Asymptotic expression of $\int_{- D}^{D} \frac{\text{tanh}(\xi)}{\xi -\omega}\mathrm{d}\xi$

How to derive the following asymptotic expression ($|\omega| \ll D $)?

$$P.V.\int_{- D}^{D} d\xi \frac{\tanh(\beta \xi)}{\xi -\omega} \approx 2 \ln\left(\frac{D}{\sqrt{\omega^2+T^2}}\right),\ \ \ \beta =1/T,\ \ \omega, T \rightarrow 0.$$

Two limiting cases:

1) $\omega=0, \lim_{T\rightarrow 0}\int_{- D}^{D} d\xi \frac{\text{tanh}(\beta \xi)}{\xi} \approx 2 \ln\left(\frac{D}{T}\right) + C \approx 2 \ln\left(\frac{D}{T}\right)$;

2) $T=0, \lim_{\omega \rightarrow 0} \int_{- D}^{D} d\xi \frac{\theta(\xi)-\theta( -\xi)}{\xi -\omega} \approx 2 \ln\left(\frac{D}{\omega}\right).$

Some write $2 \ln\left(\frac{D}{\text{max}(\omega, T)}\right)$ instead of $ 2 \ln\left(\frac{D}{\sqrt{\omega^2+T^2}}\right).$

Can anyone elaborate Jack's answer? Thank you.

Reference.

Actually, this is an important expression from physics, which is directly related to the Kondo problem, the divergence at the third order of perturbation theory.

We may just exploit the identity: $$\tanh x = \sum_{n\geq 0}\frac{8x}{(2n+1)^2\pi^2+4x^2}\tag{1}$$ that comes from the logarithmic derivative of the Weierstrass product for the (hyperbolic) cosine function: $$\cosh x = \prod_{n\geq 0}\left(1+\frac{4x^2}{(2n+1)^2\pi^2}\right).\tag{2}$$ After that, through: $$ \int_{A}^{B}\frac{8\,dx}{(2n+1)^2\pi^2+4x^2}=\left.\frac{4}{(2n+1)\pi}\,\arctan\left(\frac{2x}{(2n+1)\pi}\right)\right|_{A}^{B} \tag{3}$$ we have everything we need to estimate our integrals, through: $$\forall x\geq 0,\qquad \arctan(x)<\frac{\pi x}{1+2\sqrt{1+x^2}}<\min\left(x,\frac{\pi}{2}\right).\tag{4}$$ The sharper inequality is the Shafer-Fink inequality (see my note, for instance).

In other terms, $(1)$ just gives that $\frac{\tanh z}{z}$ is a series of Lorentzian functions, pretty easy to estimate.