Trace and Det of Laplacian on the rectangle I consider the eigenvalue problem $\Delta \varphi = \lambda \varphi$ with the Dirichlet boundary condition $\varphi|_{ \partial \Omega}=0$ on the rectangle $\Omega= [0,l] \times [0,m]$. 
By using separation of variable method, that is, look for solution of the form $\varphi(x,y)=f(x)g(y)$. The eigenfunctions are $\varphi_{jk}(x,y)=\sin(\frac{j \pi}{l}x) \sin(\frac{k\pi}{m}y)$ for $j,k \geq 1$ and have eigenvalues 
$$\lambda_{jk}=(\frac{j\pi}{l})^2+(\frac{k \pi}{m})^2$$ 
Now i want to compute the Trace and Det of Laplacian by using (zeta regularization) formula
$$\zeta_{\Delta}(s)=\sum \lambda_k^{-s}=\frac{1}{\Gamma(s)} \int_{0}^\infty t^{s-1} Trace(e^{-t\Delta})$$
$$\det(\Delta)=e^{-\zeta'_{\Delta}(0)}$$
In the simpler case, on the interval $[0,l]$, the eigenvalues are $\varphi_k=\sin(\frac{k\pi}{l}x)$ with eigenvalues $\lambda_k=(\frac{k \pi}{l})^2$ for $k \geq 1$. We can compute the Trace and Det of Laplacian in this situation 
$$\zeta_{\Delta}(s)= \sum \lambda_k^{-s}=\sum (\frac{k \pi}{l})^{-2s}=(\frac{\pi}{l})^{-2s} \zeta(2s)$$
where $\zeta(s)= \sum k^{-s}$ is the Riemann zeta function. Hence
$$\det(\Delta)=e^{-\zeta'_{\Delta}(0)}=e^{-2 \zeta'(0)}=e^{\log(2 \pi)}=2\pi$$
 A: $$\Gamma(s)\zeta_\Delta(s)=\Gamma(s)\sum_{(n,m) \ne (0,0)} (\pi a^2 n^2+\pi  b^2 m^2)^{-s}=\int_0^\infty t^{s-1} (\Theta(a t)\Theta(bt)-1)dt$$ where (from the Poisson summation formula) $$\Theta(t) = \sum_n e^{-\pi t^2} = t^{-1/2} \Theta(1/t)$$ As $ t \to 0$, $$\Theta(t) =t^{-1/2} \Theta(1/t)= t^{-1/2}(1+O(e^{-\epsilon/t}))= t^{-1/2}+O(e^{-\epsilon/t}), \\ \Theta(a t)\Theta(bt)-1 = ( (at)^{-1/2}+O(e^{-\epsilon/t})( (bt)^{-1/2}+O(e^{-\epsilon/t}))-1= (ab)^{-1/2}t^{-1}-1+O(e^{-\epsilon/t})$$ and hence
$$\int_0^\infty t^{s-1} (\Theta(a t)\Theta(bt)-1 + 1_{x < 1}-(ab)^{-1/2}t^{-1} 1_{x < 1})dt = \Gamma(s)\zeta_\Delta(s) + \frac1s - \frac{(ab)^{-1/2}}{s-1}$$ is entire, which means $\zeta_\Delta(s)$ has a simple pole at $s=1$ of residue $(ab)^{-1/2}$ and $\zeta_\Delta(0) = 1$ and $\zeta_\Delta(-k)=0$.
With $\zeta_\Delta = \zeta_{a,b}$ the change of variable $t=1/u$ yields the functional equation $$\Gamma(s)\zeta_{a,b}(s) = (ab)^{-1/2} \Gamma(1-s)\zeta_{1/a,1/b}(1-s)$$
For $a\ne b$ I doubt you can find a closed-form for $\zeta_{a,b}'(0)$.
