}{k^{2s+1} \sin(k \pi \sqrt{2})}
$$
Prudnikov Vol.1 , gives a result for this sum in 5.4.16.13 for s=1
$$
F(1)=-\frac{13 \pi^3}{360 \sqrt{2}}
$$
but no reference. How do you prove this result?
 A: Thanks Carlo for pointing out the solution @ math.stackexchange from Ron Gordon. Yes, this works. Indeed, one can use the same technique to solve my series for arbitrary positive integer $s$.
For this, start with the the contour pointed out by Carlo above, which vanishes:
$$
\oint_{C} dz \frac{1}{z^{2s+1} \sin(\pi z)\sin(\pi z (\sqrt{2}-1))}=0\,.
$$
The contour has simple poles for all integers except zero: $z=n$, $n=\pm 1,\pm 2,\pm 3,\ldots$; simple pose for all integers times $\sqrt{2}+1$: $z=(\sqrt{2}+1)n$ with the same n's and finally a pole of order $2s+3$ at $z=0$.
The residues of the poles near $z=n$ are found to be given by
$$
Res_{z=n}=\frac{(\sqrt{2}-1)^{2s}}{n^{2s+1}\pi \sin{\pi n \sqrt{2}}}.
$$
The residues of the poles near $z=n (\sqrt{2}+1)$ are given by
$$
Res_{z=n(\sqrt{2}+1)}=\frac{(\sqrt{2}-1)^{2s}}{n^{2s+1}\pi \sin(\pi n \sqrt{2})}.
$$
To find the residue of the pole at $z=0$, one can use the expansion of the cosecans function in terms of Bernoulli numbers:
$$
\csc(\pi x)=\sum_{n=0}^\infty \frac{B_{2n} (2^{2n}-2) (-1)^{n+1}}{(2n)!} (\pi x)^{2n-1},
$$
for both cosecans functions in the denominator. Here $B_{2n}$ are the Bernoulli numbers. The residue is the coefficient of the $1/z$ term in the resulting double sum, which is then given by
$$
Res_{z=0}=\sum_{n=0}^{s+1} \frac{(-1)^{s+1} (2^{2n}-2)(2^{2n-2s+2}-2) B_{2n} B_{2s-2n+2} \pi^{2s} (\sqrt{2}-1)^{2n-1}}{(2n)!(2s-2n+2)!}.
$$
Since the contour integral vanishes, the sums of the residue must also vanish, and one finds
$$
\sum_{n=1}^\infty \frac{1}{n^{2s+1}\sin (n \pi \sqrt{2})}=\frac{\pi^{2s+1} (-1)^{s}}{2 (1+(\sqrt{2}-1)^{2s})} \sum_{n=0}^{s+1} \frac{(2^{2n}-2)(2^{2s-2n+2}-2) B_{2n}B_{2s-2n+2} (\sqrt{2}-1)^{2n-1}}{(2n)!(2s-2n+2)!},
$$
proving Prudnikov's equation 5.4.16.12. Using the relation of the Bernoulli numbers and the Riemann zeta function, one can rewrite this as
$$
\sum_{n=1}^\infty \frac{1}{n^{2s+1}\sin (n \pi \sqrt{2})}=-\frac{2^{-2s-1}}{\pi (1+(\sqrt{2}-1)^{2s})} \sum_{n=0}^\infty \zeta(2n)\zeta(2s-2n+2) (2^{2n}-2)(2^{2s-2n+2}-2) (\sqrt{2}-1)^{2n-1}.
$$
I wonder if this could be used to define the sum away from positive integer $s$.
