$\log(t)$ term in the small time expansion of $\mathrm{Tr}( A e^{-tB} )$ Assume $A$ is an operator on a Hilbert space with discrete spectrum. Assume $B$ is a positive operator on the same Hilbert space also with a discrete spectrum. Also assume $A$ and $B$ commute.
I'm looking at the small $t$ expansion of the expression,
$$W(t) = \mathrm{Tr} \left( A e^{-tB} \right) = \sum_{n=0}^\infty D_n A_n e^{-t B_n}$$
where $A_n$ and $B_n$ are the eigenvalues of $A$ and $B$, $D_n$ is the dimension of the eigenspace corresponding to $A_n$ and $B_n$, and we can further assume that $B_n$ grows quadratically with $n$, $D_n$ grows at most polynomially with $n$ and $A_n$ is such that $W(t)$ is finite for $t>0$, i.e. $A_n$ does not grow too fast.
I'd like to get the $\log(t)$ term in the small $t$ expansion of $W(t)$ i.e. not the leading polynomial ones. I mean on general grounds I'll have, for $t\to 0$,
$$W(t) = \sum_j \frac{W_j}{t^{\alpha_j}} + w \log(t) + O(t\log(t)) + \ldots$$
with exponents $\alpha_j > 0$ and what I'd like to have is the coefficient $w$.
In principle I can successively find all polynomial terms, subtract them, and then the leading term will be the $\log(t)$ term, but I suppose there is a way to get directly the $\log(t)$ term, what would it be?
 A: The answer by Carlo Beenakker is correct that any contribution to $D_n A_n \sim n^p$ with $p\ge 0$ does not generate any logarithmic asymptotic terms. Instead of numerical checks, one can also examine the Taylor-Maclaurin asymptotic series for $\sum_{n=0}^\infty n^p e^{-t n^2}$ (no $\log(t)$ terms are generated by differentiating the summand w.r.t $n$).
What is left to cover is the case $p=-1$. Then
$$
  \sum_{n=1}^\infty n^{-1} e^{-t n^2} \sim \int_1^\infty n^{-1} e^{-t n^2} dn
    = \frac{1}{2} \int_t^\infty \frac{e^{-x}}{x} dx
    = -\frac{1}{2} \log(t) + O(1) .
$$
The subleading terms from the Euler-Maclaurin formula and in $O(1)$ above will also not be logarithmic (compare with the series expansion of the exponential integral $E_1(x)$ DLMF §6.6.2).
So to get the logarithmic contribution to $W(t)$, you can directly look at the $O(n^{-1})$ term in the asymptotic expansion of $D_n A_n$ (provided the expansion itself doesn't have funny terms involving $\log(n)$ or something like that).
A: I don't think there will be a logarithmic term. For definiteness, let me take $B_n=n^2$, $D_nA_n=n^p$, and approximate the sum over $n$ by an integral,
$$I_p(t)=\int_0^\infty n^p e^{-tn^2}=\tfrac{1}{2} t^{-(p+1)/2} \Gamma \left(\frac{p+1}{2}\right).$$
The $t^{-(p+1)/2}$ small-$t$ asymptote agrees quite well with a numerical evaluation of the sum $S_p(t)=\sum_{n=0}^\infty n^p e^{-tn^2}$, without any logarithmic correction.
For $p=0$ I can do better, then $S_1$ is an elliptic function,
$$S_1(t)=\tfrac{1}{2}+\tfrac{1}{2} \vartheta _3\left(0,e^{-t}\right)\rightarrow \tfrac{1}{2}\sqrt{\pi/t}+{\cal O}(t^0),$$
without a logarithmic correction.
