Help computing an integral for Green's function of a discrete Laplacian on a square lattice

I need to calculate the following integral: $$ \int_0^1 \int_0^1 \frac{1-\cos(2 \pi k_1 x) \cos(2 \pi k_2 y)}{4 \sin(\pi k_1)^2 + 4 \sin( \pi k_2)^2} dk_1 dk_2 $$ I have tried to use some contour integral techniques, but without much success. Any help will be appreciated.

Solution 1:

According to these two papers: Lattice Green's functions in all dimensions, and Discrete scattering theory: Green’s function for a square lattice, the entry of the Laplacian matrix of the two dimensional square lattice: $$ G_{xy} = \frac{1}{2}\int_0^1 \int_0^1 \frac{1-\cos(2 \pi k_1 x) \cos(2 \pi k_2 y)}{2 - \cos(2 \pi k_1 ) - \cos(2 \pi k_2 )} dk_1 dk_2\\ = \frac{1}{4}\int_0^1 \int_0^1 \frac{1-\cos(2 \pi k_1 x) \cos(2 \pi k_2 y)}{ \sin^2(\pi k_1) + \sin^2( \pi k_2) } dk_1 dk_2. $$ OP's formula looks a bit misleading because the square of sine is on the parentheses.

When $x = y = n$, $$ G_{xy} =\frac{1}{4}\int_0^1 \int_0^1 \frac{1-\cos(2 \pi k_1 n) \cos(2 \pi k_2 n)}{ \sin^2(\pi k_1) + \sin^2( \pi k_2) } dk_1 dk_2 \\ = \frac{1}{4}\int_0^1 \int_0^1 \frac{2- \cos\big(2 n\pi (k_1-k_2)\big)- \cos\big(2 n\pi (k_1+k_2) \big)}{ 2 - \cos(2 \pi k_1 ) - \cos(2 \pi k_2 ) } dk_1 dk_2 \\ \frac{1}{4}\int_0^1 \int_0^1 \frac{2- \cos\big(2 n\pi (k_1-k_2)\big)- \cos\big(2 n\pi (k_1+k_2) \big)}{ 2 - 2 \cos\big(\pi (k_1+k_2) \big) \cos\big( \pi (k_1-k_2) \big) } dk_1 dk_2. $$ Let $u = k_1 + k_2\in (0,2)$, $v = k_1- k_2\in (-1,1)$, $|\det \partial(k_1,k_2)/\partial(u,v)| = 1/2 $, hence $$ \begin{aligned} G_{n,n} &= \frac{1}{16} \int^2_0 \left(\int^{1}_{-1}\frac{2- \cos(2 n\pi v)- \cos(2 n\pi u)}{ 1 - \cos (\pi u) \cos( \pi v ) } dv\right)du \\ &= \frac{1}{16} \int^{1}_{-1} \left(\int^{1}_{-1}\frac{2- \cos(2 n\pi v)- \cos(2 n\pi u)}{ 1 + \cos (\pi u) \cos( \pi v ) } dv\right)du \\ &=\frac{1}{16} \int^{1}_{-1} \left(\int^{1}_{-1}\frac{2- 2\cos(2 n\pi u)}{ 1 + \cos (\pi u) \cos( \pi v ) } dv\right)du && \text{(By symmetry)} \\ &=\frac{1}{8} \int^{1}_{-1}\frac{1- \cos(2 n\pi u)}{ \sin(\pi u)} du \\ &=\frac{1}{4\pi} \int^{1}_{-1}\frac{ \sin^2( n\pi w)}{ \sin^2(\pi w)} d(\cos\pi w ) && \text{(Let $w= -u$)} \\ &= \frac{1}{4\pi} \int^{1}_{-1} \left(\frac{\sin\big(n \arccos x \big)}{\sqrt{1-x^2}} \right)^2 dx. \end{aligned} $$ And this Chebyshev polynomial of the second kind $U_{n-1}(x)$. By $$ \begin{aligned} &U^2_n(x) - U^2_{n-1}(x) = \big(U_n(x) + U_{n-1}(x)\big)\big(U_n(x) - U_{n-1}(x)\big) \\ =&\frac{\sin\big((n+1)w\big) + \sin\big(nw\big)}{\sin w}\cdot \frac{\sin\big((n+1)w\big) - \sin\big(nw\big)}{\sin w} \\ =&\frac{\sin\big((2n+1)w/2\big)\cos (w/2 )}{\cos (w/2 )\sin (w/2 )}\cdot \frac{\sin (w/2 )\cos\big((2n+1)w/2\big)}{\cos (w/2 )\sin (w/2 )} \\ =& \frac{\sin\big((2n+1)w/2\big)\cos\big((2n+1)w/2\big)}{\cos (w/2 )\sin (w/2 )} \\ =& \frac{\sin\big((2n+1)w/2\big)}{\sin w} = U_{2n}(x). \end{aligned} $$ Hence $$ G_{nn} = \frac{1}{2\pi} \int^{1}_{0} \sum^{n}_{k=1} U_{2k-2}(x)dx = \frac{1}{2\pi} \sum^{n}_{k=1} \frac{1}{2k-1}. $$ Now by recurrence relation we can move on to compute $x = n,y =n+1$ case. Please refer to the reference that the second paper I listed points to: Green’s Functions in Quantum Physics. My library allows me to have a free preview (or Springer just has it), please refer to formula (5.39) in the link above, the author uses an elliptic integral to represent $G_{n,n+1}$ by $G_{n,n}$ he called $G_{1,z}$ for $y-x=1$, so here I basically tried to derive a more explicit formula for $G_{n,n}$.