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In the paper Vacuum $\langle \phi^2 \rangle$ in Schwarzschild Spacetime by Candalas and Howard, they say that for each non-zero $\epsilon$ it is true that $$ \sum_{n=1}^\infty \cos\left( n \kappa \epsilon \right) \ = \ - \frac{1}{2} $$

This is equation (2.7) in the paper, where $\kappa$ is a constant (later set as the surface gravity for the black hole) and $\epsilon \to 0^{+}$ is taken as a regulator.

In what sense is this true? As some kind of distributional statement? Because this doesn't converge in the strict sense.

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    $\begingroup$ Compute $\sum_{n=1}^\infty e^{-n\rho} \cos (nx)$ instead and then take $\rho \to 0^+$ which gives $-\frac{1}{2}$. At this stage you can take $\epsilon \to 0^+$. $\endgroup$
    – Prahar
    Commented May 15, 2019 at 2:28

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Note that $$III(x)~=~ \delta(x-\mathbb{Z})~=~ \sum_{m\in\mathbb{Z}} \delta(x-m)~=~\sum_{n\in\mathbb{Z}}e^{2\pi i xn}~=~1+2\sum_{n\in\mathbb{N}}\cos(2\pi xn)$$ is the Dirac comb/Shah distribution.

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  • $\begingroup$ So for $x \notin \mathbb{Z}$ (which is going to be true for $ x = \kappa \epsilon/(2\pi)$ if $\epsilon$ is small enough), then we have $\delta(x - \mathbb{Z}) =0$ and from which the identity follows? $\endgroup$
    – QuantumEyedea
    Commented May 15, 2019 at 18:55
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    $\begingroup$ $\uparrow$ Yes. $\endgroup$
    – Qmechanic ♦
    Commented May 15, 2019 at 18:57

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