$\int\limits_{-\infty}^\infty \left(f_T(\frac{x-\mu}{1+\Psi/2})-f_T(\frac{x+\mu}{1-\Psi/2})\right)\frac{x\gamma}{(x-x_{0})^{2}+\gamma^2/4}dx$

The step in the edit is indeed the only mistake, the rest of your calculation seems to be correct. In order to see what the problem is we write $$ \mathcal{J}^{A/R} = \frac{\gamma}{2} \int \limits_{-\infty}^{\infty} \left[f_T \left(\frac{x - \mu}{1 + \Psi/2}\right) - f_T \left(\frac{x + \mu}{1 - \Psi/2}\right)\right] \, \frac{\mathrm{d} x}{x - x_0 \mp \mathrm{i} \frac{\gamma}{2}} \, .$$ You correctly noted that these integrals are convergent due to the difference of the two Fermi functions. If there was only one Fermi function, they would diverge logarithmically at the lower limit.

However, you then proceed to perform two different changes of variables ($x = \pm \mu + (1 \pm \Psi/2) z$) for the first and the second term. This amounts to splitting the convergent integral into two divergent integrals, doing the substitutions and then combining them again, which is not allowed and changes the result!

There are two simple ways to fix this:

1. We can integrate by parts to obtain an integral with two convergent parts, split it up, do the substitutions and then reverse the integration by parts (the missing term will appear in the last step).
2. We can introduce a variable but finite lower limit (thus making both parts of the integral convergent individually), follow your steps and send the lower limit to infinity at the end.

I will show the second possibility here, since it illustrates nicely how the error arises. We have \begin{align} \mathcal{J}^{A/R} &= \lim_{r \to \infty} \frac{\gamma}{2} \int \limits_{-r}^{\infty} \left[f_T \left(\frac{x - \mu}{1 + \Psi/2}\right) - f_T \left(\frac{x + \mu}{1 - \Psi/2}\right)\right] \, \frac{\mathrm{d} x}{x - x_0 \mp \mathrm{i} \frac{\gamma}{2}} \\ &= \lim_{r \to \infty} \frac{\gamma}{2} \left[~\int \limits_{-r}^{\infty} \, \frac{f_T \left(\frac{x - \mu}{1 + \Psi/2}\right)}{x - x_0 \mp \mathrm{i} \frac{\gamma}{2}} \, \mathrm{d} x - \int \limits_{-r}^{\infty} \, \frac{f_T \left(\frac{x + \mu}{1 - \Psi/2}\right)}{x - x_0 \mp \mathrm{i} \frac{\gamma}{2}} \, \mathrm{d} x \right] \\ &= \lim_{r \to \infty} \frac{\gamma}{2} \left[~\int \limits_{- \frac{r + \mu}{1 + \Psi/2}}^{\infty} \, \frac{f_T \left(z\right)}{z - \frac{x_0 - \mu \pm \mathrm{i} \frac{\gamma}{2}}{1 + \Psi/2}} \, \mathrm{d} z - \int \limits_{- \frac{r - \mu}{1 - \Psi/2}}^{\infty} \, \frac{f_T \left(z\right)}{z - \frac{x_0 + \mu \pm \mathrm{i} \frac{\gamma}{2}}{1 - \Psi/2}} \, \mathrm{d} z\right] \\ &= \lim_{r \to \infty} \frac{\gamma}{2} \left[~\int \limits_{- \frac{r + \mu}{1 + \Psi/2}}^{\infty} f_T (z) \left(\frac{1}{z - \frac{x_0 - \mu \pm \mathrm{i} \frac{\gamma}{2}}{1 + \Psi/2}} - \frac{1}{z - \frac{x_0 + \mu \pm \mathrm{i} \frac{\gamma}{2}}{1 - \Psi/2}}\right) \, \mathrm{d} z + \int \limits_{- \frac{r + \mu}{1 + \Psi/2}}^{- \frac{r - \mu}{1 - \Psi/2}} \frac{f_T \left(z\right)}{z - \frac{x_0 + \mu \pm \mathrm{i} \frac{\gamma}{2}}{1 - \Psi/2}} \, \mathrm{d} z \right]. \end{align} Letting $r \to \infty$ in the first integral we simply obtain your result ($\mathcal{I}^{A/R}$). The second term is the correction we are looking for: \begin{align} \mathcal{J}^{A/R} - \mathcal{I}^{A/R} &= \lim_{r \to \infty} \frac{\gamma}{2} \int \limits_{- \frac{r + \mu}{1 + \Psi/2}}^{- \frac{r - \mu}{1 - \Psi/2}} \frac{f_T \left(z\right)}{z - \frac{x_0 + \mu \pm \mathrm{i} \frac{\gamma}{2}}{1 - \Psi/2}} \, \mathrm{d} z \\ &= \lim_{r \to \infty} \frac{\gamma}{2} \int \limits_{- \frac{r + \mu}{1 + \Psi/2}}^{- \frac{r - \mu}{1 - \Psi/2}} \frac{\mathrm{d} z}{z - \frac{x_0 + \mu \pm \mathrm{i} \frac{\gamma}{2}}{1 - \Psi/2}} - \lim_{r \to \infty} \frac{\gamma}{2} \int \limits_{- \frac{r + \mu}{1 + \Psi/2}}^{- \frac{r - \mu}{1 - \Psi/2}} \frac{1 - f_T \left(z\right)}{z - \frac{x_0 + \mu \pm \mathrm{i} \frac{\gamma}{2}}{1 - \Psi/2}} \, \mathrm{d} z \, . \end{align} Since $1 - f_T(z) \leq \mathrm{e}^{z/T}$, the second limit vanishes and we are left with \begin{align} \mathcal{J}^{A/R} - \mathcal{I}^{A/R} &= \lim_{r \to \infty} \frac{\gamma}{2} \int \limits_{- \frac{r + \mu}{1 + \Psi/2}}^{- \frac{r - \mu}{1 - \Psi/2}} \frac{\mathrm{d} z}{z - \frac{x_0 + \mu \pm \mathrm{i} \frac{\gamma}{2}}{1 - \Psi/2}} = \lim_{r \to \infty} \frac{\gamma}{2} \log \left(\frac{- \frac{r - \mu}{1 - \Psi/2} - \frac{x_0 + \mu \pm \mathrm{i} \frac{\gamma}{2}}{1 - \Psi/2}}{- \frac{r + \mu}{1 + \Psi/2} - \frac{x_0 + \mu \pm \mathrm{i} \frac{\gamma}{2}}{1 - \Psi/2}}\right) \\ &= \frac{\gamma}{2} \log \left(\frac{1 + \Psi/2}{1-\Psi/2}\right) = \gamma \operatorname{artanh} \left(\frac{\Psi}{2}\right) \, . \end{align}

With this addition the correct final result is \begin{align} W &= \int\limits_{-\infty}^\infty \left[f_T\left(\frac{x-\mu}{1+\Psi/2}\right)-f_T\left(\frac{x+\mu}{1-\Psi/2}\right)\right] \frac{\gamma x}{(x-x_0)^2 + \gamma^2/4} \, \mathrm{d} x \\ &=2 \gamma \operatorname{artanh} \left(\frac{\Psi}{2}\right) \!+ \!(\gamma \operatorname{Re} {} \! -\! 2 x_0 \operatorname{Im}) \left[\operatorname{\psi} \left(\frac{1}{2} \! + \! \frac{\gamma/2 \! + \! \mathrm{i} (x_0 \! - \! \mu)}{2 \pi T (1 \! + \! \Psi/2)}\right) \! - \! \operatorname{\psi} \left(\frac{1}{2} \! + \! \frac{\gamma/2 \! + \! \mathrm{i} (x_0 \! + \! \mu)}{2 \pi T (1 \! - \! \Psi/2)} \right) \right] . \end{align}