# 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 $$\det(\Delta)=e^{-\zeta'_{\Delta}(0)}=e^{-2 \zeta'(0)}=e^{\log(2 \pi)}=2\pi$$

$$\Gamma(s)\zeta_\Delta(s)=\Gamma(s)\sum_{(n,m) \ne (0,0)} (\pi a^2 n^2+\pi b^2 m^2)^{-s}=\int_0^\infty t^{s-1} (\Theta(a t)\Theta(bt)-1)dt$$ 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 $$\zeta_{a,b}'(0)$$.