Evaluate $\int_0^\infty\frac{dx}{1+x^2}\prod_i\arctan a_ix$ (product of arctangents and Lorentzian)

Question

Define $$I(a_1,\dots,a_n)=\int_0^\infty\frac{dx}{1+x^2}\prod_{i=1}^n\arctan a_ix$$ with $a_i>0$. By this answer $\newcommand{Li}{\operatorname{Li}_2}$ $$I(a,b)= \frac\pi4\left(\frac{\pi^2}6 -\Li\left(\frac{1-a}{1+a}\right) -\Li\left(\frac{1-b}{1+b}\right) +\Li\left(\frac{1-a}{1+a}\frac{1-b}{1+b}\right)\right)$$ By taking the limit as $b\to\infty$ or by here we get $$I(a)= \frac12\left(\frac{\pi^2}4 -\Li\left(\frac{1-a}{1+a}\right) +\Li\left(-\frac{1-a}{1+a}\right)\right)$$ But I have not been able to reduce $I(a,b,c)$ to polylogarithms like the previous two equations.

1. Can $I(a_1,\dots,a_n)$ be reduced to elementary functions, polylogarithms and possibly other special functions appearing in the literature for general $a_i$ and $n$? Contour integration might help for even $n$.
2. If not, then what about the special case where $a_i\in\{1,a\}$ for a given $a$? Such integrals occur in the probability discussed here.

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Here is my attempt at calculating $I(a,b,c)$. Define the self-inverse transformation $f:v\mapsto\frac{1-v}{1+v}$, $x=f(a),y=f(b),z=f(c)$ and $J(a,b,c)=I(f(a),f(b),f(c))$. Then $I(a,b,c)=J(x,y,z)$ with $-1<x,y,z<1$ and we can get the Maclaurin series expansion of $J$ in $x,y,z$ by differentiating under the integral sign. Numerical results strongly suggest that the $x^iy^jz^k$ coefficient of this series is the sum of

• $\frac{2^{p-2}}{(i+j+k)(i-j-k)(j-k-i)(k-i-j)}$ if $i+j+k$ is odd, where $p$ of the exponents $i,j,k$ are positive
• $\frac{\pi^2}{16i^2}$ if $p=2$ and (without loss of generality) $i=j>0,k=0$
• $-\frac{\pi^2}{8i^2}$ if $p=1$ and (without loss of generality) $i>0,j=k=0$
• $\frac{\pi^4}{64}$ if $p=0$ (i.e. $i=j=k=0$)

This would mean $\newcommand{Li}{\operatorname{Li}_2}$

$$I(a,b,c)=\frac{\pi^4}{64}-\frac{\pi^2}8(\Li(x)+\Li(y)+\Li(z))+\frac{\pi^2}{16}(\Li(xy)+\Li(zx)+\Li(yz))$$

$$+\sum_{i,j,k=0,2\nmid i+j+k}^\infty\frac{2^{p-2}}{(i+j+k)(i-j-k)(j-k-i)(k-i-j)}x^iy^jz^k$$ But an observed pattern is not a proof. How can this last equation be proved and can the last sum be reduced to special functions in the literature?

Answer 1

Not sure if it helps, but you can derive a recurrence for $$I_n(a_1,\dots,a_n)=\int_0^\infty\frac{dx}{1+x^2}\prod_{i=1}^n\arctan (a_ix) \, .$$ $I_n$ is symmetric with respect to permutation of its arguments and anti-symmetric in $a_i$ for all $i$. As a function over $\mathbb{C}$ it converges for all complex values $a_i$, since the singularities are integrable. Since $\arctan(x)$ has cuts on $(-i\infty,-i) \cup (i,i\infty)$, $I_n$ will have discontinuities along the imaginary axis with the exception at $a_i=0$ where $I_n$ vanishes and is differentiable off the imaginary axis. In the following we can assume $a_i \in \mathbb{R}^+$ for which $I_n$ is well defined and a continuously increasing function in $a_i$. We can now do the following manipulations $$I_n(a_1,\dots,a_n)=\int_0^{a_n}dt\int_0^{a_{n-1}} ds \int_0^\infty dx \, \frac{x^2}{(1+x^2)(1+s^2x^2)(1+t^2x^2)} \prod_{i=1}^{n-2}\arctan (a_ix) \\ =\int_0^{a_n}dt\int_0^{a_{n-1}} ds \int_0^\infty dx \prod_{i=1}^{n-2}\arctan (a_ix) \\ \quad \times \left( \frac{s}{(1-s^2)(s^2-t^2)} \frac{s}{1+s^2x^2} + \frac{t}{(1-t^2)(t^2-s^2)} \frac{t}{1+t^2x^2} - \frac{1}{(1-s^2)(1-t^2)} \frac{1}{1+x^2}\right) \\ \stackrel{\substack{u=sx\\v=tx}}{=} \int_0^{a_n}dt\int_0^{a_{n-1}} ds \left( \frac{s}{(1-s^2)(s^2-t^2)} \, I_{n-2}(a_1/s,\dots,a_{n-2}/s) \\ + \frac{t}{(1-t^2)(t^2-s^2)} \, I_{n-2}(a_1/t,\dots,a_{n-2}/t) - \frac{1}{(1-s^2)(1-t^2)} \, I_{n-2}(a_1,...,a_{n-2})\right) \, .$$ The integrand appears to have singularities, but expanding about $s=t=1$ or $s=t$ it is seen to be continuous. Nevertheless, each individual term does have singularities. Before doing any further manipulations, we can now fix a contour for the $s$- and $t$-integral. We choose the line of integration for the $t$-integral to be slightly above the real axis (in the upper half-plane) and for $s$ to be below the real axis (if necessary). This way the paths of integration do not cross. Because the integrand is continuous, we have the choice of picking an inner integral ($t$) and for that the principal value $PV$ before we break it up into individual terms. Finally we let $a_n \to \infty$ $$I_n(a_1,\dots,a_{n-1},\infty)= \frac{\pi}{2} \, I_{n-1}(a_1,\dots,a_{n-1}) \\ =PV \int_0^{a_{n-1}} ds \int_0^{\infty}dt \left( \frac{s}{(1-s^2)(s^2-t^2)} \, I_{n-2}(a_1/s,\dots,a_{n-2}/s) \\ + \frac{t}{(1-t^2)(t^2-s^2)} \, I_{n-2}(a_1/t,\dots,a_{n-2}/t) - \frac{1}{(1-s^2)(1-t^2)} \, I_{n-2}(a_1,...,a_{n-2})\right) \\ =\int_0^{a_{n-1}} ds \, PV \int_0^{\infty}dt \, \frac{t}{(1-t^2)(t^2-s^2)} \, I_{n-2}(a_1/t,\dots,a_{n-2}/t) \, .$$ In the last line we broke up the integrand and used the fact that $$PV \int_0^\infty \frac{dt}{s^2-t^2}=0$$ for all $s>0$. Thus (after the shift $n\to n+1$) $$I_n(a_1,\dots,a_n) = \frac2\pi \int_0^{a_n} ds \, PV \int_0^\infty dt \, \frac{t}{(1-t^2)(t^2-s^2)} \, I_{n-1}(a_1/t,\dots,a_{n-1}/t) \, . \tag{1}$$ This integral is well defined for all $a_n>0$. However, the principal value doesn't allow the naive interchange of the order of integration. This can be remedied by considering again the (complex) line-contour (as above) and adding (+subtracting) back the missing infinitesimal semi-circle (in the upper half-plane) at $t=s$. This gives an additional term by the residue theorem and the principal value now only takes into account the singularity at $t=1$ $$I_n(a_1,\dots,a_n) = \frac2\pi \int_0^{a_n} ds \, PV_{t=1} \int_0^\infty dt \, \frac{t}{(1-t^2)(t^2-s^2)} \, I_{n-1}(a_1/t,\dots,a_{n-1}/t) \\ \qquad \qquad + i \int_0^{a_n} ds \frac{1}{(1-s^2)} \, I_{n-1}(a_1/s,\dots,a_{n-1}/s) \, .$$ The $s$ and $t$ integral can now be interchanged, since there is no discontinuity issue at $s=t$ $$I_n(a_1,\dots,a_n) = \frac2\pi \, PV_{t=1} \int_0^\infty dt \, \frac{1}{(1-t^2)} \, I_{n-1}(a_1/t,\dots,a_{n-1}/t) \left[ PV\int_0^{a_n} ds \, \frac{t}{t^2-s^2} \\ \qquad \qquad - \frac{i\pi}{2} \left[t\in(0,a_n)\right] \right] + i \int_0^{a_n} ds \frac{1}{(1-s^2)} \, I_{n-1}(a_1/s,\dots,a_{n-1}/s) \, .$$ The $s$ integral was splitted into the principal value and the positively oriented infinitesimal semi-circle about $s=t$ evaluated using the residue, which only occurs if $t\in(0,a_n)$. $[\dots]$ is the iversion bracket. The previous expression then becomes $$I_n(a_1,\dots,a_n)= \frac1\pi \, PV \int_0^\infty dt \, \frac{1}{1-t^2} \, \log\left|\frac{a_n+t}{a_n-t}\right| \, I_{n-1}(a_1/t,\dots,a_{n-1}/t) \\ -i \, PV\int_0^{a_n} \frac{dt}{1-t^2} \, I_{n-1}(a_1/t,\dots,a_{n-1}/t) + i \int_0^{a_n} ds \frac{1}{(1-s^2)} \, I_{n-1}(a_1/s,\dots,a_{n-1}/s) \, .$$ Now remember that the $t$-line-contour was above the real line in the upper complex plane, while $s$ followed a path below the real line. Therefore the second line essentially cancels (i.e. the principal values cancel), except the positively oriented infinitesimal semi-circle at $s=1$, which also occurs only in the case $a_n>1$ and the result $$I_n(a_1,\dots,a_n)= \frac1\pi \, PV \int_0^\infty dt \, \frac{1}{1-t^2} \, \log\left|\frac{a_n+t}{a_n-t}\right| \, I_{n-1}(a_1/t,\dots,a_{n-1}/t) \\ \qquad\qquad + [a_n>1] \, \frac\pi2 \, I_{n-1}(a_1,\dots,a_{n-1})$$ follows.

For example for $n=1$ we have $$I_0=\frac{\pi}{2}$$ and by (1) $$I_1(a)=PV\int_0^{a} ds \int_0^\infty dt \, \frac{t}{(1-t^2)(t^2-s^2)} = PV \int_0^a ds \, \frac{\log(s)}{s^2-1} \\ = \frac12 \, \log(a) \log\left(\frac{1-a}{1+a}\right) + \frac12 \int_0^a \frac{\log(1+s)-\log(1-s)}{s} \, ds \\ =\frac12 \left[ \log(a) \log\left(\frac{1-a}{1+a}\right) + {\rm Li_2}(a) - {\rm Li_2}(-a) \right] \\ =\frac12 \left[ \frac{\pi^2}{4} - {\rm Li_2}\left(\frac{1-a}{1+a}\right) + {\rm Li_2}\left(-\frac{1-a}{1+a}\right) \right] $$ by Abel's and Euler's reflection identity.