# Abel–Plana formula with fractional offset

The Abel–Plana formula compares the sum $$\sum_{n=0}^\infty f(n)$$ to the integral $$\int_0^\infty f(x)\,dx$$, $$$$\sum_{n=0}^{\infty}f\left(n\right)-\int_{0}^{\infty}f\left(x\right)dx=\frac{1}{2}f(0)+i\int_{0}^{\infty}\frac{f\left(ix\right)-f\left(-ix\right)}{e^{2\pi x}-1}\,dx.\tag{1}\label{1}$$$$ In applications to the Casimir force between two mirrors one is interested in the effect of different boundary conditions on the mirrors, producing a fractional offset $$\nu\in(0,1)$$ of the index $$n$$, so one wishes to compare $$\sum_{n=0}^\infty f(n+\nu)$$ to the same integral $$\int_0^\infty f(x)\,dx$$.

The fractional offset $$0<\nu<1$$ can be accounted for by a phase shift in the Abel–Plana formula, $$$$\sum_{n=0}^\infty f(n+\nu)-\int_{0}^{\infty}f(x)\,dx=i\int_0^\infty \left(\frac{f(ix)}{e^{2\pi (x+ i\nu)}-1}-\frac{f(-ix)}{e^{2\pi (x- i\nu)}-1}\right)\,dx.\tag{2}\label{2}$$$$ Note that there is no $$f(0)$$ term on the r.h.s.

Q: I would like to be able to refer to \eqref{2}, is it in the literature? Alternatively, I would be happy to refer to an MO answer that derives it.

Notes:

Ramanujan gives several generalizations of the Abel-Plana formula \eqref{1}, including the case of a $$\nu=1/2$$ offset, and a formula that follows from \eqref{2} by subtracting the offsets $$\nu=(1\pm\alpha)/2$$, with $$0<\alpha<1$$: $$\sum_{n=0}^\infty[f(2n+1-\alpha)-f(2n+1+\alpha)]=\frac{\sin\pi\alpha}{2}\int_0^\infty\frac{f(ix)+f(-ix)}{\cosh\pi x + \cos \pi\alpha}\,dx.\tag{3}\label{3}$$ See Ramanujan and Koshliakov meet Abel and Plana by B.C. Berndt et al.

I could not find the formula \eqref{2} itself in Ramanujan's notebook (although I don't doubt that he knew it).

A more modern derivation for the case of a $$\nu=1/2$$ offset can be found in Appendix B.2 of Phys.Rev.Research 3, 023201 (2021).

• Might also be interesting to look at a fractional offset generalization of the Euler-Maclaurin formula, in the same sense as equation $(2)$ Commented Nov 30, 2023 at 21:43