I am not satisfied with the explanations of $$\sum_n \log \lambda_n = - \frac{d}{ds} \sum_n \lambda_n^{-s}\bigg|_{s=0}$$ "trick" used in zeta function regularization, discussed here and here, or the references therein. If someone could give a step-by-step explanation or a reference if the explananation is more complicated than I'm guessing, that would be greatly appreciated.

  • 1
    $\begingroup$ For fun, an other basic trick, used for entropy, is $S = - Tr(\rho \log \rho) = -\frac{d}{ds} Tr(\rho^s)_{|s=1}$ $\endgroup$
    – Trimok
    Commented Oct 17, 2013 at 16:02

1 Answer 1


Consider a function of $s$ as $$ f(s) = \frac{1}{a^s} $$ Now take the derivative with respect to $s$: $$ f'(s)=\frac{d}{ds}a^{-s}=-\frac{\log(a)}{a^s} $$ What happens at $s=0$? $$ f'(s=0)=-\frac{\log(a)}{a^0}=-\log(a) $$

What happens if $a=a_0+a_1+a_2+\cdots$ (i.e., $s$-independent constants)? We simply stick in a sum: $$ f(s) =\sum_n \frac{1}{a_n^s} $$ What does this do to the final answer? Nothing, we still can stick in the sum: $$ f'(s=0)=-\sum_n\frac{\log(a_n)}{a_n^0}=-\sum_n\log(a_n) $$

Thus, we get our relation, $$ \left.\frac{d}{ds}\sum_n\lambda_n^{-s}\right|_{s=0}=-\sum_n\log\left(\lambda_n\right) $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.