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

I am looking at the integral:

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

where $$\tau_1$$ is real and $$a, b$$ real positive. So far I was only able to solve the following special case:

$$I(\tau_1,a,a) = \int_{\tau_1}^\infty d\tau_2\ \frac{1}{a^2 + \tau_2^2} \left(\pi - 2 \tan^{-1} \frac{\tau_2}{a} \right)^2 = \frac{1}{6 a} \left(\pi - \tan^{-1} \frac{\tau_1}{a} \right)^3, \tag{2}$$

but I cannot find a way to crack $$(1)$$. I am mostly interested by the case $$b=1$$.

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:

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

• Thank you very much for this work! I already had $I_1$, and to be honest I am not sure what you suggest I should do with $I_2$ and $I_3$?
– Pxx
May 27, 2020 at 12:43

The given integral can be presented in the form of $$I(\tau,a,b) = 4\int\limits_\tau^\infty \operatorname{arccot}^2\dfrac{\tau_2}a \,\dfrac{\mathrm d\tau_2}{\tau_2^2+b^2}.\tag1$$

Substitution $$\varphi=\operatorname{arccot} \dfrac{\tau_2}a,\quad \tau_2 = a\cot\varphi,\quad\mathrm d\tau_2=-a(\cot^2\varphi+1)\,\mathrm d\varphi,$$ allows to write $$I(\tau,a,b) = 4a\int\limits_0^{\operatorname{arccot}{\Large\frac\tau a}} \dfrac{\cot^2\varphi+1}{a^2\cot^2\varphi+b^2}\,\varphi^2\,\mathrm d\varphi = 4a\int\limits_0^{\operatorname{arccot}{\Large\frac\tau a}} \dfrac{\varphi^2\,\mathrm d\varphi}{a^2\cos^2\varphi+b^2\sin^2\varphi}.\tag2$$ Then $$I(\tau,a,a) = \dfrac4{3a}\varphi^3\bigg|_0^{{\Large\frac\pi2}-\arctan\Large\frac\tau a} = \dfrac1{6a}\left(\pi-2\arctan\frac\tau a\right)^3\tag3$$ (see Wolfram Alpha test).

At the same time, the antiderivative of $$(2)$$ is \color{brown}{\mathbf{\begin{align} &J(\varphi,a.b)=4a\int \dfrac{\varphi^2}{a^2\cos^2\varphi+b^2\sin^2\varphi}\,\mathrm d\varphi = \dfrac{2\varphi}b \big(\operatorname{Li_2}(r e^{2i\varphi})-\operatorname{Li_2}(^1\!/_{\large r}\, e^{2i\varphi})\big)\\[4pt] &+\dfrac{i}{b}\big(\operatorname{Li_3}(re^{2i\varphi})-\operatorname{Li_3}(^1\!/_{\large r}\, e^{2i\varphi})\big) +\dfrac{2i}{b}\varphi^2\ln\dfrac{1-re^{2i\varphi}}{1-\,^1\!/_{\large r}\,e^{2i\varphi}} +\operatorname{const}, \end{align}}}\tag4 (see also Wolfram Alpha calculations), where $$\operatorname{Li_n}$$ is the Polylogarithm, $$r=\dfrac{b-a}{a+b}.\tag5$$

Therefore, \begin{align} &\color{brown}{\mathbf{I(\tau,a,b)= J\left(\operatorname{arccot}\frac \tau a,a,b\right) -J(0,a,b).}}\tag6 \end{align}

If $$a=11,\ b=17,\ \tau = 5,$$ then $$r = \frac3{14},$$ $$I(\tau,a,b)\approx 0.10429\,46124\,85634,$$ (see also Wolfram Alpha result), $$J\left(\operatorname{arccot}\frac \tau a,a,b\right)\approx 0.32355\,66131\,49807 -0.26227\,19119\,51703\,i$$ (see also Wolfram Alpha result), $$J(0,a,b)\approx0.21926\,20006\,64173 - 0.26227\,19119\,51703\,i$$ (see also Wolfram Alpha result), $$J\left(\operatorname{arccot}\frac \tau a,a,b\right)-J(0,a,b)\approx 0.10429\,46124\,85634\approx I(\tau,a,b).$$

Test results confirm obtained closed form for the given integral.

• That’s awesome, thanks a lot!
– Pxx
May 30, 2020 at 16:29
• @Jxx You are welcome! May 30, 2020 at 20:17