# Matsubara sum arising from QFT and contour integral

In the lecture of E. Fradkin on quantum field theory, an example of Matsubara sum is performed using contour integration (see eq. 5.214 in the lecture). It reads $$\sum_{n=-\infty}^{\infty} \frac{e^{in\tau}}{n^2+x^2} = \oint_{C_{+}\cup C_{-}} \frac{dz}{2\pi i} \frac{e^{iz\tau}}{z^2+x^2} \pi\cot(\pi z)\quad(\tau, x>0),$$ where $$C_{\pm}$$ denotes the lines $$z\pm i\epsilon$$ followed counter clockwise (see this figure). I see that above equation can be deduced applying residue theorem to the finite box, obtained by truncating $$C_{+}\cup C_{-}$$, and taking a limit.

To evaluate the right hand side, the author deforms the contour into $$C^{+}\cup C^{-}$$ and calculate instead residues at $$z=\pm ix$$. I am curious about the mathematical justification of this process.

To me, equating $$\oint_{C_{-}}(\text{integrand})$$ with residue at $$-ix$$ seems illegitimate since $$e^{iz\tau}$$ blows up as $$z\rightarrow -i\infty$$. I don't get what 'deforming $$C_{-}$$ to $$C^{-}$$' even means.

One approach that I tried is the change of variable $$z\rightarrow 1/z$$, which amounts to passing to the other coordinate chart of Riemann sphere. It goes like $$(RHS)=\oint_{\Gamma} \frac{dz}{2\pi i} \frac{e^{i\tau/z}}{1+z^2x^2} \pi\cot(\pi/z),$$ where $$\Gamma$$ is the image of $$C_{+}\cup C_{-}$$ followed counter clockwise. The problem here is twofold; first, $$\Gamma$$ passes through the essential singularity at $$z=0$$. Second, if $$C_{+}\cup C_{-}$$ is taken to be the limit of boxes $$B_n$$ and $$\Gamma$$ the limit of their images $$B'_n$$, each contour $$B'_n$$ encircles an infinite subset of the poles $$\{\pm 1/n: n\in\mathbb{Z}\}$$. I have no idea if I can apply the residue theorem and take a limit in this case.

## edit

Following Svyatoslav's suggestion, I obtained (for $$0<\tau<2\pi$$) $$\sum_{n=-\infty}^{\infty} \frac{e^{in\tau}}{n^2+x^2} = \oint_{C_{+}\cup C_{-}} \frac{dz}{2\pi i} \frac{e^{iz\tau}}{z^2+x^2} \frac{\pi e^{-i\pi z}}{\sin\pi z} = \frac{\pi}{x} \frac{\cosh(\pi-\tau)x}{\sinh(\pi x)} \\ = \frac{\pi}{x} \left[ \cosh(\tau x) \coth(\pi x) - \sinh(\tau x) \right].$$ However, it is claimed in (5.215) of the lecture that $$\sum_{n=-\infty}^{\infty} \frac{e^{in\tau}}{n^2+x^2} \approx \frac{\pi}{2x}\coth(\pi x) e^{-|\tau|x} \quad \text{for small } \tau.$$ Did the author make a mistake?

• You are right - $e^{iz\tau}$ blows up at $z\to-i\infty$ if we use $\pi\cot(\pi/z)$. It is more convenient here to use $\pi\frac{e^{-\pi{i}z}}{\sin(\pi{z})}$ instead of $\pi\cot(\pi z)$ (which also has the residual $=1$ at integer $z$), if the contour goes below and above axis $X$ counter clockwise. The condition $r\in(-\pi;\pi)$ provides convergence at $z\to\pm{i}\infty$ Feb 22, 2021 at 16:06
• could you elaborate on what $r$ is and how your modification deals with both limits $z\rightarrow\pm i\infty$? Feb 22, 2021 at 17:21
• Sorry for mistake - I designate $r$ your $\tau$, and $r\in(0;2\pi)$ At $z=-it$ ($t$ - a real positive number) and $t\to\infty$ we get $e^{izr}\frac{e^{-\pi{i}z}}{\sin(\pi{z})}$$\sim\frac{\exp(-(\pi-r)t)}{exp(\pi{t})}=\exp(-(2\pi-r)t)\to0. If z=it and t\to+\infty e^{izr}$$\frac{e^{-\pi{i}z}}{\sin(\pi{z})}\sim\frac{\exp(-(r-\pi)t)}{exp(\pi{t})}=\exp(-rt)\to0$. Feb 22, 2021 at 18:30
• As soon as $r>2\pi$ or $r<-0$ due to periodicity of $\exp(inr)$ the value of the sum $\sum_{n=-\infty}^{\infty} \frac{e^{inr}}{n^2+x^2}$ reduces to the case $r\in(0;2\pi)$. For example $\exp(-\pi{i}n/2)=\exp(3\pi{i}n/2)$, so the case $r=-\pi/2$ is identical to $r=3\pi/2$ Feb 22, 2021 at 18:32
• To get a closed form of the sum you should then deform the contour to the big circle $R\to\infty$, "catching" the poles at $z=\pm{i}x$ Feb 22, 2021 at 18:55

This is a duplicate of what I wrote on Physics Stack exchange. I hope this is OK because there is a separate readership here.

You can also derive @Svyatoslav's correct expression by Poisson Summation: $$\frac 1{2\pi} \sum_{n=-\infty}^\infty \frac{e^{in\tau}}{n^2+M^2}= \sum_{n=-\infty}^\infty \frac 1{2|M |} e^{-|M||\tau+2\pi n|}, \quad \hbox{(Poisson Summation)}\\ = \frac 1 {2M} \frac{\cosh(\pi -\tau)M}{\sinh \pi M}, \quad 0<\tau<2\pi,\nonumber$$ The first line come from applying Poisson summation to the zero temperature expression
$$\int_{-\infty}^{\infty} \frac{dk}{2\pi}\frac{e^{ik\tau}} {k^2+M^2}=\frac 1 {2|M|}e^{-|\tau||M|}$$ and has the physical interpretation as the method-of-images sum over the $$n$$-fold winding of the particle trajectory around the periodic imaginary time direction. The passage from the first to second lines is just summing the two geometric series from $$n=0$$ to $$\infty$$ and $$n=-\infty$$ to $$-1$$.

This is @Svyatoslav's answer.

That the integrand blows up on the lower half-plane is correct. For a legitimate calculation, put $$\pi e^{-i\pi z}$$ instead of $$\pi\cot(\pi z)$$:

$$\sum_{n=-\infty}^{\infty} \frac{e^{in\tau}}{n^2+x^2} = \oint_{C_{+}\cup C_{-}} \frac{dz}{2\pi i} \frac{e^{iz\tau}}{z^2+x^2} \frac{\pi e^{-i\pi z}}{\sin\pi z} = \frac{\pi}{x} \frac{\cosh(\pi-\tau)x}{\sinh(\pi x)} \\ = \frac{\pi}{x} \left[ \cosh(\tau x) \coth(\pi x) - \sinh(\tau x) \right].$$

Here we assume, without loss of generality, $$0\leq \tau<2\pi$$. Note that as $$t\rightarrow\infty,$$

$$e^{iz\tau}\frac{\pi e^{-i\pi z}}{\sin \pi z} \sim \begin{cases} e^{-\tau t}, &\quad z=it \\ e^{(\tau-2\pi)t}, &\quad z=-it \end{cases}$$

so we can close the contour around a large circle toward both the upper and lower half-planes.