I am currently studying the XXX Heisenberg spin chain using the Bethe ansatz. I am working in the string hypothesis and I am having troubles deriving a simple expression for Fourier transformation of the string-string derivative kernel $A_{mn}(\omega)$. In particular (looking at the screenshot I have added below) i cannot find a way to move from equation $(155)$ to equation $(156)$. I have been stuck on this for a while and I was hoping someone could nudge me toward the right direction.
Formula (156) is obtained by direct substitution of expression (154) into equality (155). For convenience, I will take $n\geq m$ and $\omega \geq 0$, then $|n-m| = n-m$ and $|\omega| = \omega$. These conditions are not restrictive due to the following properties: $a_n(-\omega) = a_n(\omega)$ and $A_{nm} = A_{mn}$. I'll also take $m\geq 2$. Now substituting (154) into (155) gives $$ A_{nm}(\omega) = (1-\delta_{nm})e^{-\frac\omega2(n-m)} + 2\left(e^{-\frac\omega2(n-m+2)} + e^{-\frac\omega2(n-m+4)} + \ldots + e^{-\frac\omega2(n+m-2)}\right) + e^{-\frac\omega2(n+m)} $$ The first term on the right side is equal to $$ (1-\delta_{nm})e^{-\frac\omega2(n-m)} = e^{-\frac\omega2(n-m)} - \delta_{nm}. $$ The sum of the finite geometric progression in the middle is $$ e^{-\frac\omega2(n-m+2)} + e^{-\frac\omega2(n-m+4)} + \ldots + e^{-\frac\omega2(n+m-2)} = e^{-\frac\omega2(n-m+2)}\left(1 + e^{-\omega} + \ldots + e^{-\omega(m-2)} \right) = $$ $$ = e^{-\frac\omega2(n-m+2)}\ \frac{1-e^{-\omega(m-1)}}{1-e^{-\omega}} $$ Now we have $$ A_{nm}(\omega) = -\delta_{nm} + e^{-\frac\omega2(n-m)} + 2 e^{-\frac\omega2(n-m+2)}\ \frac{1-e^{-\omega(m-1)}}{1-e^{-\omega}} + e^{-\frac\omega2(n+m)} $$ Simple algebraic transformations lead to the following expression $$ A_{nm}(\omega) = -\delta_{nm} + \frac{1+e^{-\omega}}{1-e^{-\omega}} \left( e^{-\frac\omega2(n-m)} - e^{-\frac\omega2(n+m)}\right) $$ which exactly coincides with (156).
