Functional determinant

where tr stands for the functional trace: the determinant is then defined by

This path integral is only well defined up to some divergent multiplicative constant. To give it a rigorous meaning it must be divided by another functional determinant, thus effectively cancelling the problematic 'constants'.

The problem is to find a way to make sense of the determinant of an operator S on an infinite dimensional function space. One approach, favored in quantum field theory, in which the function space consists of continuous paths on a closed interval, is to formally attempt to calculate the integral

where N is an infinite constant that needs to be dealt with by some regularization procedure. The product of all eigenvalues is equal to the determinant for finite-dimensional spaces, and we formally define this to be the case in our infinite-dimensional case also. This results in the formula

and so if the functional determinant is well-defined, then it should be given by

Since the analytic continuation of the zeta function is regular at zero, this can be rigorously adopted as a definition of the determinant.

from which a similar formula for the hyperbolic sine function can be derived: