I need to evaluate the following summation to closed form:
$$\sum _{k=1}^{\infty } \frac{\sin (e k n \pi ) \sin (e k (1+n) \pi )}{k \pi \sin (e k \pi )}$$
where: $n>1$ and $n\in \mathbb{Z}$, $e \approx 2.718281828459$
By using the Abel-Plana formula, I obtain: $$ \sum_{n = 0}^{\infty}f(n) = \int \limits_{0}^{\infty}f(x)dx + \frac{1}{2} f(0) + i\int \limits_{0}^{\infty}\frac{f(ix) - f(-ix)}{e^{2\pi x}-1}dx $$
$$\sum _{k=0}^{\infty } \frac{\sin (e (1+k) n \pi ) \sin (e (1+k) (1+n) \pi )}{(1+k) \pi \sin (e (1+k) \pi )}=\int_0^{\infty } \frac{\sin (e (1+k) n \pi ) \sin (e (1+k) (1+n) \pi )}{(1+k) \pi \sin (e (1+k) \pi )} \, dk+\frac{1}{2} \left(\lim_{k\to 0} \, \frac{\sin (e (1+k) n \pi ) \sin (e (1+k) (1+n) \pi )}{(1+k) \pi \sin (e (1+k) \pi )}\right)+0$$
$$\sum _{k=0}^{\infty } \frac{\sin (e (1+k) n \pi ) \sin (e (1+k) (1+n) \pi )}{(1+k) \pi \sin (e (1+k) \pi )}=\int_0^{\infty } \frac{\sin (e (1+k) n \pi ) \sin (e (1+k) (1+n) \pi )}{(1+k) \pi \sin (e (1+k) \pi )} \, dk+\frac{\csc (e \pi ) \sin (e n \pi ) \sin (e (1+n) \pi )}{2 \pi }$$
dosen't seems to work,but I put $n=2$ and calculate numerical the results do not match.
$$\sum _{k=0}^{\infty } \frac{\sin (2 e (1+k) \pi ) \sin (3 e (1+k) \pi )}{(1+k) \pi \sin (e (1+k) \pi )} \approx -0.154747$$
by Abel-Plana formula I got: $\approx -0.107417$