Consider on the circle $S^1$ the operator
$$L := - \frac{\partial^2}{\partial \theta^2} + c$$
for some constant $c \in \mathbb{R}$. 


**What is its $\zeta$-regularized determinant?**


This should be well-known, I suppose, but I didn't find a reference.


*Some background:* The eigenvalues of $L$ are $n^2+c$ for $n \in \mathbb{Z}$ (with multiplicity two each), and therefore the zeta-function for positive $c$ is given by
$$\zeta\_c(s) = 2\sum\_{n=0}^\infty \frac{1}{(n^2+c)^s}.$$
Now the $\zeta$-regularized determinant is defined by
$$\det(L) = e^{-\zeta^\prime(0)}.$$
How does one compute such a thing?


\Edit: The hint to look for "thermal zeta functions" was good: Here is a paper that computes the determinant in question: <http://arxiv.org/pdf/hep-th/9505154v1.pdf>



Zeta-Determinant for shifted Laplacians on the circle
Let me write a bit of what seems to me that natural/naive approach, and perhaps the questioner can comment on the direction...


From $\int\_0^\infty y^s\,e^{-ty}\;{dy\over y}=t^{-s}\,\Gamma(s)$, writing $Z(s)$ for your zeta, 
$$
\pi^{-s/2}\,\Gamma(s/2)\,(Z(s)-{1\over c^s})
\;=\; \int\_0^\infty y^{s/2}\,\sum\_{n\in \mathbb Z} e^{-\pi (n^2+c)y}\;{dy\over y}
$$ 
Unlike Riemann's argument, the presence of $c>0$ will cause Poisson summation to produce an expression that behaves well, without breaking the integral into two pieces. Namely, the Fourier transform of $x\to e^{-\pi (x^2+c)y}$ is ${1\over \sqrt{y}} e^{-\pi (x^2/y +cy)}$, so this becomes
$$
\sum\_{n\in \mathbb Z} \int\_0^\infty y^{{s-1\over 2}} e^{-\pi (n^2/y+cy)}\;{dy\over y}
$$
The $n=0$ term should be taken out, and is elementary. For the rest, replace $y$ by $\sqrt{c} y/n$, to get
$$
\pi^{-s/2}\,\Gamma(s/2)\,(Z(s)-{1\over c^s})
\;=\; (n=0) + \sum\_{n\not=0}(\sqrt{c}/n)^{(s-1)/2} \int\_0^\infty y^{{s-1}\over 2} e^{-\pi \sqrt{c}\,n\,(y+1/y)}\;{dy\over y}
$$
The right-hand side seems to be entire in $s$, so the factor of $\Gamma(s/2)$ on the left-hand side means that the factor $Z(s)-c^{-s}$ vanishes at $s=0$. Thus, the right-hand side is close to computing the derivative of $Z(s)-c^{-s}$ at $s=0$.


But/and this makes me wonder whether about that extra term $c^{-s}$. It is needed to make Poisson summation work, but then it doesn't seem to me that $Z(s)=0$, so $Z'(0)$ would not be the leading term, which I would have thought would have been the object of interest.


In any case, the right-hand side is a sum of values of Bessel functions, but not obviously (to me) further simplifiable at $s=0$. At $s=1$, the sum over $n$ admits summing as geometric series, so at least the outcome is just an integral in $y$... though it seems not elementary, still.


Comment?


