Integral $I(\tau_1,a,b) = \int_{\tau_1}^\infty d\tau_2\ \frac{1}{b^2 + \tau_2^2} \left(\pi - 2 \tan^{-1} \frac{\tau_2}{a} \right)^2$

Some work, too large for a comment.

Well, we have the following function:

$$\text{y}\left(\text{k},\text{m},\text{n},\text{p},x\right):=\frac{1}{\text{n}+x^2}\left(\text{m}-\text{k}\arctan\left(\frac{x}{\text{p}}\right)\right)^2=$$ $$\frac{\text{m}^2}{\text{n}+x^2}-\frac{2\text{k}\text{m}\arctan\left(\frac{x}{\text{p}}\right)}{\text{n}+x^2}+\frac{\text{k}^2\arctan^2\left(\frac{x}{\text{p}}\right)}{\text{n}+x^2}\tag1$$

So, when we integrate:

$$\mathcal{I}_\epsilon\left(\text{k},\text{m},\text{n},\text{p}\right):=\int_\epsilon^\infty\text{y}\left(\text{k},\text{m},\text{n},\text{p},x\right)\space\text{d}x=$$ $$\underbrace{\int_\epsilon^\infty\frac{\text{m}^2}{\text{n}+x^2}\space\text{d}x}_{\text{I}_1}-\underbrace{\int_\epsilon^\infty\frac{2\text{k}\text{m}\arctan\left(\frac{x}{\text{p}}\right)}{\text{n}+x^2}\space\text{d}x}_{\text{I}_2}+\underbrace{\int_\epsilon^\infty\frac{\text{k}^2\arctan^2\left(\frac{x}{\text{p}}\right)}{\text{n}+x^2}\space\text{d}x}_{\text{I}_3}\tag2$$

Now, for $\text{I}_1$ we get:

$$\text{I}_1=\int_\epsilon^\infty\frac{\text{m}^2}{\text{n}+x^2}\space\text{d}x=\frac{\text{m}^2}{\text{n}}\int_\epsilon^\infty\frac{1}{1+\frac{x^2}{\text{n}}}\space\text{d}x\tag3$$

Let $\text{u}=\frac{x}{\sqrt{\text{n}}}$, so we get:

$$\text{I}_1=\frac{\text{m}^2}{\sqrt{\text{n}}}\lim_{x\to\infty}\int_\frac{\epsilon}{\sqrt{\text{n}}}^\frac{x}{\sqrt{\text{n}}}\frac{1}{1+\text{u}^2}\space\text{du}=\frac{\text{m}^2}{\sqrt{\text{n}}}\lim_{x\to\infty}\left[\arctan\left(\text{u}\right)\right]_\frac{\epsilon}{\sqrt{\text{n}}}^\frac{x}{\sqrt{\text{n}}}=$$ $$\frac{\text{m}^2}{\sqrt{\text{n}}}\lim_{x\to\infty}\left(\arctan\left(\frac{x}{\sqrt{\text{n}}}\right)-\arctan\left(\frac{\epsilon}{\sqrt{\text{n}}}\right)\right)\tag4$$

Knowing that $\text{n}>0$ implies that $\sqrt{\text{n}}>0$, so:

$$\text{I}_1=\frac{\text{m}^2}{\sqrt{\text{n}}}\left(\frac{\pi}{2}-\arctan\left(\frac{\epsilon}{\sqrt{\text{n}}}\right)\right)\tag5$$

Now, for $\text{I}_2$ we get:

$$\text{I}_2=\int_\epsilon^\infty\frac{2\text{k}\text{m}\arctan\left(\frac{x}{\text{p}}\right)}{\text{n}+x^2}\space\text{d}x=2\text{k}\text{m}\int_\epsilon^\infty\frac{\arctan\left(\frac{x}{\text{p}}\right)}{\text{n}+x^2}\space\text{d}x\tag6$$

Now, let's find:

$$\frac{\partial\text{I}_2}{\partial\text{p}}=-2\text{k}\text{m}\int_\epsilon^\infty\frac{x}{\left(\text{n}+x^2\right)\left(\text{p}^2+x^2\right)}\space\text{d}x\tag7$$

Using partial fractions it is not difficult to see that:

$$\frac{\partial\text{I}_2}{\partial\text{p}}=\frac{\text{km}\ln\left(\frac{\text{p}^2+\epsilon^2}{\text{n}+\epsilon^2}\right)}{\text{p}^2-\text{n}}\tag8$$