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We know that (G&R): $$\int_0^{\infty } \log \left(\frac{a^2}{x^2}+1\right) \log \left(\frac{b^2}{x^2}+1\right) \, dx=2 \pi ((a+b) \log (a+b)-a \log (a)-b \log (b)) $$ Where $a, b>0$. It can be proved by using Feynman's trick (i.e. differentiate w.r.t parameters) twice. The problem is: What are the closed-forms of following generalized integral: $$I=\int_0^{\infty } \log \left(\frac{a^2}{x^2}+1\right) \log \left(\frac{b^2}{x^2}+1\right) \log \left(\frac{c^2}{x^2}+1\right) \, dx $$ I've weakened the original problem and would like you to give some suggestions on it. Also this is related. Thank you.

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    $\begingroup$ Not that this is your fault, but it is inappropriate to attribute this technique (differentiation under the integral) to Feynman. It was known and widely used before him, and he himself says he learned it from a textbook (as did I from a textbook that pre-dates him). However for whatever reason a lot of his contemporaries had textbooks that didn't mention it, or at least didn't develop it enough for them to remember. So it only came to their attention from his autobiography where he mentions it. $\endgroup$ Jan 5 '20 at 20:19
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    $\begingroup$ Isn't Feynman's trick just Leibniz integral rule? $\endgroup$ Mar 11 '20 at 14:39
  • $\begingroup$ @crystal_math - not quite. Leibniz's rule is used for the trick but the trick itself is that sometimes after differentiation by some parameter (not the variable-of-integration), the differentiated integrand has a known closed form solution. You can then integrate that solution with respect to the parameter to get a solution to the original problem. $\endgroup$ Jul 26 at 14:23
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We have for $a,b,c>0$

  • $\scriptsize \int_0^{\infty } \log \left(\frac{a^2}{x^2}+1\right) \log \left(\frac{b^2}{x^2}+1\right) \log \left(\frac{c^2}{x^2}+1\right) \, dx=\Re\left(2 c \pi \log ^2(-c)+2 c \pi (-\log (-c)+\log (c)-2) \log (-c)+2 a \pi \log (-c)+2 c \pi \log ^2(c)-2 (a-c) \pi \log \left(1-\frac{a}{a-c}\right)+2 (a-c) \pi \log \left(1-\frac{a+b}{a-c}\right)+2 c \pi \log (b) \log \left(1-\frac{b}{c}\right)-2 c \pi \log \left(1-\frac{b}{c}\right)-2 c \pi \log (a) \log \left(-\frac{c}{a-c}\right)+2 c \pi \log (a+b) \log \left(-\frac{b+c}{a-c}\right)-2 (a+b) \pi \log (-b-c)+2 (b-c) \pi (\log (b-c)-1)-2 c \pi \log (b) \log \left(\frac{b+c}{c}\right)+2 c \pi \log \left(\frac{b+c}{c}\right)+2 c \pi (\log (-c)-1)+2 c \pi (\log (-c)-\log (c)+2)-2 c \pi \log (b-c) (\log (c)-1)+6 c \pi (\log (c)-1)+2 a \pi \log (c)-4 c \pi \log (c)-2 a \pi \log (a) \log (c-a)+2 a \pi \log (a+b) \log (c-a)+2 c \pi \log (a) \log \left(\frac{c}{a+c}\right)-2 c \pi \log (a+b) \log \left(\frac{c-b}{a+c}\right)-2 (a+c) \pi (\log (a+c)-1)-2 a \pi \log (a) \log (a+c)+2 a \pi \log (a+b) \log (a+c)+2 \pi (\log (b-c) c-c-b \log (b-c)) \log (a+c)-2 c \pi (\log (-c)-1) \log (a+c)-2 c \pi (\log (c)-1) \log (a+c)+2 a \pi \log (a+c)-2 b \pi \log (b) \log (c-b)+2 b \pi \log (a+b) \log (c-b)-2 (a+b) \pi \log (c-b)-2 b \pi \log (b) \log (b+c)+2 b \pi \log (a+b) \log (b+c)-2 c \pi (\log (c)-1) \log (b+c)+2 \pi (\log (a+c) c-c+a \log (a+c)) \log (b+c)+2 b \pi \log (b+c)-2 (a+c) \pi \log \left(1-\frac{a}{a+c}\right)+2 (a+c) \pi \log \left(1-\frac{a+b}{a+c}\right)-\frac{2}{3} \pi \left(-3 a \log ^2(a)-3 a \log (-a) \log (a)-3 a \log \left(-\frac{b}{a-b}\right) \log (a)+3 b \log \left(-\frac{b}{a-b}\right) \log (a)+3 a \log (b) \log (a)-3 b \log (b) \log (a)+3 a \log (a+b) \log (a)+a \pi ^2-3 b \log ^2(b)+3 a-3 b-3 a \log \left(\frac{1}{a}\right)-3 a \log \left(-\frac{1}{a}\right)+6 a \log \left(-\frac{b}{a}\right)-6 a \log (-b)-6 b \log (-b)+6 b \log (b)-3 b \log (-b) \log (b)+3 a \log (b) \log \left(\frac{a}{a+b}\right)+3 b \log (b) \log \left(\frac{a}{a+b}\right)+3 a \log (-a) \log (a+b)+3 b \log (-b) \log (a+b)+3 b \log (b) \log (a+b)-6 a \text{Li}_2\left(\frac{a+b}{a}\right)-3 (a-b) \text{Li}_2\left(\frac{a}{a-b}\right)+3 (a+b) \text{Li}_2\left(\frac{b}{a+b}\right)+3 a i \pi \right)-2 a \pi \left(\log \left(\frac{c}{a}+1\right) \log (-c)+\text{Li}_2\left(-\frac{c}{a}\right)\right)-2 a \pi \left(\log \left(\frac{c}{a}+1\right) \log (c)+\text{Li}_2\left(-\frac{c}{a}\right)\right)-2 a \pi \left(\log \left(1-\frac{a+c}{a-b}\right) \log (a+c)+\text{Li}_2\left(\frac{a+c}{a-b}\right)\right)+2 b \pi \left(\log \left(1-\frac{a+c}{a-b}\right) \log (a+c)+\text{Li}_2\left(\frac{a+c}{a-b}\right)\right)+2 b \pi \left(\log \left(1-\frac{c}{b}\right) \log (c)+\text{Li}_2\left(\frac{c}{b}\right)\right)-2 b \pi \left(\log \left(\frac{c}{b}+1\right) \log (c)+\text{Li}_2\left(-\frac{c}{b}\right)\right)+2 a \pi \left(\log \left(1-\frac{b-c}{a+b}\right) \log (b-c)+\text{Li}_2\left(\frac{b-c}{a+b}\right)\right)+2 b \pi \left(\log \left(1-\frac{b-c}{a+b}\right) \log (b-c)+\text{Li}_2\left(\frac{b-c}{a+b}\right)\right)+2 (a-c) \pi \text{Li}_2\left(\frac{a}{a-c}\right)-2 (a-c) \pi \text{Li}_2\left(\frac{a+b}{a-c}\right)+2 c \pi \text{Li}_2\left(\frac{b}{c}\right)-2 c \pi \text{Li}_2\left(-\frac{b}{c}\right)+2 (a+c) \pi \text{Li}_2\left(\frac{a}{a+c}\right)-2 (a+c) \pi \text{Li}_2\left(\frac{a+b}{a+c}\right)\right)$

For instance, taking $(a,b,c)=(1,2,3)$

  • $\small \int_0^{\infty } \log \left(\frac{1}{x^2}+1\right) \log \left(\frac{4}{x^2}+1\right) \log \left(\frac{9}{x^2}+1\right) \, dx=4 \pi \Re( \text{Li}_2(3))+2 \pi \Re\left(\text{Li}_2(4)+2 \text{Li}_2\left(\frac{3}{2}\right)\right)+2 \pi \text{Li}_2(-4)-2 \pi \text{Li}_2(-3)-4 \pi \text{Li}_2\left(-\frac{1}{2}\right)+6 \pi \text{Li}_2\left(-\frac{1}{3}\right)-6 \pi \text{Li}_2\left(-\frac{2}{3}\right)+8 \pi \text{Li}_2\left(\frac{1}{4}\right)-8 \pi \text{Li}_2\left(\frac{3}{4}\right)-\frac{5 \pi ^3}{6}+8 \pi \log ^2(2)+12 \pi \log ^2(3)-18 \pi \log (3) \log (2)+10 \pi \log (5) \log (2)$
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    $\begingroup$ This is really cool! $\endgroup$ Mar 30 '20 at 11:53
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    $\begingroup$ Awesome! I don't think that I will try to verify either result. $\endgroup$ Apr 1 '20 at 4:49
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Define the function $\mathcal{I}:\mathbb{R}_{>0}^{3}\rightarrow\mathbb{R}$ via the improper integral

$$\mathcal{I}{\left(a,b,c\right)}:=\int_{0}^{\infty}\mathrm{d}x\,\ln{\left(1+\frac{a^{2}}{x^{2}}\right)}\ln{\left(1+\frac{b^{2}}{x^{2}}\right)}\ln{\left(1+\frac{c^{2}}{x^{2}}\right)}.\tag{1}$$

Our objective is to derive a closed-form expression for $\mathcal{I}$ in terms of polylogarithms and elementary functions.

This function $\mathcal{I}$ has two important properties that follow almost immediately from definition $(1)$. First, it is symmetric under any permutation of its three parameters. Second, it obeys the scaling relation

$$\mathcal{I}{\left(a,b,c\right)}=d\,\mathcal{I}{\left(d^{-1}a,d^{-1}b,d^{-1}c\right)};~~~\small{\left(a,b,c,d\right)\in\mathbb{R}_{>0}^{4}}.$$


Consider the following derivative, which holds for fixed but arbitrary $x\in\mathbb{R}_{>0}$:

$$\frac{\partial}{\partial y}\ln{\left(1+\frac{y^{2}}{x^{2}}\right)}=\frac{2y}{x^{2}}\cdot\frac{1}{\left(1+\frac{y^{2}}{x^{2}}\right)}=\frac{2y}{\left(x^{2}+y^{2}\right)}.$$

Integrating both sides, we obtain the following integral representation:

$$\ln{\left(1+\frac{a^{2}}{x^{2}}\right)}=\int_{0}^{a}\mathrm{d}y\,\frac{2y}{\left(x^{2}+y^{2}\right)};~~~\small{a\in\mathbb{R}_{>0}\land x\in\mathbb{R}_{>0}}.\tag{2}$$


Suppose $\left(a,b,c\right)\in\mathbb{R}_{>0}^{3}$. Using the integral representation in $(4)$ above to rewrite $\mathcal{I}$ as a multiple integral and changing the order of integration, we find that $\mathcal{I}$ can be expressed as a triple integral as follows:

$$\begin{align} \mathcal{I}{\left(a,b,c\right)} &=\int_{0}^{\infty}\mathrm{d}x\,\ln{\left(1+\frac{a^{2}}{x^{2}}\right)}\ln{\left(1+\frac{b^{2}}{x^{2}}\right)}\ln{\left(1+\frac{c^{2}}{x^{2}}\right)}\\ &=\int_{0}^{\infty}\mathrm{d}x\int_{0}^{a}\mathrm{d}p\,\frac{2p}{\left(x^{2}+p^{2}\right)}\int_{0}^{b}\mathrm{d}q\,\frac{2q}{\left(x^{2}+q^{2}\right)}\int_{0}^{c}\mathrm{d}r\,\frac{2r}{\left(x^{2}+r^{2}\right)}\\ &=\int_{0}^{\infty}\mathrm{d}x\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\int_{0}^{c}\mathrm{d}r\,\frac{8pqr}{\left(x^{2}+p^{2}\right)\left(x^{2}+q^{2}\right)\left(x^{2}+r^{2}\right)}\\ &=\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\int_{0}^{c}\mathrm{d}r\int_{0}^{\infty}\mathrm{d}x\,\frac{8pqr}{\left(x^{2}+p^{2}\right)\left(x^{2}+q^{2}\right)\left(x^{2}+r^{2}\right)}\\ &=\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\int_{0}^{c}\mathrm{d}r\int_{0}^{\infty}\mathrm{d}x\,\frac{\left(-1\right)8pqr}{\left(p^{2}-q^{2}\right)\left(q^{2}-r^{2}\right)\left(r^{2}-p^{2}\right)}\bigg{[}\frac{\left(q^{2}-r^{2}\right)}{\left(x^{2}+p^{2}\right)}\\ &~~~~~+\frac{\left(r^{2}-p^{2}\right)}{\left(x^{2}+q^{2}\right)}+\frac{\left(p^{2}-q^{2}\right)}{\left(x^{2}+r^{2}\right)}\bigg{]}\\ &=\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\int_{0}^{c}\mathrm{d}r\,\frac{\left(-1\right)8pqr}{\left(p^{2}-q^{2}\right)\left(q^{2}-r^{2}\right)\left(r^{2}-p^{2}\right)}\bigg{[}\int_{0}^{\infty}\mathrm{d}x\,\frac{\left(q^{2}-r^{2}\right)}{\left(x^{2}+p^{2}\right)}\\ &~~~~~+\int_{0}^{\infty}\mathrm{d}x\,\frac{\left(r^{2}-p^{2}\right)}{\left(x^{2}+q^{2}\right)}+\int_{0}^{\infty}\mathrm{d}x\,\frac{\left(p^{2}-q^{2}\right)}{\left(x^{2}+r^{2}\right)}\bigg{]}\\ &=\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\int_{0}^{c}\mathrm{d}r\,\frac{\left(-1\right)8pqr}{\left(p^{2}-q^{2}\right)\left(q^{2}-r^{2}\right)\left(r^{2}-p^{2}\right)}\bigg{[}\frac{\left(q^{2}-r^{2}\right)}{p}\\ &~~~~~+\frac{\left(r^{2}-p^{2}\right)}{q}+\frac{\left(p^{2}-q^{2}\right)}{r}\bigg{]}\int_{0}^{\infty}\mathrm{d}y\,\frac{1}{\left(y^{2}+1\right)}\\ &=\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\int_{0}^{c}\mathrm{d}r\,\frac{8pqr}{\left(p^{2}-q^{2}\right)\left(q^{2}-r^{2}\right)\left(r^{2}-p^{2}\right)}\bigg{[}-\frac{\left(q^{2}-r^{2}\right)}{p}\\ &~~~~~-\frac{\left(r^{2}-p^{2}\right)}{q}-\frac{\left(p^{2}-q^{2}\right)}{r}\bigg{]}\frac{\pi}{2}\\ &=\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\int_{0}^{c}\mathrm{d}r\,\frac{4\pi\left(p+q+r\right)}{\left(p+q\right)\left(q+r\right)\left(r+p\right)}.\tag{3}\\ \end{align}$$


Suppose $\left(a,b\right)\in\mathbb{R}_{>0}^{2}$, and consider the case where the third parameter of $\mathcal{I}$ is set equal to one. The integrand of our triple integral is a rational function, so at least one of the three integrations can be done in elementary terms. We find

$$\begin{align} \mathcal{I}{\left(a,b,1\right)} &=\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\int_{0}^{1}\mathrm{d}r\,\frac{4\pi\left(p+q+r\right)}{\left(p+q\right)\left(q+r\right)\left(r+p\right)}\\ &=\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\,\frac{4\pi}{\left(p-q\right)\left(p+q\right)}\int_{0}^{1}\mathrm{d}r\,\frac{\left(p-q\right)\left(p+q+r\right)}{\left(q+r\right)\left(r+p\right)}\\ &=\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\,\frac{4\pi}{\left(p-q\right)\left(p+q\right)}\int_{0}^{1}\mathrm{d}r\,\left[\frac{p}{\left(q+r\right)}-\frac{q}{\left(r+p\right)}\right]\\ &=\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\,\frac{4\pi}{\left(p-q\right)\left(p+q\right)}\left[p\int_{0}^{1}\mathrm{d}r\,\frac{1}{\left(q+r\right)}-q\int_{0}^{1}\mathrm{d}r\,\frac{1}{\left(r+p\right)}\right]\\ &=\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\,\frac{4\pi}{\left(p-q\right)\left(p+q\right)}\left[p\ln{\left(\frac{1+q}{q}\right)}-q\ln{\left(\frac{1+p}{p}\right)}\right]\\ &=2\pi\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\,\frac{2}{\left(p-q\right)\left(p+q\right)}\left[p\ln{\left(\frac{1+q}{q}\right)}-q\ln{\left(\frac{1+p}{p}\right)}\right]\\ &=2\pi\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\,\bigg{[}\frac{\ln{\left(\frac{1+q}{q}\right)}+\ln{\left(\frac{1+p}{p}\right)}}{\left(p+q\right)}+\frac{\ln{\left(\frac{1+q}{q}\right)}-\ln{\left(\frac{1+p}{p}\right)}}{\left(p-q\right)}\bigg{]}\\ &=2\pi\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\,\bigg{[}\frac{\ln{\left(\frac{p}{1+p}\right)}-\ln{\left(\frac{q}{1+q}\right)}}{\left(p-q\right)}-\frac{\ln{\left(\frac{p}{1+p}\right)}+\ln{\left(\frac{q}{1+q}\right)}}{\left(p+q\right)}\bigg{]}.\tag{4}\\ \end{align}$$


Set $A:=\frac{a}{1+a}\in\left(0,1\right)\land B:=\frac{b}{1+b}\in\left(0,1\right)$. Continuing from the double integral in the last line of $(4)$ above,

$$\begin{align} \frac{\mathcal{I}{\left(a,b,1\right)}}{2\pi} &=\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\,\bigg{[}\frac{\ln{\left(\frac{p}{1+p}\right)}-\ln{\left(\frac{q}{1+q}\right)}}{\left(p-q\right)}-\frac{\ln{\left(\frac{p}{1+p}\right)}+\ln{\left(\frac{q}{1+q}\right)}}{\left(p+q\right)}\bigg{]}\\ &=\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\,\frac{\ln{\left(\frac{p}{1+p}\right)}-\ln{\left(\frac{q}{1+q}\right)}}{\left(p-q\right)}-\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\,\frac{\ln{\left(\frac{p}{1+p}\right)}+\ln{\left(\frac{q}{1+q}\right)}}{\left(p+q\right)}\\ &=\int_{0}^{\frac{a}{1+a}}\mathrm{d}x\,\frac{1}{\left(1-x\right)^{2}}\int_{0}^{\frac{b}{1+b}}\mathrm{d}y\,\frac{1}{\left(1-y\right)^{2}}\cdot\frac{\ln{\left(x\right)}-\ln{\left(y\right)}}{\left(\frac{x}{1-x}-\frac{y}{1-y}\right)};~~~\small{\left[p=\frac{x}{1-x}\land q=\frac{y}{1-y}\right]}\\ &~~~~~-\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\,\frac{\ln{\left(\frac{p}{1+p}\right)}}{\left(p+q\right)}-\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\,\frac{\ln{\left(\frac{q}{1+q}\right)}}{\left(p+q\right)}\\ &=\int_{0}^{A}\mathrm{d}x\int_{0}^{B}\mathrm{d}y\,\frac{\ln{\left(x\right)}-\ln{\left(y\right)}}{\left(1-x\right)\left(x-y\right)\left(1-y\right)}\\ &~~~~~-\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\,\frac{\ln{\left(\frac{p}{1+p}\right)}}{\left(p+q\right)}-\int_{0}^{b}\mathrm{d}q\int_{0}^{a}\mathrm{d}p\,\frac{\ln{\left(\frac{q}{1+q}\right)}}{\left(p+q\right)}\\ &=\int_{0}^{A}\mathrm{d}x\,\frac{1}{\left(1-x\right)^{2}}\int_{0}^{B}\mathrm{d}y\,\frac{\left(1-x\right)}{\left(x-y\right)\left(1-y\right)}\left[\ln{\left(x\right)}-\ln{\left(y\right)}\right]\\ &~~~~~-\int_{0}^{a}\mathrm{d}p\,\ln{\left(\frac{p}{1+p}\right)}\int_{0}^{b}\mathrm{d}q\,\frac{1}{\left(p+q\right)}-\int_{0}^{b}\mathrm{d}q\,\ln{\left(\frac{q}{1+q}\right)}\int_{0}^{a}\mathrm{d}p\,\frac{1}{\left(p+q\right)}\\ &=\int_{0}^{A}\mathrm{d}x\,\frac{1}{\left(1-x\right)^{2}}\int_{0}^{B}\mathrm{d}y\,\left[\frac{1}{\left(x-y\right)}-\frac{1}{\left(1-y\right)}\right]\left[\ln{\left(x\right)}-\ln{\left(y\right)}\right]\\ &~~~~~-\int_{0}^{a}\mathrm{d}p\,\ln{\left(\frac{p}{1+p}\right)}\ln{\left(\frac{p+b}{p}\right)}\\ &~~~~~-\int_{0}^{b}\mathrm{d}q\,\ln{\left(\frac{q}{1+q}\right)}\ln{\left(\frac{a+q}{q}\right)}\\ &=\int_{0}^{A}\mathrm{d}x\,\frac{1}{\left(1-x\right)^{2}}\int_{0}^{B}\mathrm{d}y\,\left[\frac{\ln{\left(x\right)}-\ln{\left(y\right)}}{\left(x-y\right)}-\frac{\ln{\left(x\right)}-\ln{\left(y\right)}}{\left(1-y\right)}\right]\\ &~~~~~+\int_{0}^{a}\mathrm{d}p\,\ln{\left(\frac{1+p}{p}\right)}\ln{\left(\frac{p+b}{p}\right)}\\ &~~~~~+\int_{0}^{b}\mathrm{d}q\,\ln{\left(\frac{1+q}{q}\right)}\ln{\left(\frac{a+q}{q}\right)}.\\ \end{align}$$

Given $0<x<1\land0<B<1$,

$$\begin{align} \int_{0}^{B}\mathrm{d}y\,\frac{\ln{\left(x\right)}-\ln{\left(y\right)}}{\left(x-y\right)} &=\int_{0}^{\frac{B}{x}}\mathrm{d}t\,x\frac{\ln{\left(x\right)}-\ln{\left(xt\right)}}{\left(x-xt\right)};~~~\small{\left[y=xt\right]}\\ &=\int_{0}^{\frac{B}{x}}\mathrm{d}t\,\frac{(-1)\ln{\left(t\right)}}{\left(1-t\right)}\\ &=\int_{1-\frac{B}{x}}^{1}\mathrm{d}u\,\frac{(-1)\ln{\left(1-u\right)}}{u};~~~\small{\left[t=1-u\right]}\\ &=\operatorname{Li}_{2}{\left(1\right)}-\operatorname{Li}_{2}{\left(1-\frac{B}{x}\right)},\\ \end{align}$$

and

$$\begin{align} \int_{0}^{B}\mathrm{d}y\,\frac{\ln{\left(x\right)}-\ln{\left(y\right)}}{\left(1-y\right)} &=\int_{1-B}^{1}\mathrm{d}t\,\frac{\ln{\left(x\right)}-\ln{\left(1-t\right)}}{t};~~~\small{\left[y=1-t\right]}\\ &=\operatorname{Li}_{2}{\left(1\right)}-\operatorname{Li}_{2}{\left(1-B\right)}-\ln{\left(1-B\right)}\ln{\left(x\right)}.\\ \end{align}$$

Then,

$$\begin{align} \frac{\mathcal{I}{\left(a,b,1\right)}}{2\pi} &=\int_{0}^{A}\mathrm{d}x\,\frac{1}{\left(1-x\right)^{2}}\int_{0}^{B}\mathrm{d}y\,\left[\frac{\ln{\left(x\right)}-\ln{\left(y\right)}}{\left(x-y\right)}-\frac{\ln{\left(x\right)}-\ln{\left(y\right)}}{\left(1-y\right)}\right]\\ &~~~~~+\int_{0}^{a}\mathrm{d}p\,\ln{\left(\frac{1+p}{p}\right)}\ln{\left(\frac{p+b}{p}\right)}+\int_{0}^{b}\mathrm{d}q\,\ln{\left(\frac{1+q}{q}\right)}\ln{\left(\frac{a+q}{q}\right)}\\ &=\int_{0}^{A}\mathrm{d}x\,\frac{1}{\left(1-x\right)^{2}}\left[\operatorname{Li}_{2}{\left(1-B\right)}-\operatorname{Li}_{2}{\left(1-\frac{B}{x}\right)}+\ln{\left(1-B\right)}\ln{\left(x\right)}\right]\\ &~~~~~+\int_{0}^{a}\mathrm{d}p\,\ln{\left(\frac{1+p}{p}\right)}\ln{\left(\frac{p+b}{p}\right)}+\int_{0}^{b}\mathrm{d}q\,\ln{\left(\frac{1+q}{q}\right)}\ln{\left(\frac{a+q}{q}\right)}\\ &=\left(\frac{A}{1-A}\right)\left[\operatorname{Li}_{2}{\left(1-B\right)}-\operatorname{Li}_{2}{\left(1-\frac{B}{A}\right)}+\ln{\left(1-B\right)}\ln{\left(A\right)}\right]\\ &~~~~~-\lim_{x\to0}\left(\frac{x}{1-x}\right)\left[\operatorname{Li}_{2}{\left(1-B\right)}-\operatorname{Li}_{2}{\left(1-\frac{B}{x}\right)}+\ln{\left(1-B\right)}\ln{\left(x\right)}\right]\\ &~~~~~-\int_{0}^{A}\mathrm{d}x\,\left(\frac{x}{1-x}\right)\frac{d}{dx}\left[\operatorname{Li}_{2}{\left(1-B\right)}-\operatorname{Li}_{2}{\left(1-\frac{B}{x}\right)}+\ln{\left(1-B\right)}\ln{\left(x\right)}\right];~~~\small{I.B.P.s}\\ &~~~~~+\int_{0}^{a}\mathrm{d}p\,\ln{\left(\frac{1+p}{p}\right)}\ln{\left(\frac{p+b}{p}\right)}+\int_{0}^{b}\mathrm{d}q\,\ln{\left(\frac{1+q}{q}\right)}\ln{\left(\frac{a+q}{q}\right)}\\ &=\left(\frac{A}{1-A}\right)\left[\operatorname{Li}_{2}{\left(1-B\right)}-\operatorname{Li}_{2}{\left(1-\frac{B}{A}\right)}+\ln{\left(1-B\right)}\ln{\left(A\right)}\right]\\ &~~~~~+\int_{0}^{A}\mathrm{d}x\,\left[\frac{B\ln{\left(\frac{B}{x}\right)}}{\left(1-x\right)\left(B-x\right)}-\frac{\ln{\left(1-B\right)}}{\left(1-x\right)}\right]\\ &~~~~~+\int_{0}^{a}\mathrm{d}p\,\ln{\left(\frac{1+p}{p}\right)}\ln{\left(\frac{p+b}{p}\right)}+\int_{0}^{b}\mathrm{d}q\,\ln{\left(\frac{1+q}{q}\right)}\ln{\left(\frac{a+q}{q}\right)}\\ &=\frac{A}{1-A}\left[\operatorname{Li}_{2}{\left(1-B\right)}-\operatorname{Li}_{2}{\left(1-\frac{B}{A}\right)}+\ln{\left(1-B\right)}\ln{\left(A\right)}\right]\\ &~~~~~-\frac{B}{1-B}\int_{0}^{A}\mathrm{d}x\,\frac{\left(1-B\right)\ln{\left(\frac{x}{B}\right)}}{\left(1-x\right)\left(B-x\right)}-\int_{0}^{A}\mathrm{d}x\,\frac{\ln{\left(1-B\right)}}{\left(1-x\right)}\\ &~~~~~+\int_{0}^{a}\mathrm{d}p\,\ln{\left(\frac{1+p}{p}\right)}\ln{\left(\frac{p+b}{p}\right)}+\int_{0}^{b}\mathrm{d}q\,\ln{\left(\frac{1+q}{q}\right)}\ln{\left(\frac{a+q}{q}\right)}\\ &=\frac{A}{1-A}\left[\operatorname{Li}_{2}{\left(1-B\right)}-\operatorname{Li}_{2}{\left(1-\frac{B}{A}\right)}+\ln{\left(1-B\right)}\ln{\left(A\right)}\right]\\ &~~~~~-\frac{B}{1-B}\int_{0}^{A}\mathrm{d}x\,\left[\frac{\ln{\left(\frac{x}{B}\right)}}{\left(B-x\right)}-\frac{\ln{\left(\frac{x}{B}\right)}}{\left(1-x\right)}\right]\\ &~~~~~+\ln{\left(1-A\right)}\ln{\left(1-B\right)}\\ &~~~~~+\int_{0}^{a}\mathrm{d}p\,\ln{\left(\frac{1+p}{p}\right)}\ln{\left(\frac{p+b}{p}\right)}+\int_{0}^{b}\mathrm{d}q\,\ln{\left(\frac{1+q}{q}\right)}\ln{\left(\frac{a+q}{q}\right)}\\ &=\frac{A}{1-A}\left[\operatorname{Li}_{2}{\left(1-B\right)}-\operatorname{Li}_{2}{\left(1-\frac{B}{A}\right)}+\ln{\left(1-B\right)}\ln{\left(A\right)}\right]\\ &~~~~~+\frac{B}{1-B}\left[-\int_{0}^{A}\mathrm{d}x\,\frac{\ln{\left(\frac{x}{B}\right)}}{\left(B-x\right)}-\int_{0}^{A}\mathrm{d}x\,\frac{\ln{\left(B\right)}-\ln{\left(x\right)}}{\left(1-x\right)}\right]\\ &~~~~~+\ln{\left(1-A\right)}\ln{\left(1-B\right)}\\ &~~~~~+\int_{0}^{a}\mathrm{d}p\,\ln{\left(\frac{1+p}{p}\right)}\ln{\left(\frac{p+b}{p}\right)}+\int_{0}^{b}\mathrm{d}q\,\ln{\left(\frac{1+q}{q}\right)}\ln{\left(\frac{a+q}{q}\right)}\\ &=\frac{A}{1-A}\left[\operatorname{Li}_{2}{\left(1-B\right)}-\operatorname{Li}_{2}{\left(1-\frac{B}{A}\right)}+\ln{\left(A\right)}\ln{\left(1-B\right)}\right]\\ &~~~~~+\frac{B}{1-B}\left[\operatorname{Li}_{2}{\left(1-A\right)}-\operatorname{Li}_{2}{\left(1-\frac{A}{B}\right)}+\ln{\left(B\right)}\ln{\left(1-A\right)}\right]\\ &~~~~~+\ln{\left(1-A\right)}\ln{\left(1-B\right)}\\ &~~~~~+\int_{0}^{a}\mathrm{d}p\,\ln{\left(\frac{1+p}{p}\right)}\ln{\left(\frac{p+b}{p}\right)}+\int_{0}^{b}\mathrm{d}q\,\ln{\left(\frac{1+q}{q}\right)}\ln{\left(\frac{a+q}{q}\right)},\\ \end{align}$$

and then,

$$\begin{align} \frac{\mathcal{I}{\left(a,b,1\right)}}{2\pi} &=\ln{\left(1+a\right)}\ln{\left(1+b\right)}\\ &~~~~~+a\left[\operatorname{Li}_{2}{\left(\frac{1}{1+b}\right)}-\operatorname{Li}_{2}{\left(\frac{a-b}{a(1+b)}\right)}-\ln{\left(\frac{a}{1+a}\right)}\ln{\left(1+b\right)}\right]\\ &~~~~~+b\left[\operatorname{Li}_{2}{\left(\frac{1}{1+a}\right)}-\operatorname{Li}_{2}{\left(\frac{b-a}{b(1+a)}\right)}-\ln{\left(\frac{b}{1+b}\right)}\ln{\left(1+a\right)}\right]\\ &~~~~~+\int_{0}^{a}\mathrm{d}p\,\ln{\left(\frac{1+p}{p}\right)}\ln{\left(\frac{b+p}{p}\right)}+\int_{0}^{b}\mathrm{d}q\,\ln{\left(\frac{1+q}{q}\right)}\ln{\left(\frac{a+q}{q}\right)}\\ &=\ln{\left(1+a\right)}\ln{\left(1+b\right)}\\ &~~~~~+a\left[\operatorname{Li}_{2}{\left(\frac{1}{1+b}\right)}-\operatorname{Li}_{2}{\left(\frac{a-b}{a(1+b)}\right)}-\ln{\left(\frac{a}{1+a}\right)}\ln{\left(1+b\right)}\right]\\ &~~~~~+b\left[\operatorname{Li}_{2}{\left(\frac{1}{1+a}\right)}-\operatorname{Li}_{2}{\left(\frac{b-a}{b(1+a)}\right)}-\ln{\left(\frac{b}{1+b}\right)}\ln{\left(1+a\right)}\right]\\ &~~~~~+a\int_{0}^{1}\mathrm{d}x\,\ln{\left(\frac{1+ax}{ax}\right)}\ln{\left(\frac{b+ax}{ax}\right)}\\ &~~~~~+b\int_{0}^{1}\mathrm{d}x\,\ln{\left(\frac{1+bx}{bx}\right)}\ln{\left(\frac{a+bx}{bx}\right)}\\ &=\ln{\left(1+a\right)}\ln{\left(1+b\right)}\\ &~~~~~+a\left[\operatorname{Li}_{2}{\left(\frac{1}{1+b}\right)}-\operatorname{Li}_{2}{\left(\frac{a-b}{a(1+b)}\right)}-\ln{\left(\frac{a}{1+a}\right)}\ln{\left(1+b\right)}\right]\\ &~~~~~+b\left[\operatorname{Li}_{2}{\left(\frac{1}{1+a}\right)}-\operatorname{Li}_{2}{\left(\frac{b-a}{b(1+a)}\right)}-\ln{\left(\frac{b}{1+b}\right)}\ln{\left(1+a\right)}\right]\\ &~~~~~+a\,\mathcal{J}{\left(a,\frac{a}{b}\right)}\\ &~~~~~+b\,\mathcal{J}{\left(b,\frac{b}{a}\right)},\\ \end{align}$$

where in the last line above we've defined another function $\mathcal{J}:\mathbb{R}_{>0}\rightarrow\mathbb{R}$ via the integral

$$\mathcal{J}{\left(p,q\right)}:=\int_{0}^{1}\mathrm{d}x\,\ln{\left(\frac{1+px}{px}\right)}\ln{\left(\frac{1+qx}{qx}\right)}.$$

For $p>0\land q>0$, we obtain

$$\begin{align} \mathcal{J}{\left(p,q\right)} &=\int_{0}^{1}\mathrm{d}x\,\ln{\left(\frac{1+px}{px}\right)}\ln{\left(\frac{1+qx}{qx}\right)}\\ &=\ln{\left(\frac{1+p}{p}\right)}\ln{\left(\frac{1+q}{q}\right)}+\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(\frac{1+qx}{qx}\right)}}{\left(1+px\right)}+\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(\frac{1+px}{px}\right)}}{\left(1+qx\right)};~~~\small{I.B.P.s}\\ &=\ln{\left(\frac{1+p}{p}\right)}\ln{\left(\frac{1+q}{q}\right)}\\ &~~~~~+\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1+qx\right)}}{\left(1+px\right)}+\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1+px\right)}}{\left(1+qx\right)}\\ &~~~~~-\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(x\right)}}{\left(1+px\right)}-\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(x\right)}}{\left(1+qx\right)}\\ &~~~~~-\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(q\right)}}{\left(1+px\right)}-\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(p\right)}}{\left(1+qx\right)}\\ &=\ln{\left(\frac{1+p}{p}\right)}\ln{\left(\frac{1+q}{q}\right)}\\ &~~~~~+\frac{\operatorname{Li}_{2}{\left(\frac{p-q}{p+1}\right)}-\operatorname{Li}_{2}{\left(\frac{p}{1+p}\right)}-\operatorname{Li}_{2}{\left(-q\right)}}{p}+\frac{\operatorname{Li}_{2}{\left(\frac{q-p}{q+1}\right)}-\operatorname{Li}_{2}{\left(\frac{q}{1+q}\right)}-\operatorname{Li}_{2}{\left(-p\right)}}{q}\\ &~~~~~-\frac{\operatorname{Li}_{2}{\left(-p\right)}}{p}-\frac{\operatorname{Li}_{2}{\left(-q\right)}}{q}\\ &~~~~~-\frac{\ln{\left(q\right)}\ln{\left(1+p\right)}}{p}-\frac{\ln{\left(p\right)}\ln{\left(1+q\right)}}{q}.\\ \end{align}$$

And with that, our general evaluation of $\mathcal{I}$ is in principle complete, though the final expression is too cumbersome to bother writing out explicitly.

Cheers!


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Each residue in the upper half-plane is of the form

$$\dfrac 1{2ia_j\prod_{k\ne j}(a_k^2 + a_j^2)}$$

assuming that $a_k > 0$ for all $k$. By the method of partial fractions you can rewrite this in the form $$\frac {A_j}{a_j} + \sum_{k\ne j}\left(\frac {A_k}{a_j + ia_k} + \frac {B_k}{a_j - ia_k}\right)$$ for appropriate constants $A_k, B_k$. Elementary anti-derivatives with respect to $a_j$ exist for this expression. So you should be able to extend this to any finite number of factors.

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  • $\begingroup$ Since this method fully covers what you asked in your question, I don't see how you would expect something to be more "general". $\endgroup$ Jan 6 '20 at 17:15
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    $\begingroup$ @MHZ Have you tried looking at the $A_k, B_k$ for the first few cases to see if you can find a general form for them? $\endgroup$ Mar 13 '20 at 3:41
  • $\begingroup$ @MHZ - The ability to edit other people's posts is intended only for correcting minor issues such as typos, spelling errors, formatting issues. To change the actual content of someone else's post is beyond rude. $\endgroup$ Mar 30 '20 at 11:54

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