How to determine the trace and determinant of a differential operator? How to determine the trace and determinant of the operator like $\Box$ or $\nabla^2$ etc. But first of all how to find the same for the simpler operator $\frac{d}{dx}$? I proceeded as follows. What basis functions should I choose- $\{e^{ikx}\}$ or $\{\delta(x-x^\prime)\}$? Since, the first basis is the diagonal basis the calculation of trace and determinant will bee easier. The $kk^\prime$-matrix element is given by $$\int\limits_{-\infty}^{\infty} e^{-ikx}\frac{d}{dx}e^{ik^\prime x}=2\pi ik\delta(k-k^\prime).$$ Now the task is to determine the trace and the determinant. Since, in discrete case $$\sum\limits_{i,j} A_{ij}\delta_{ij}=\sum\limits_{i} A_{ii}$$ gives the trace of a matrix A. Then, can we find the trace as $$\int dk\int dk^\prime 2\pi ik\delta(k-k^\prime).$$ But the result is infinite! Is this approach correct? 
Edit : If not please indicate the correct method and expected result.
 A: In infinite dimensions, you can define the trace only for a special class of compact operators: the so-called trace-class operators. Given an Hilbert space $\mathscr{H}$, the space of trace class operators $\mathscr{I}_1(\mathscr{H})$ is a two-sided ideal of the bounded operators $\mathscr{L}(\mathscr{H})$.
The two operations $\mathrm{Tr}$ and $\mathrm{Det}(1+\cdot)$ defined as follows:
$$\mathrm{Tr}: A\mapsto \mathrm{Tr}(A)\; ,\; \mathrm{Det}(1+\cdot): A\mapsto \mathrm{Det}(1+A)\; ,$$
have the following properties:


*

*the first is a linear bounded functional on $\mathscr{I}_1(\mathscr{H})$;

*the second is a continuous function on $\mathscr{I}_1(\mathscr{H})$.
The operators you cite (all of them), are unbounded. So you cannot expect in any way that their trace (or determinant) should be finite, for it is like expecting that $\sum_{n=0}^\infty n$ is finite.
A: Renormalization of a determinant $\det M$ of an unbounded operator $M$ 
is invariably of the following type: 
It essentially consists of writing the operator $A$ as a product $M=M_0(1+A)$ where $A$ is trace class and $M_0$ is known. Then $\det M=\det M_0\det(1+A)$ in any regularization (i.e., representation by limits of bounded operators). The (in the limit infinite) value $\det M_0$ must cancel rigorously in all observable quantities computed from this, in order that the regularization procedure is mathematically consistent.
Typically, this infinite factor is the determinant of an exactly solvable system of which the determinant of interest can be considered as a (relatively compact) perturbation. 
The book ''Trace ideals and their applications'' by B. Simon 
discusses the regularization procedure in some detail in Chapter 9. 
(But for field theories in more than 2 dimensions, more stringent renormalization is needed than discussed there, to make $A$ trace class.)
