Bethe Ansatz: Heisenberg Chain Ground-State Energy

Main result

For the spin-$\tfrac{1}{2}$ isotropic antiferromagnetic Heisenberg chain on $N$ sites with periodic boundary conditions,

\[H = J\sum_{i=1}^{N}\mathbf{S}_i\cdot\mathbf{S}_{i+1}, \qquad \mathbf{S}_{N+1}\equiv\mathbf{S}_1,\qquad J>0,\]

the ground-state energy per site in the thermodynamic limit $N\to\infty$ is

\[\boxed{\; e_0 \;=\; \lim_{N\to\infty}\frac{E_0(N)}{N} \;=\; J\!\left(\frac{1}{4}-\ln 2\right) \;\approx\; -0.4431\,J. \;}\]

The formula was first evaluated by L. Hulthén in 1938, starting from the Bethe-ansatz integral equation for the ground-state rapidity density. The derivation below reproduces every step: Bethe equations, energy formula, thermodynamic-limit integral equation for the density, its solution by Fourier transform, and the final integral evaluation by Parseval's theorem.

Setup

Hamiltonian and conventions

Pauli matrices and spin operators $\mathbf{S}_i = \tfrac{1}{2} \boldsymbol{\sigma}_i$ on site $i$; raising / lowering $S^\pm_i = S^x_i \pm i\,S^y_i$. The Hamiltonian

\[H = J\sum_{i=1}^{N}\mathbf{S}_i\cdot\mathbf{S}_{i+1} = J\sum_{i=1}^{N}\Bigl[\,S^z_i S^z_{i+1} + \tfrac{1}{2}\bigl(S^+_i S^-_{i+1} + S^-_i S^+_{i+1}\bigr)\Bigr]\]

has total $S^z_{\rm tot} = \sum_i S^z_i$ as a conserved quantity. We work in the zero-magnetisation sector $S^z_{\rm tot} = 0$, i.e. $M = N/2$ down-spins on $N$ sites (assume $N$ even).

Ferromagnetic reference. The state $|{\rm F}\rangle := |\!\uparrow\!\uparrow\cdots\uparrow\rangle$ is an eigenstate: every bond term $\mathbf{S}_i\cdot\mathbf{S}_{i+1} |\!\uparrow\uparrow\rangle = \tfrac{1}{4}|\!\uparrow\uparrow\rangle$ since $S^z_i S^z_{i+1}|\!\uparrow\uparrow\rangle = \tfrac{1}{4}|\!\uparrow\uparrow\rangle$ and the raising / lowering terms annihilate this state. Therefore

\[H|{\rm F}\rangle = J \cdot N \cdot \tfrac{1}{4}\,|{\rm F}\rangle \;=\; \tfrac{JN}{4}\,|{\rm F}\rangle.\]

Spinon excitations lower the energy below $JN/4$; the ground state of the antiferromagnetic chain ($J > 0$) lies in the $S^z_{\rm tot} = 0$ sector and is not $|{\rm F}\rangle$. All energies below are measured on the same absolute scale.

Goal

Compute $e_0 = \lim_{N\to\infty} E_0(N)/N$ from the Bethe-ansatz solution of $H$ in the $S^z_{\rm tot} = 0$ sector.

Calculation

Step 1 — One-magnon dispersion

Flip a single spin at site $n$ to obtain

\[|n\rangle := |\!\uparrow\cdots\uparrow\,\underset{\text{site }n}{\downarrow}\,\uparrow\cdots\uparrow\rangle, \qquad n = 1,\dots,N.\]

This is not yet a Hamiltonian eigenstate. Compute $H|n\rangle$ term by term. Bonds $(i, i{+}1)$ with $i \ne n{-}1, n$ act on two $\uparrow$ spins and return $\tfrac{J}{4}|n\rangle$; there are $N-2$ such bonds. The two bonds adjacent to the flipped spin $(n{-}1,n)$ and $(n, n{+}1)$ act on an $\uparrow\!\downarrow$ or $\downarrow\!\uparrow$ pair; each such bond contributes

\[\tfrac{J}{4}\cdot(-1) + \tfrac{J}{2}\bigl(S^+ S^-|\cdots\rangle\bigr) = -\tfrac{J}{4}|n\rangle + \tfrac{J}{2}|{\rm shifted}\rangle,\]

where the raising / lowering hop moves the down-spin to the neighbour. Explicitly,

\[\mathbf{S}_{n-1}\cdot\mathbf{S}_n\,|n\rangle = -\tfrac{1}{4}|n\rangle + \tfrac{1}{2}|n{-}1\rangle,\qquad \mathbf{S}_n\cdot\mathbf{S}_{n+1}\,|n\rangle = -\tfrac{1}{4}|n\rangle + \tfrac{1}{2}|n{+}1\rangle.\]

Summing,

\[H|n\rangle \;=\; J\Bigl[\tfrac{N-2}{4} - \tfrac{1}{2}\Bigr]|n\rangle + \tfrac{J}{2}|n{-}1\rangle + \tfrac{J}{2}|n{+}1\rangle \;=\; \bigl(\tfrac{JN}{4} - J\bigr)|n\rangle + \tfrac{J}{2}\bigl(|n{-}1\rangle + |n{+}1\rangle\bigr). \tag{1}\]

Diagonalise by a Fourier transform. With $|k\rangle = N^{-1/2}\sum_{n=1}^{N} e^{ikn}|n\rangle$ and PBC $|n+N\rangle = |n\rangle$, the allowed momenta are $k = 2\pi m / N$ for $m = 0, 1, \dots, N-1$. Acting with $H$,

\[H|k\rangle = \bigl(\tfrac{JN}{4} - J\bigr)|k\rangle + \tfrac{J}{2}(e^{ik} + e^{-ik})|k\rangle = \Bigl[\tfrac{JN}{4} - J(1 - \cos k)\Bigr]|k\rangle.\]

Thus the one-magnon eigenenergy is

\[E_{1}(k) \;=\; \tfrac{JN}{4} - J(1-\cos k) \;=\; \tfrac{JN}{4} - 2 J\sin^{2}(k/2). \tag{2}\]

The quantity $\varepsilon(k) \equiv -2 J\sin^{2}(k/2)$ is the single- magnon dispersion relative to the ferromagnetic reference. It is negative for every $k \ne 0$ — magnons lower the energy of the AF chain.

Step 2 — Two-magnon Bethe ansatz and phase shift

For $M = 2$ down-spins at positions $(n_1, n_2)$ with $n_1 < n_2$, let $|n_1 n_2\rangle$ denote the corresponding basis state and write the eigenstate ansatz

\[|\psi\rangle = \sum_{1 \le n_1 < n_2 \le N} \Psi(n_1, n_2)\,|n_1 n_2\rangle.\]

The same kinetic-energy bookkeeping as in Step 1 gives, when the two down-spins are not adjacent ($n_2 > n_1 + 1$),

\[[H\Psi](n_1, n_2) = \bigl(\tfrac{JN}{4} - 2 J\bigr)\Psi(n_1, n_2) + \tfrac{J}{2}\bigl[\Psi(n_1{-}1, n_2) + \Psi(n_1{+}1, n_2) + \Psi(n_1, n_2{-}1) + \Psi(n_1, n_2{+}1)\bigr]. \tag{3}\]

When the two down-spins are adjacent ($n_2 = n_1 + 1$), the bond $(n_1, n_2)$ has two down-spins, so its diagonal contribution is $+\tfrac{J}{4}$ (not $-\tfrac{J}{4}$) and its off-diagonal part vanishes. This modifies (3) into a boundary condition on $\Psi(n_1, n_2)$ when $n_2 = n_1 + 1$. Bethe's idea is to absorb the boundary condition into a single "2-body phase shift" by writing

\[\Psi(n_1, n_2) = A\,e^{i(k_1 n_1 + k_2 n_2)} + A'\,e^{i(k_2 n_1 + k_1 n_2)}. \tag{4}\]

Substituting (4) into (3) diagonalises the bulk operator with eigenvalue

\[E_2 = \tfrac{JN}{4} - J(1 - \cos k_1) - J(1 - \cos k_2) = \tfrac{JN}{4} + \varepsilon(k_1) + \varepsilon(k_2). \tag{5}\]

The adjacency boundary condition determines the ratio $A'/A$. Carrying out the algebra (see e.g. Karbach–Muller 1997 eqs. (7)–(10); the calculation is standard and amounts to matching the two sides of the scattering equation at $n_2 = n_1 + 1$) gives

\[\frac{A'}{A} \;=\; -\,\frac{e^{i(k_1 + k_2)} + 1 - 2\,e^{i k_1}} {e^{i(k_1 + k_2)} + 1 - 2\,e^{i k_2}} \;\equiv\; e^{i\theta(k_1, k_2)}. \tag{6}\]

Taking the log of (6) after simplifying the trigonometry yields the Bethe 2-body phase shift

\[\theta(k_1, k_2) \;=\; 2\arctan\!\bigl(\cot(k_1/2) - \cot(k_2/2)\bigr) \tag{7}\]

(a unique branch choice is fixed by requiring $\theta \to 0$ when $k_1 \to k_2$). Changing variables to rapidities

\[\lambda_j \;\equiv\; \tfrac{1}{2}\cot(k_j/2), \qquad \Longleftrightarrow\qquad e^{i k_j} = \frac{\lambda_j + i/2}{\lambda_j - i/2}, \tag{8}\]

equation (7) becomes

\[\theta(k_1, k_2) = 2\arctan(\lambda_1 - \lambda_2) = -i\ln\frac{\lambda_1 - \lambda_2 + i}{\lambda_1 - \lambda_2 - i}. \tag{9}\]

Step 3 — Bethe equations for $M$ magnons

For a state with $M$ magnons (down-spins at positions $n_1 < \dots < n_M$), the ansatz generalises to a superposition over all $M!$ permutations of $\{k_1, \dots, k_M\}$:

\[\Psi(n_1,\dots,n_M) = \sum_{P \in S_M} A_P\, \exp\!\Bigl(i\sum_{j=1}^{M} k_{P(j)} n_j\Bigr), \tag{10}\]

with the ratios $A_P/A_{P'}$ fixed by products of the 2-body phase shifts (9) across transpositions — this is the Yang–Baxter factorisation specific to integrable models. Imposing periodic boundary conditions $\Psi(n_1,\dots,n_M) = \Psi(n_2,\dots,n_M, n_1 + N)$ on the ansatz gives, for each $j = 1,\dots,M$,

\[e^{i k_j N} = \prod_{\ell \ne j}(-e^{-i\theta(k_j, k_\ell)}).\]

Substituting (8) and (9) and simplifying,

\[\boxed{\; \left(\frac{\lambda_j + i/2}{\lambda_j - i/2}\right)^{N} = \prod_{\ell \ne j}^{M}\frac{\lambda_j - \lambda_\ell + i} {\lambda_j - \lambda_\ell - i}, \qquad j = 1,\dots, M. \;} \tag{11}\]

These are the Bethe equations. Any solution $\{\lambda_j\}$ of (11) defines a Hamiltonian eigenstate via (10) with eigenvalue

\[E(\{\lambda_j\}) \;=\; \tfrac{JN}{4} + \sum_{j=1}^{M}\varepsilon(k_j). \tag{12}\]

Using the rapidity map (8), $\sin^2(k_j/2) = 1/(1 + \cot^2(k_j/2)) = 1/(1 + 4\lambda_j^2) = 1/[4(\lambda_j^2 + 1/4)]$, hence

\[\varepsilon(k_j) = -2 J\sin^{2}(k_j/2) = -\frac{J}{2(\lambda_j^{2} + 1/4)},\]

and (12) takes its rapidity form

\[\boxed{\; E(\{\lambda_j\}) \;=\; \tfrac{JN}{4} \;-\; \frac{J}{2}\sum_{j=1}^{M}\frac{1}{\lambda_j^{2} + 1/4}. \;} \tag{13}\]

The factor $\tfrac{1}{2}$ in front of the sum is essential — it is the source of the factor of $2$ that distinguishes the Hulthén answer $1/4 - \ln 2$ from the naive $1/4 - 2\ln 2$ one would get by omitting it.

Step 4 — Ground-state rapidity distribution

Taking the logarithm of (11) and using $\tfrac{\lambda + i/2}{\lambda - i/2} = e^{-i\,2\arctan(2\lambda)} \cdot e^{i\cdot(\text{integer multiple of } 2\pi)}$,

\[-2\arctan(2\lambda_j) \cdot N \;=\; -\sum_{\ell \ne j}^{M} 2\arctan(\lambda_j - \lambda_\ell) + 2\pi I_j, \tag{14}\]

where $I_j$ are distinct half-integers (their parity fixed by $M$). Divide by $N$ and define the counting function

\[Y_N(\lambda) \;:=\; -\frac{1}{\pi}\arctan(2\lambda) + \frac{1}{\pi N}\sum_{\ell=1}^{M}\arctan(\lambda - \lambda_\ell).\]

Equation (14) reads $Y_N(\lambda_j) = I_j/N$. In the ground state the quantum numbers $I_j$ fill consecutively the smallest possible integers symmetric about zero, so as $N \to \infty$ the $\{\lambda_j\}$ become a dense, symmetric set on the real axis with density $\rho(\lambda) > 0$ defined by

\[\rho(\lambda)\,d\lambda \;=\; \lim_{N\to\infty} \#\{j\;:\;\lambda_j \in [\lambda, \lambda + d\lambda]\}/N.\]

Differentiating $Y_N(\lambda_j) = I_j/N$ w.r.t. $j$ gives $Y_N'(\lambda) \rho_N(\lambda) = 1/N$ on one side, and on the other — after replacing the sum by an integral $(1/N)\sum_\ell \to \int \rho(\mu) d\mu$ in the $N \to \infty$ limit — the derivative of $Y_N$ reads

\[Y'(\lambda) = -\frac{1}{\pi}\cdot\frac{2}{1 + 4\lambda^2} + \frac{1}{\pi}\int\frac{1}{1 + (\lambda - \mu)^2}\rho(\mu)\,d\mu.\]

Matching $Y'(\lambda) \cdot \rho \to \rho \cdot \rho$ at leading order in $1/N$ (for the ground-state sector this simplifies to the linear equation below; a careful derivation of the sign and factor of two is in Takahashi 1999 §3.2) gives the linear integral equation for $\rho$:

\[\boxed{\; 2\pi\,\rho(\lambda) \;=\; \frac{1}{\lambda^{2} + 1/4} \;-\; 2\int_{-\infty}^{\infty}\frac{\rho(\mu)}{(\lambda-\mu)^{2} + 1}\,d\mu. \;} \tag{15}\]

Equation (15) is the cornerstone of the Hulthén calculation. The normalisation constraint is the half-filling condition

\[\int_{-\infty}^{\infty}\rho(\lambda)\,d\lambda \;=\; \frac{M}{N}\Bigr|_{M = N/2} \;=\; \frac{1}{2}. \tag{16}\]

Step 5 — Solution by Fourier transform

Adopt the convention

\[\hat{f}(q) \;=\; \int_{-\infty}^{\infty} e^{-i q\lambda}\,f(\lambda)\,d\lambda, \qquad f(\lambda) \;=\; \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{i q\lambda}\,\hat{f}(q)\,dq.\]

Two Fourier transforms we shall need (both are elementary contour integrals; see e.g. Whittaker–Watson 1927 §6):

\[\int_{-\infty}^{\infty}\frac{e^{-i q\lambda}}{\lambda^{2}+a^{2}}\,d\lambda \;=\; \frac{\pi}{a}\,e^{-a|q|}, \qquad a > 0. \tag{17}\]

With $a = \tfrac{1}{2}$, the Fourier transform of the inhomogeneous term in (15) is $2\pi\,e^{-|q|/2}$. With $a = 1$, the kernel $K(\lambda) := 2/(\lambda^2 + 1)$ transforms to $\hat{K}(q) = 2\pi\, e^{-|q|}$. Fourier-transforming (15) and using the convolution theorem on the $\rho \star K$ term,

\[2\pi\,\hat{\rho}(q) \;=\; 2\pi\,e^{-|q|/2} \;-\; 2\pi\,e^{-|q|}\,\hat{\rho}(q).\]

Solve for $\hat{\rho}$:

\[\hat{\rho}(q)\,\bigl(1 + e^{-|q|}\bigr) \;=\; e^{-|q|/2} \quad\Longrightarrow\quad \hat{\rho}(q) \;=\; \frac{e^{-|q|/2}}{1 + e^{-|q|}} \;=\; \frac{1}{e^{|q|/2} + e^{-|q|/2}} \;=\; \frac{1}{2\cosh(q/2)}. \tag{18}\]

Normalisation check. $\hat{\rho}(0) = 1/(2\cosh 0) = 1/2$, and $\int\rho\,d\lambda = \hat{\rho}(0) = 1/2$. Equation (16) is satisfied.

Inverting (18) uses the companion Fourier transform

\[\int_{-\infty}^{\infty}\frac{e^{-i q\lambda}}{\cosh(a\lambda)}\,d\lambda \;=\; \frac{\pi}{a\,\cosh(\pi q/(2 a))}, \qquad a > 0. \tag{19}\]

This is the standard result obtained by closing the contour in the lower half-plane and summing residues over $\lambda = -i(n+1/2)\pi/a$; the sign pattern $(-1)^n$ telescopes to the $\cosh$ in the denominator. Applied with $a = \pi/2$,

\[\int_{-\infty}^{\infty}\frac{e^{-i q\lambda}}{\cosh(\pi\lambda/2)}\,d\lambda \;=\; \frac{2}{\cosh q}.\]

Rescaling $\lambda \to 2\lambda$ and $q \to q/2$ converts this to the inverse Fourier transform of (18):

\[\rho(\lambda) \;=\; \frac{1}{2\pi}\int_{-\infty}^{\infty} \frac{e^{i q\lambda}}{2\cosh(q/2)}\,dq \;=\; \frac{1}{2\cosh(\pi\lambda)}.\]

Hence

\[\boxed{\; \rho(\lambda) \;=\; \frac{1}{2\cosh(\pi\lambda)}, \qquad -\infty < \lambda < \infty. \;} \tag{20}\]

Step 6 — Evaluation of the energy integral

From (13), the ground-state energy per site in the thermodynamic limit is

\[e_0 \;=\; \frac{E_0(N)}{N}\bigg|_{N\to\infty} \;=\; \frac{J}{4} \;-\; \frac{J}{2}\cdot \lim_{N\to\infty}\frac{1}{N}\sum_{j=1}^{M}\frac{1}{\lambda_j^2 + 1/4}.\]

The continuum replacement $(1/N)\sum_j \to \int \rho(\lambda)\, d\lambda$ converts the sum to

\[e_0 \;=\; \frac{J}{4} - \frac{J}{2}\int_{-\infty}^{\infty} \frac{\rho(\lambda)}{\lambda^{2} + 1/4}\,d\lambda. \tag{21}\]

Write $I := \int \rho(\lambda)/(\lambda^2 + 1/4)\,d\lambda$ and evaluate it by Parseval's theorem,

\[\int_{-\infty}^{\infty} f(\lambda)\,g(\lambda)\,d\lambda \;=\; \frac{1}{2\pi}\int_{-\infty}^{\infty} \hat{f}(q)\,\overline{\hat{g}(q)}\,dq,\]

with $f = \rho$ (so $\hat{f}(q) = 1/(2\cosh(q/2))$ by (18)) and $g(\lambda) = 1/(\lambda^2 + 1/4)$ (so $\hat{g}(q) = 2\pi\,e^{-|q|/2}$ by (17)). Both are real and even; complex conjugation leaves them unchanged. Therefore

\[I \;=\; \frac{1}{2\pi}\int_{-\infty}^{\infty} \frac{1}{2\cosh(q/2)}\cdot 2\pi\,e^{-|q|/2}\,dq \;=\; \int_{-\infty}^{\infty}\frac{e^{-|q|/2}}{2\cosh(q/2)}\,dq.\]

The integrand is even in $q$, so

\[I \;=\; 2\int_{0}^{\infty}\frac{e^{-q/2}}{2\cosh(q/2)}\,dq \;=\; \int_{0}^{\infty}\frac{e^{-q/2}}{\cosh(q/2)}\,dq.\]

Using $\cosh(q/2) = (e^{q/2} + e^{-q/2})/2$,

\[\frac{e^{-q/2}}{\cosh(q/2)} \;=\; \frac{2 e^{-q/2}}{e^{q/2} + e^{-q/2}} \;=\; \frac{2}{e^{q} + 1}.\]

\[I \;=\; 2\int_{0}^{\infty}\frac{dq}{e^{q} + 1}.\]

Substitute $u = e^{-q}$, $du = -e^{-q}\,dq$, $dq = -du/u$; limits $q\in[0,\infty)$ map to $u\in(0,1]$. In the new variable, $1/(e^{q} + 1) = u/(1 + u)$ and $dq = -du/u$, so

\[\int_{0}^{\infty}\frac{dq}{e^{q} + 1} \;=\; \int_{1}^{0}\frac{u}{1 + u}\cdot\Bigl(-\frac{du}{u}\Bigr) \;=\; \int_{0}^{1}\frac{du}{1 + u} \;=\; \bigl[\ln(1 + u)\bigr]_{0}^{1} \;=\; \ln 2.\]

Therefore

\[\boxed{\;I \;=\; 2\ln 2.\;} \tag{22}\]

Step 7 — Assembly of $e_0$

Substitute (22) into (21):

\[e_0 \;=\; \frac{J}{4} - \frac{J}{2}\cdot 2\ln 2 \;=\; \frac{J}{4} - J\ln 2 \;=\; J\!\left(\frac{1}{4} - \ln 2\right).\]

This is the Main-result value. Numerically $1/4 - \ln 2 \approx 0.25 - 0.6931 = -0.4431$, i.e. $e_0 \approx -0.4431\,J$.

Step 8 — Limiting-case and finite-size checks

(i) Dimer $N = 2$. With $N = 2$ and PBC, the two bonds are identical, $H = 2 J\,\mathbf{S}_1\cdot\mathbf{S}_2$. The two-spin Hilbert space decomposes into singlet ($S_{\rm tot} = 0$, energy $2 J\cdot(-3/4) = -3 J/2$) and triplet ($S_{\rm tot} = 1$, energy $+ J/2$); see the self-contained calculation at Heisenberg dimer: singlet–triplet splitting. The per-bond ground-state energy is $-3J/4 = -0.75\,J$. For $N = 2$ this overshoots the thermodynamic value $-0.4431\,J$: finite-size corrections scale as $\sim -\pi^2/(6 N^2)$, and at $N = 2$ this estimate is of order unity, consistent with the large overshoot.

(ii) Finite-size PBC trend. Numerical diagonalisation of the $2^N$-dimensional chain for $N = 4, 6, 8, 10, \dots$ converges to $e_0 = J(1/4 - \ln 2)$ as $1/N^2$ (for PBC, in the singlet sector). The test suite verifies this in test/verification/test_universality_cross_check.jl: ED at $N = 4, 6, 8$ PBC cross-checks that $|E_0(N)/N - e_0| < 5\%$ already at $N = 8$.

(iii) Independent route — XXZ $\Delta\to 1^-$ limit. The same Main-result value is the $\Delta = 1$ endpoint of the XXZ Luttinger- parameter family derived in XXZ Luttinger parameters: the Luttinger velocity $u(\Delta) = J(\pi/2) \sin\gamma/\gamma$ with $\gamma = \arccos\Delta$ tends to $\pi J/2$ as $\gamma\to 0$ (the des Cloizeaux–Pearson spinon velocity), and the same Fourier machinery that produces (18) in the isotropic case produces the XXZ dressed- charge $Z^2 = \pi/(2(\pi - \gamma)) \to 1/2$ at $\Delta = 1$. The ground-state energy per site in the Heisenberg limit of that XXZ calculation reproduces $J(1/4 - \ln 2)$ (Takahashi 1999 §4.3).

References

H. Bethe, Zur Theorie der Metalle. I. Eigenwerte und Eigenfunktionen der linearen Atomkette, Z. Physik 71, 205 (1931). Original Bethe ansatz.
L. Hulthén, Über das Austauschproblem eines Kristalles, Ark. Mat. Astron. Fys. 26A, No. 11 (1938). Evaluation of the ground-state energy per site.
C. N. Yang and C. P. Yang, One-dimensional chain of anisotropic spin-spin interactions I, Phys. Rev. 150, 321 (1966). Systematic treatment of the rapidity-density equation.
M. Karbach and G. Müller, Introduction to the Bethe ansatz I, Comput. Phys. 11, 36 (1997). Pedagogical derivation of (11) and (13) including the 2-body phase shift.
M. Takahashi, Thermodynamics of One-Dimensional Solvable Models (Cambridge University Press, 1999), §3.2 (derivation of (15)) and §4.3 (Heisenberg limit as $\Delta = 1$ of XXZ).
E. T. Whittaker and G. N. Watson, A Course of Modern Analysis, 4th ed. (Cambridge University Press, 1927), §6 (contour integrals (17) and (19)).

Used by

Heisenberg model page — thermodynamic-limit ground-state energy density.
XXZ model page — $\Delta = 1$ isotropic point.
XXZ Luttinger parameters note — Heisenberg limit as a cross-check of the same Fourier machinery.