Krdytkyu

Not having a great series background, I turn to the wider world... :)

I need to find two sums:

$$sum_{m=1}^{infty}frac{sin^2(mpi)}{(m^2-a^2)^2}$$

and

$$sum_{m=1}^{infty}frac{m^2}{(m^2-a^2)^2}$$

I've seen some related series on here, but not close enough that I can figure out how to get these sums from the ones posted. Any help would be much appreciated.

The first one is $0$ because $sin mpi = 0$ for any $min Bbb Z$.

For the second one, we split it into two sums,
begin{align*}
S=sum_{m=1}^inftyfrac{m^2}{(m^2-a^2)^2}&=sum_{m=1}^inftyfrac{m^2-a^2+a^2}{(m^2-a^2)^2}\
&=underbrace{sum_{m=1}^inftyfrac{1}{m^2-a^2}}_{S_1}+a^2underbrace{sum_{m=1}^inftyfrac{1}{(m^2-a^2)^2}}_{S_2}.
end{align*}
From this answer, we know
$$S_1=frac1{2a^2}-frac{picot,pi a}{2a}.tag{*}$$
Differentiate $(*)$ with respect to $a$ (we can do that because $S_1$ uniformly converges),
begin{align*}S_1'=frac d{da}sum_{m=1}^inftyfrac{1}{m^2-a^2}&=2asum_{m=1}^inftyfrac{1}{(m^2-a^2)^2}=2aS_2\
implies S_2&=frac{1}{2a}S_1'.
end{align*}
$$therefore S=S_1+frac{a}{2}S_1'=frac{pi^2}{4}csc^2pi a-frac{picot pi a}{4a}.$$

I did find an answer to the second one:

$$sum_{m=1}^{infty} frac{m^2}{(m^2-a^2)^2}= frac{pi[2pi a-sin(2pi a)]}{8asin^2(pi a)}$$

According to a Dodonov, Klimov, and Nikonov "Quantum Particle in a Box with Moving Walls" J. Math. Phys. 34 (8) August 1993