This question looks quite lengthy because I'm sketching the proof in the lecture - the two questions (look out for something bold) are actually relatively short.
I need some help with a proof in our statistical mechanics course (for $\mathbb R^2$) that goes like this:
- Find that it suffices to prove positive definiteness of the "matrix" $[(x_1+y_1-1)^2+(x_2-y_2)]^{-p/2}, x_i,y_i\in\mathbb Z, x_1,y_1\leq 0$.
- To show that, write $$[(x_1+y_1-1)^2+(x_2-y_2)]^{-p/2}=c_p\int_{\mathbb R^2}\frac{dk}{|k|^{2-p}}e^{ik_1(x_1+y_1-1)}e^{ik_2(x_2-y_2)}.$$ (This follows from the fact that $|\cdot|^{-p}$ is radial and homogeneous of degree $-p$; using scaling arguments one can see that its Fourier transform has to be of the form given above.)
- Considering only the integration over $k_1$ (i.e. the integral $\int_\mathbb{R}\frac{dk_1}{(k_1^2+k_2^2)^{(2-p)/2}}e^{ik_1(x_1+y_1-1)}$) we want to rewrite it in terms of a contour integral along the branch cut of the square root, which we choose to be the one going from the singularity $-ik_2$ to $-i\infty$.
- To do this think of the contour
that goes like $[-R,R]$, then (from $+R$) like a semicircle to the lower halfplane until it reaches $-iR+\varepsilon$, then goes up till $-ik_2+\varepsilon$, makes an almost closed circle around this singularity (going to $-ik_2-\varepsilon$) and then the same way back to $-R$.
- because of the exponential decay the "semicircle" parts will not contribute as $R\rightarrow\infty$.
- because of Darboux's inequality the "almost" circle around $-ik_2$ will not contribute as $\varepsilon\rightarrow 0$.
- hence (because of Cauchy $\oint=0$) the integrals $\int_{-iR+\varepsilon}^{-ik_2+\varepsilon}+\int_{-ik_2-\varepsilon}^{-iR-\varepsilon}$ are (in the limits) equal to $-\int_\mathbb{R}$.
- Using those results we are supposed to get $\int_\mathbb{R}\frac{dk_1}{(k_1^2+k_2^2)^{(2-p)/2}}e^{ik_1(x_1+y_1-1)}=2\sin(\pi(1-p/2))\int_{k_2}^\infty \frac{d\tau}{(\tau^2-k_2^2)^{1-p/2}}e^{\tau(x_1+y_1-1)}$.
My first question: We choose the branch to be along the imaginary axis (going downwards), which, I thought, would be achieved by writing $\sqrt z=\sqrt{|z|}e^{i\theta/2}$ for $\theta$ being the argument and being in $[-\pi/2,3\pi/2)$. So shouldn't we get $e^{i\pi/2(-(1-p/2))}$ and $e^{i3\pi/2(-(1-p/2))}$ (one of which gets multiplied by $-1$ in order to interchange the bounds of integration so that the two integrals can be added up). But this clearly doesn't sum up to $2\sin(1-p/2)$ (which would require to choose the negative real axis as branch of the square root).
- In the proof we continued to write (for $x,y$ as above, arbitrary $f$ having finite support on $\{(x_1,x_2)\in\mathbb Z^2|x_1\leq 0\}$) $$\sum_{x,y}f(x)f(y)[(x_1+y_1-1)^2+(x_2-y_2)^2]^{-p/2}=$$ $$c_p 2\sin(\pi(1-p/2))\int_\mathbb{R}dk_2\int_{|k_2|}^\infty \frac{d\tau}{(\tau^2-k_2^2)^{1-p/2}}e^{-\tau}\left|\sum_x f(x)e^{\tau x_1+ik_2x_2}\right|^2\geq 0,$$ which proves the claim.
My second question: Is this equality supposed to obvious? Also, what justifies the absolute values around the sum?