TITLE: Help computing an integral for Green's function of a discrete Laplacian on a square lattice
QUESTION [10 upvotes]: 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.

REPLY [3 votes]: 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 $G_{n,n}$.