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Yes, I have verified that this is the case. However, I obtained it in a different way as a particular case of an even more general identity (see P.S. in the question).
@ Iosif Pinelis I think prefactor in your solution should be $(k\pi)^4$, not $k^4$. What is $c$ in the general case and why you cannot use $|x-k\pi|<1/k$ in this case also?
Richtmyer says it is square-integrable, but it seems you are right (I have done some numerical calculations). The same statement is reproduced in arxiv.org/abs/quant-ph/9907069. Maybe there was a typo in Richtmyer.
