Let $M$ be an $n$ dimensional closed manifold and consider an elliptic, pseudodifferential operator $P$ of order $-n$. Here are some facts which I had learned so far:

1. There exists a density defined in terms of $-n$-th homogenous part of the symbol of $P$ such that integrating this density one obtains a trace on the algebra of all (classical) pseudodifferential operators. This is the so called *Wodzicki residue*, to be denoted by $Wres$.
2. If $n>1$ this is unique such trace.

3. If $P$ is as above that $P$ belongs to the Dixmier trace, is measurable (i.e. the Dixmier trace does not depend from the choice of a state) and thus Dixmier trace $Tr^+P$ is defined.

4. We have and equality $Tr^+P=Res_{s=1}(Trace(P^s))$ and thus $P \mapsto Res_{s=1} (Trace(P^s))$ is also a trace.

5. Also the function $P \mapsto Res_{s=0}(Trace(P\Delta^{-\frac{s}{2}}))$ is a trace.

Finally the last fact:
6. There is a universal constant $c>0$ (depending only on the dimension of $M$) such that $Res_{s=0}(Trace(P\Delta^{-\frac{s}{2}}))=c Wres(P)$ (note that this is slightly more which can be deduced from the uniqueness of the trace since this constant does not depend on the choice of $M$ provided that the dimension remains the same).

To have the full picture of these issues I would like to know

- How to determine (in the easiest way, provided we know all the above facts) this universal constant $c$ (it should be $\frac{1}{n(2\pi)^n}$).
- Is it clear that $Res_{s=1}(Trace(P^s))$ coincides with $Res_{s=0}(Trace(P\Delta^{-\frac{s}{2}}))$? Is it clear that the former is a trace?

I would be happy in any solution to these two problems (i.e. answering 2. once we know the answer for 1. etc).