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Integrated kernel and spectral zeta function

  1. Aug 25, 2010 #1
    I was looking at a paper about strong-coupling expansion (N. F. Svaiter, Physica (Amsterdam) 345A, 517 (2005) ) and it claims that
    [tex] -\int d^d x \int d^d y (-\Delta + m^2)\delta^d(x-y) = \textbf{Tr} I + \left.\frac{d}{ds}\zeta(s)\right|_{s=0} [/tex]
    where [tex]\zeta(s)[/tex] is the spectral zeta function, and [tex]I[/tex] is the identity matrix.

    It is clear to me that the derivative of the zeta function is related to the logarithm of the determinant of the operator in the left-hand side. What is *not* clear is how that double integral is related to that.
  2. jcsd
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