Getting an equivalent integral equation from a given one I'm reading a paper and don't understand some of the calculations. We are given an integral equation with asymptotic boundary conditions
$\rho_+(u)=\frac{1}{2\pi} \int\limits_{|v|>\mu}^{}\mathrm{d}v\,\frac{2\hbar \rho_+(v)}{(u-v)^2+\hbar^2}$
$\rho_+(u)=\ln(|u|)-\frac{1}{2} \ln\left(\Xi\right)+O(u-1),\,u\rightarrow\infty$
$\rho_+$ is a density of zeros an $\ln(\Xi)$ can be seen as a density of particles. But the exact meaning is not important for further calculations. The first integral ranges over  $\mathbb{R}$ exceot the interval $[-\mu,\mu]$, where we have $\rho_+=0$.
Now a new function $\rho_-$ eis defined as
$\rho_-(u)=\begin{cases} 0 &\mbox{if } |u|\geq\mu \\
-\frac{1}{\pi \hbar}[\rho_+(u)-\frac{1}{\pi}\int\limits_{|v|>\mu}^{}\mathrm{d}v\,\frac{\hbar \rho_+(v)}{(u-v)^2+\hbar^2} & \mbox{if } |u|<\mu \end{cases}$
with which we can rewrite the first integral equation as
$\pi\hbar\rho_- +(1-K_+)\rho_+=0$
with $K_+\rho(u)=\int\limits_{-\infty}^{\infty} k_+(u-v)\rho(v) \mathrm{d} v$ and $k_+(u)=\frac{1}{\pi} \frac{\hbar}{u^2+\hbar^2}$.
So far everything is clear. Obviously a constant function is an eigenfunction fo $K_+$ with eigenvalue 1 and so $(1-K_+)$ ist invertible. It is the said, that an operator $K_-$ can be definied on functions vanishing at $\infty$
 by on functions vanishing at $\infty$
$1-K_-=(1-K_+)^{-1}$
This is still quite clear although I don't really understand, why the functions must vanish at $\infty$.
Then it ist said, that due to the degeneracy of the operator $K_+$, $K_-$ ist just defined up to a constant. What ist meant by this? Is it just due to the fact that every constant function is an eigenfunction?
Now without any calculations it is said, that the kernel $k_-(u)$ is given by 
$k_-(u)=\frac{1}{\pi\hbar} \Psi(1+i\frac{u}{\hbar})+\Psi(1-i\frac{u}{\hbar})+\Delta$
where $\Psi(u)=\Gamma\,'(u)/\Gamma(u)$ is the digamma-function.
It is only said, that this can easily obtaind by fourier transform. I don't have really much knowledge about integral-equations and do not see how this result can be found. Could please someone show me?
I have some other questions concerning the following calculations, but maybe I stop here until this problem is solved.
 A: We start with the Lorentzian distribution
$$ \tag{4.11} k_{+}(u)~:=~ \frac{1}{\pi} \frac{\eta}{u^2+\eta^2}; $$
with Fourier transform 
$$ \tag{4.11'} \tilde{k}_{+}(x)
~:=~\int_{\mathbb{R}}\!du ~e^{-ixu} k_{+}(u)~=~e^{-\eta|x|}; $$
with integral operator
$$ \tag{4.10} (K_{\pm}\rho)(u)~:=~\int_{\mathbb{R}}\!dv~ k_{\pm}(u-v)\rho(v)
~=~(k_{\pm}\ast\rho)(u). $$
The Fourier transformed operator is a multiplication operator
$$ \tag{4.10'} (K_{\pm}\rho)^{\sim}(x)~:=~\tilde{k}_{\pm}(x) \tilde{\rho}(x). $$
Repeated application of the integral operator leads to repeated convolutions
$$ \tag{A}   (K_{\pm}^n\rho)(u)
~=~(\underbrace{k_{\pm}\ast\ldots \ast k_{\pm}}_{n\text{ factors}}\ast\rho)(u), \qquad  (K_{\pm}^n\rho)^{\sim}(x)~:=~\tilde{k}_{\pm}^n(x) \tilde{\rho}(x). $$
Note that for a constant function $\rho \propto 1$ is an eigenfunction with eigenvalue 1 for the integral operator $K_{+}$:
$$\tag{B}  (K_{+} 1)(u) ~=~\int_{\mathbb{R}}\!dv~ k_{+}(u-v)~=~1.
$$ 
Since we want the operator $1-K_{+}$ to be invertible, we will from now on only consider integrable functions $\rho:\mathbb{R}\to \mathbb{C}$, such that $\int_{\mathbb{R}}\!du~\rho(u)=0$. In particular, we will exclude constant functions $\rho \propto 1$. This means that the kernels $k_{\pm}(u)\to k_{\pm}(u)+\Delta_{\pm}$ are only defined modulo additive constants $\Delta_{\pm}$.
Furthermore,
$$\tag{C} 1-K_{-} ~=~ \frac{1}{1-K_{+}}~=~\sum_{n=0}^{\infty}K_{+}^n  ,$$
and therefore, naively, 
$$\tag{D}  \tilde{k}_{-}(x)~=~\left(1- \frac{1}{\tilde{k}_{+}(x)}\right)^{-1}
~=~\frac{1}{1-e^{\eta|x|}} . $$
However the expression (D) is not integrable at $x=0$. The solution is to regularize $k_{-}(u)$ with an infinite additive constant $\Delta_{-}=\int_{\mathbb{R}}\!\frac{dx}{2\pi}~\frac{e^{-\eta|x|}}{\eta|x|}=\infty$, so that
$$ k_{-}(u)~=~ \int_{\mathbb{R}}\!\frac{dx}{2\pi}
\left[\frac{e^{-\eta|x|}}{\eta|x|}+
\frac{e^{ixu}}{1-e^{\eta|x|}}\right]
~=~\int_{0}^{\infty}\!\frac{dx}{\pi} \left[\frac{e^{-\eta x}}{\eta x}
+\frac{\cos(xu)}{1-e^{\eta x}} \right]$$
$$\tag{E} ~\stackrel{t=\eta x}{=}~\sum_{\pm}\int_{0}^{\infty}\!\frac{dt}{2\pi\eta} \left[\frac{e^{-t}}{t}+\frac{\exp\left(\pm\frac{ iut}{\eta}\right) }{1-e^{t}}\right]~=~\frac{1}{2\pi\eta}\sum_{\pm}\psi(1\pm\frac{iu}{\eta}),
$$
cf. the Digamma function and Abramowitz & Stegun eq. (6.3.21). Note that the final formula (E) differs from Sklyanin's formula (4.12) by a factor 2 in the overall normalization.
References:


*

*E.K. Sklyanin, The quantum Toda chain, Lecture Notes in Physics, 226 (1985) 196.

