Asymptotics of $\int_1^L\int_1^L\int_1^L\int_1^L \frac{\mathrm dx_1~\mathrm dx_2 ~ \mathrm dx_3 ~ \mathrm dx_4}{(x_1+x_2)(x_2+x_3)(x_3+x_4)(x_4+x_1)}$

Let $L>1$. I am looking for the value, or the leading asymptotics for $L\to\infty$, of $$\int_1^L\int_1^L\int_1^L\int_1^L \dfrac{\mathrm dx_1~\mathrm dx_2 ~ \mathrm dx_3 ~ \mathrm dx_4}{(x_1+x_2)(x_2+x_3)(x_3+x_4)(x_4+x_1)}$$ More generally, I'd like to know the leading asymptotics of an expression like this with $2n$ terms, where the above has $2n=4$.

• I don't know the answer, but the expression spontanously reminds me of en.wikipedia.org/wiki/Selberg_integral as well as en.wikipedia.org/wiki/Feynman_parametrization. The integral bounds are different but you might change them to your integral by substitution and then substract the 0 to 1 part again. Commented Mar 1, 2013 at 11:13
• Have you tried the $n=1$ case? Commented Mar 1, 2013 at 11:14
• @MichaelAlbanese : I can't write down that case, this question nicely has knitted together that n has to be at least 2. Commented Mar 1, 2013 at 11:20
• If you do them one integral at a time (let's say w.r.t $x_1$ first), you can move two of the factors outside the first integration and you're down to an integral of the form $$\int_1^L \frac{dx}{(x+a)(x+b)}$$ which should be pretty solvable with partial fractions, or just any table of integrals. While writing this comment, though, I've come to realize that the next integrand might not be so pretty, but there's only one way to find out, right? Commented Mar 1, 2013 at 11:59
• @Arthur: Yes, this one integration is easy. The square of its value isn't, though. Commented Mar 1, 2013 at 12:19

The general asymptotics as $$L \to\infty$$ of $$I_m(L) = \int_1^L \cdots \int_1^L \frac {dx_1 \cdots dx_{m}} {(x_1+x_2)(x_2+x_3) \cdots (x_{m} +x_1)}$$ for $$m\in\mathbb N$$ can be computed in the following way:

1. Show that $$I_m(L) - \int_1^L \int_1^\infty \cdots \int_1^\infty \frac {dx_1 \cdots dx_{m}} {(x_1+x_2)(x_2+x_3) \cdots (x_{m} +x_1)} = O(1)$$ as $$L\to\infty$$, i.e. you perform all integrals up to $$\infty$$ except one integral.

2. You write the integral as the trace of an operator $$\int_1^L \int_1^\infty \cdots \int_1^\infty \frac {dx_1 \cdots dx_{m}} {(x_1+x_2)(x_2+x_3) \cdots (x_{m} +x_1)} = \operatorname{tr} 1_{[1,L]} K^m 1_{[1,L]}$$ where $$K$$ is the Hankel operator $$K:L^2([0,\infty)) \to L^2([0,\infty))$$ with kernel $$K(x,y) = \frac 1 {x+y}$$, i.e. $$(Kf)(x) = \int_0^\infty \frac1{x+y}f(y)$$ where $$\operatorname{tr}$$ is the trace and $$1_{[1,L]}$$ is the orthogonal projection onto $$L^2([1,L])$$. The operator $$K$$ is also called Carleman operator.

3. The important point why this helps is that $$K$$ can be explicitly diagonalized using the Mellin transform, i.e. $$K \psi_k = f(k) \psi_k$$ with $$(\psi_k)(x) = \frac 1 {\sqrt{2\pi}}x^{-1/2 + ik}$$ and $$f(k) = \frac \pi{\cosh(\pi k)}$$ and $$k\in\mathbb R$$. Noting that the $$\psi_k$$ define a unitary transform (Mellin transform) one can write $$K = \int_{\mathbb R} f(k) |\psi_k\rangle\langle\psi_k| dk$$ and $$\langle\psi_k,\psi_m\rangle =\delta_{km}$$.

4. Inserting this in $$\operatorname{tr} 1_{[1,L]} K^m 1_{[1,L]}$$ we obtain $$\operatorname{tr} 1_{[1,L]} K^m 1_{[1,L]} = \int_1^L d x \int_{\mathbb R} d k f(k)^m |\psi_k(x)|^2$$ Now $$|\psi_k(x)|^2 = \frac 1 {2\pi x}$$ independently of $$k$$ and integrating over $$x$$ gives $$\int_1^L d x \int_{\mathbb R} d k f(k)^m |\psi_k(x)|^2 = \frac{\log L}{2\pi} \int_{\mathbb R} d k \big(\frac \pi{\cosh(\pi x)}\big)^m.$$ This implies that $$I_m(L) \sim \frac{\log L}{2\pi} \int_{\mathbb R} d k \big(\frac \pi{\cosh(\pi x)}\big)^m.$$ For $$m=4$$ computing the integral gives $$I_4(L) \sim \log L \frac 2 3 \pi^2$$ which is consistent with the previous answer.

• I think this answer's reasoning might be incorrect. The answer predicts $\int_{1}^L \int_1^\infty \frac{dx_1 dx_2}{(x_1+x_2)(x_2+x_1)}$ can be tranformed into $\frac{\log(L)}{2 \pi} \int_{-\infty}^\infty \frac{\pi^2}{\cosh(\pi x)^2} = \log(L)$, but one can check that $\int_{1}^L \int_1^\infty \frac{dx_1 dx_2}{(x_1+x_2)(x_2+x_1)}$ is equal to $\log(\frac{1+L}{2})$. I think the mistake is that $$\operatorname{tr} 1_{[1,L]} K^m 1_{[1,L]}= \int_1^L \int_0^\infty \cdots \int_0^\infty \frac {dx_1 \cdots dx_{m}} {(x_1+x_2)(x_2+x_3) \cdots (x_{m} +x_1)}.$$ Commented Jul 17, 2023 at 5:36
• To be more explicit, the projectors don't affect the bounds on the inner K's (only the outermost one), so I think the expression in this answer that $$\operatorname{tr} 1_{[1,L]} K^m 1_{[1,L]}= \int_1^L \int_1^\infty \cdots \int_1^\infty \frac {dx_1 \cdots dx_{m}} {(x_1+x_2)(x_2+x_3) \cdots (x_{m} +x_1)}$$ is incorrect and should be replaced by the expression in my previous comment. Commented Jul 17, 2023 at 5:39
• Explicit evaluation for small $n$ supports @user196574's assertion. It can also be viewed as ordinary convolution of Fourier transforms $\int_{\mathbb R^{n-1}}f(y_1-y_2)f(y_2-y_3)\cdots f(y_n-y_1)d y_2d y_3\cdots d y_n=\int_{\mathbb R}\hat f(w)^ndw$ after substitution. Commented Aug 23, 2023 at 15:06

The leading term for $n=2$ is $\frac23\pi^2\log L$.

As has been mentioned in the comments, you can perform one half of the integrations explicitly:

\begin{align} &\int_1^L\int_1^L\int_1^L\int_1^L\frac{\mathrm dx_1\mathrm dx_2\mathrm dx_3\mathrm dx_4}{(x_1+x_2)(x_2+x_3)(x_3+x_4)(x_4+x_1)} \\ =& \int_1^L\int_1^L\left(\int_1^L\frac{\mathrm dx_2}{(x_1+x_2)(x_2+x_3)}\right)\left(\int_1^L\frac{\mathrm dx_4}{(x_3+x_4)(x_4+x_1)}\right)\mathrm dx_1\mathrm dx_3 \\ =& \int_1^L\int_1^L\left(\frac{\log\frac{(x_3+L)(x_1+1)}{(x_1+L)(x_3+1)}}{x_1-x_3}\right)^2\mathrm dx_1\mathrm dx_3\;. \\ =& 2\int_1^L\int_1^{x_3}\left(\frac{\log\frac{(x_3+L)(x_1+1)}{(x_1+L)(x_3+1)}}{x_1-x_3}\right)^2\mathrm dx_1\mathrm dx_3 \\ =& 2\int_1^L\int_{1/x_3}^1\left(\frac{\log\frac{(x_3+L)(\lambda x_3+1)}{(\lambda x_3+L)(x_3+1)}}{\lambda x_3-x_3}\right)^2\mathrm d(\lambda x_3)\,\mathrm dx_3 \\ =& 2\int_1^L\frac1{x_3}\int_{1/x_3}^1\left(\frac{\log\frac{(x_3+L)(\lambda x_3+1)}{(\lambda x_3+L)(x_3+1)}}{\lambda-1}\right)^2\mathrm d\lambda\,\mathrm dx_3\;, \end{align}

The dominating factor here is $1/x_3$, whose integral goes with $\log L$. All orders of magnitude of $x_3$ contribute to that integral equally, so for large $L$ the contributions from orders of magnitude close to $1$ or close to $L$ make a negligible contribution; thus we can make the approximations $x_3+L\approx\lambda x_3+L\approx L$, $\lambda x_3+1\approx\lambda x_3$ and $x_3+1\approx x_3$, which yield

$$2\int_1^L\frac1{x_3}\int_{1/x_3}^1\left(\frac{\log\lambda}{\lambda-1}\right)^2\mathrm d\lambda\,\mathrm dx_3\;.$$

For large $L$ almost all contributions come from large $x_3$, so we can replace $1/x_3$ by $0$, which makes the inner integral independent of $x_3$ and yields

\begin{align} 2\int_1^L\frac1{x_3}\int_0^1\left(\frac{\log\lambda}{\lambda-1}\right)^2\mathrm d\lambda\,\mathrm dx_3 &= 2\log L\int_0^1\left(\frac{\log\lambda}{\lambda-1}\right)^2\mathrm d\lambda \\ &= 2\log L\int_0^1\log^2\lambda\left(\sum_{n=0}^\infty\lambda^n\right)^2\mathrm d\lambda \\ &= 2\log L\int_0^1\log^2\lambda\sum_{n=0}^\infty(n+1)\lambda^n\mathrm d\lambda \\ &= 2\log L\sum_{n=0}^\infty\int_0^1\log^2\lambda\,(n+1)\lambda^n\mathrm d\lambda \\ &= 2\log L\sum_{n=0}^\infty\frac2{(n+1)^2} \\ &= 4\zeta(2)\log L \\ &= \frac23\pi^2\log L\;. \end{align}

Note that $\frac23\pi^2$ is one third of the surface area of the unit $3$-sphere. We can express the $x_i$ in hyperspherical coordinates; then the radial integration yields $\log L$, and we can roughly regard $\frac23\pi^2$ as the integral over the angular coordinates.

Here's a logarithmic plot of essentially exact values obtained by Gaussian quadrature of the two-dimensional integral (transformed to $\log x_1$ and $\log x_3$ to improve convergence):

The green line is $\frac23\pi^2\log L-24$; the $24$ is from trial and error, and I didn't try to derive it.

The same approach can be used for $n\gt2$, though it may no longer be possible to perform the "angular" integration in closed form. For example, for $n=3$:

\begin{align} &\int_1^L\int_1^L\int_1^L\frac{\log x_1-\log x_3}{x_1-x_3}\frac{\log x_3-\log x_5}{x_3-x_5}\frac{\log x_5-\log x_1}{x_5-x_1}\mathrm dx_5\mathrm dx_3\mathrm dx_1 \\ =&3\int_1^L\frac1{x_1}\int_{1/x_1}^1\int_{1/x_1}^1\frac{-\log\lambda_3}{1-\lambda_3}\frac{\log\lambda_3-\log\lambda_5}{\lambda_3-\lambda_5}\frac{\log\lambda_5}{\lambda_5-1}\mathrm d\lambda_5\mathrm d\lambda_3\mathrm dx_1 \\ \approx&3\int_1^L\frac1{x_1}\int_0^1\int_0^1\frac{-\log\lambda_3}{1-\lambda_3}\frac{\log\lambda_3-\log\lambda_5}{\lambda_3-\lambda_5}\frac{\log\lambda_5}{\lambda_5-1}\mathrm d\lambda_5\mathrm d\lambda_3\mathrm dx_1 \\ =&3\log L\int_0^1\int_0^1\frac{-\log\lambda_3}{1-\lambda_3}\frac{\log\lambda_3-\log\lambda_5}{\lambda_3-\lambda_5}\frac{\log\lambda_5}{\lambda_5-1}\mathrm d\lambda_5\mathrm d\lambda_3 \\ \approx&51.95\log L\;, \end{align}

where the coefficient in the last line was obtained by Gaussian quadrature (after transforming to $\sqrt\lambda_3$ and $\sqrt\lambda_5$ to improve convergence).

Define the function $$\mathcal{I}:\left(1,\infty\right)\rightarrow\mathbb{R}_{>0}$$ via the quadruple integral

$$\mathcal{I}{\left(a\right)}:=\int_{\left[1,a\right]^{4}}\mathrm{d}t\,\mathrm{d}u\,\mathrm{d}v\,\mathrm{d}w\,\frac{1}{\left(t+u\right)\left(u+v\right)\left(v+w\right)\left(w+t\right)}.\tag{1}$$

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

Suppose $$a\in\left(1,\infty\right)$$. We begin by reducing $$\mathcal{I}$$ to the following double integral form:

\begin{align} \mathcal{I}{\left(a\right)} &=\int_{\left[1,a\right]^{4}}\mathrm{d}t\,\mathrm{d}u\,\mathrm{d}v\,\mathrm{d}w\,\frac{1}{\left(t+u\right)\left(u+v\right)\left(v+w\right)\left(w+t\right)}\\ &=\int_{1}^{a}\mathrm{d}t\int_{1}^{a}\mathrm{d}u\int_{1}^{a}\mathrm{d}v\int_{1}^{a}\mathrm{d}w\,\frac{1}{\left(t+u\right)\left(u+v\right)\left(v+w\right)\left(w+t\right)}\\ &=\int_{1}^{a}\mathrm{d}t\int_{1}^{a}\mathrm{d}v\int_{1}^{a}\mathrm{d}u\int_{1}^{a}\mathrm{d}w\,\frac{1}{\left(t+u\right)\left(u+v\right)\left(v+w\right)\left(w+t\right)}\\ &=\int_{1}^{a}\mathrm{d}t\int_{1}^{a}\mathrm{d}v\int_{1}^{a}\mathrm{d}u\,\frac{1}{\left(t+u\right)\left(u+v\right)}\int_{1}^{a}\mathrm{d}w\,\frac{1}{\left(v+w\right)\left(w+t\right)}\\ &=\int_{1}^{a}\mathrm{d}t\int_{1}^{a}\mathrm{d}v\,\left[\int_{1}^{a}\mathrm{d}u\,\frac{1}{\left(t+u\right)\left(u+v\right)}\right]^{2}\\ &=\int_{1}^{a}\mathrm{d}t\int_{1}^{a}\mathrm{d}v\,\left[\int_{1}^{a}\mathrm{d}u\,\frac{d}{du}\frac{\ln{\left(u+v\right)}-\ln{\left(t+u\right)}}{t-v}\right]^{2}\\ &=\int_{1}^{a}\mathrm{d}t\int_{1}^{a}\mathrm{d}v\,\left[\frac{\ln{\left(a+v\right)}-\ln{\left(t+a\right)}}{t-v}-\frac{\ln{\left(1+v\right)}-\ln{\left(t+1\right)}}{t-v}\right]^{2}\\ &=\int_{1}^{a}\mathrm{d}t\int_{1}^{a}\mathrm{d}v\,\left[\frac{\ln{\left(\frac{1+t}{a+t}\right)}-\ln{\left(\frac{1+v}{a+v}\right)}}{t-v}\right]^{2}\\ &=2\int_{1}^{a}\mathrm{d}t\int_{1}^{t}\mathrm{d}v\,\left[\frac{\ln{\left(\frac{1+t}{a+t}\right)}-\ln{\left(\frac{1+v}{a+v}\right)}}{t-v}\right]^{2};~~~\small{symmetry}\\ &=2\int_{1}^{a}\mathrm{d}t\int_{\frac{2}{a+1}}^{\frac{1+t}{a+t}}\mathrm{d}y\,\frac{\left(a-1\right)}{\left(1-y\right)^{2}}\left[\frac{\ln{\left(\frac{1+t}{a+t}\right)}-\ln{\left(y\right)}}{t-\left(\frac{ay-1}{1-y}\right)}\right]^{2};~~~\small{\left[\frac{1+v}{a+v}=y\right]}\\ &=2\int_{\frac{2}{a+1}}^{\frac{1+a}{2a}}\mathrm{d}x\,\frac{\left(a-1\right)}{\left(1-x\right)^{2}}\int_{\frac{2}{a+1}}^{x}\mathrm{d}y\,\frac{\left(a-1\right)}{\left(1-y\right)^{2}}\left[\frac{\ln{\left(x\right)}-\ln{\left(y\right)}}{\left(\frac{ax-1}{1-x}\right)-\left(\frac{ay-1}{1-y}\right)}\right]^{2};~~~\small{\left[\frac{1+t}{a+t}=x\right]}\\ &=2\int_{\frac{2}{a+1}}^{\frac{1+a}{2a}}\mathrm{d}x\int_{\frac{2}{a+1}}^{x}\mathrm{d}y\,\left[\frac{\ln{\left(x\right)}-\ln{\left(y\right)}}{x-y}\right]^{2}\\ &=2\int_{1}^{\frac{(1+a)^{2}}{4a}}\mathrm{d}x\int_{1}^{x}\mathrm{d}y\,\left[\frac{\ln{\left(x\right)}-\ln{\left(y\right)}}{x-y}\right]^{2};~~~\small{\left[(x,y)\mapsto\left(\frac{2x}{a+1},\frac{2y}{a+1}\right)\right]}\\ &=2\int_{1}^{A}\mathrm{d}x\int_{1}^{x}\mathrm{d}y\,\left[\frac{\ln{\left(x\right)}-\ln{\left(y\right)}}{x-y}\right]^{2};~~~\small{\left[A:=\frac{(1+a)^{2}}{4a}>1\right]}\\ &=2\int_{1}^{A}\mathrm{d}x\int_{1}^{x}\mathrm{d}y\,\frac{\left[\ln{\left(x\right)}-\ln{\left(y\right)}\right]^{2}}{\left(x-y\right)^{2}}.\tag{2}\\ \end{align}

Consider the following derivative, computed using the product rule:

\begin{align} \frac{d}{dy}\frac{\left[\ln{\left(x\right)}-\ln{\left(y\right)}\right]^{2}}{x-y} &=\left[\ln{\left(x\right)}-\ln{\left(y\right)}\right]^{2}\frac{d}{dy}\frac{1}{x-y}+\frac{1}{x-y}\frac{d}{dy}\left[\ln{\left(x\right)}-\ln{\left(y\right)}\right]^{2}\\ &=\frac{\left[\ln{\left(x\right)}-\ln{\left(y\right)}\right]^{2}}{\left(x-y\right)^{2}}-\frac{2\left[\ln{\left(x\right)}-\ln{\left(y\right)}\right]}{\left(x-y\right)y}.\\ \end{align}

Rearranging the terms and integrating both sides with respect to $$y$$ yields the integration by parts formula

\begin{align} \int_{1}^{x}\mathrm{d}y\,\frac{\left[\ln{\left(x\right)}-\ln{\left(y\right)}\right]^{2}}{\left(x-y\right)^{2}} &=\int_{1}^{x}\mathrm{d}y\,\frac{d}{dy}\frac{\left[\ln{\left(x\right)}-\ln{\left(y\right)}\right]^{2}}{\left(x-y\right)}+\int_{1}^{x}\mathrm{d}y\,\frac{2\left[\ln{\left(x\right)}-\ln{\left(y\right)}\right]}{\left(x-y\right)y}\\ &=\lim_{y\to x}\frac{\left[\ln{\left(x\right)}-\ln{\left(y\right)}\right]^{2}}{\left(x-y\right)}-\lim_{y\to1}\frac{\left[\ln{\left(x\right)}-\ln{\left(y\right)}\right]^{2}}{\left(x-y\right)}\\ &~~~~~+\int_{1}^{x}\mathrm{d}y\,\frac{2\left[\ln{\left(x\right)}-\ln{\left(y\right)}\right]}{\left(x-y\right)y}\\ &=-\frac{\ln^{2}{\left(x\right)}}{\left(x-1\right)}+\int_{1}^{x}\mathrm{d}y\,\frac{2\left[\ln{\left(x\right)}-\ln{\left(y\right)}\right]}{\left(x-y\right)y}.\tag{3}\\ \end{align}

\begin{align} \frac12\mathcal{I}{\left(a\right)} &=\int_{1}^{A}\mathrm{d}x\int_{1}^{x}\mathrm{d}y\,\frac{\left[\ln{\left(x\right)}-\ln{\left(y\right)}\right]^{2}}{\left(x-y\right)^{2}}\\ &=\int_{1}^{A}\mathrm{d}x\,\bigg{[}-\frac{\ln^{2}{\left(x\right)}}{\left(x-1\right)}+\int_{1}^{x}\mathrm{d}y\,\frac{2\left[\ln{\left(x\right)}-\ln{\left(y\right)}\right]}{\left(x-y\right)y}\bigg{]};~~~\small{I.B.P.}\\ &=-\int_{1}^{A}\mathrm{d}x\,\frac{\ln^{2}{\left(x\right)}}{\left(x-1\right)}+\int_{1}^{A}\mathrm{d}x\int_{1}^{x}\mathrm{d}y\,\frac{2\left[\ln{\left(x\right)}-\ln{\left(y\right)}\right]}{\left(x-y\right)y}\\ &=-\int_{A^{-1}}^{1}\mathrm{d}t\,\frac{\ln^{2}{\left(t\right)}}{t\left(1-t\right)};~~~\small{\left[x=\frac{1}{t}\right]}\\ &~~~~~+\int_{1}^{A}\mathrm{d}x\int_{1}^{x}\mathrm{d}y\,\frac{2\left[\ln{\left(x\right)}-\ln{\left(y\right)}\right]}{xy}+\int_{1}^{A}\mathrm{d}x\int_{1}^{x}\mathrm{d}y\,\frac{2\left[\ln{\left(x\right)}-\ln{\left(y\right)}\right]}{x\left(x-y\right)}\\ &=-\int_{A^{-1}}^{1}\mathrm{d}t\,\frac{\ln^{2}{\left(t\right)}}{t}-\int_{A^{-1}}^{1}\mathrm{d}t\,\frac{\ln^{2}{\left(t\right)}}{\left(1-t\right)}\\ &~~~~~+\int_{1}^{A}\mathrm{d}x\,\frac{\ln^{2}{\left(x\right)}}{x}+\int_{1}^{A}\mathrm{d}x\int_{x^{-1}}^{1}\mathrm{d}t\,\frac{2\left[-\ln{\left(t\right)}\right]}{x\left(1-t\right)};~~~\small{\left[y=xt\right]}\\ &=-\int_{1}^{A}\mathrm{d}x\,\frac{\ln^{2}{\left(x\right)}}{x};~~~\small{\left[t=\frac{1}{x}\right]}\\ &~~~~~+\int_{1}^{A}\mathrm{d}x\,\frac{\ln^{2}{\left(x\right)}}{x}-\int_{A^{-1}}^{1}\mathrm{d}t\,\frac{\ln^{2}{\left(t\right)}}{\left(1-t\right)}\\ &~~~~~-\int_{A^{-1}}^{1}\mathrm{d}u\int_{u}^{1}\mathrm{d}t\,\frac{2\ln{\left(t\right)}}{u\left(1-t\right)};~~~\small{\left[x=\frac{1}{u}\right]}\\ &=-\int_{A^{-1}}^{1}\mathrm{d}t\,\frac{\ln^{2}{\left(t\right)}}{\left(1-t\right)}-2\int_{A^{-1}}^{1}\mathrm{d}t\int_{A^{-1}}^{t}\mathrm{d}u\,\frac{\ln{\left(t\right)}}{u\left(1-t\right)}\\ &=-\int_{A^{-1}}^{1}\mathrm{d}t\,\frac{\ln^{2}{\left(t\right)}}{\left(1-t\right)}-2\int_{A^{-1}}^{1}\mathrm{d}t\,\frac{\ln{\left(t\right)}}{\left(1-t\right)}\int_{1}^{At}\mathrm{d}v\,\frac{1}{v};~~~\small{\left[u=\frac{v}{A}\right]}\\ &=-\int_{A^{-1}}^{1}\mathrm{d}t\,\frac{\ln^{2}{\left(t\right)}}{\left(1-t\right)}-2\int_{A^{-1}}^{1}\mathrm{d}t\,\frac{\ln{\left(At\right)}\ln{\left(t\right)}}{\left(1-t\right)}\\ &=-3\int_{A^{-1}}^{1}\mathrm{d}t\,\frac{\ln^{2}{\left(t\right)}}{\left(1-t\right)}-2\ln{\left(A\right)}\int_{A^{-1}}^{1}\mathrm{d}t\,\frac{\ln{\left(t\right)}}{\left(1-t\right)}\\ &=-3\int_{z}^{1}\mathrm{d}t\,\frac{\ln^{2}{\left(t\right)}}{\left(1-t\right)}+2\ln{\left(z\right)}\int_{z}^{1}\mathrm{d}t\,\frac{\ln{\left(t\right)}}{\left(1-t\right)};~~~\small{\left[z:=A^{-1}\in(0,1)\right]}\\ &=-6\int_{0}^{1-z}\mathrm{d}u\,\frac{\ln^{2}{\left(1-u\right)}}{2u}-2\ln{\left(z\right)}\int_{0}^{1-z}\mathrm{d}u\,\frac{(-1)\ln{\left(1-u\right)}}{u};~~~\small{\left[t=1-u\right]}\\ &=-6S_{1,2}{\left(1-z\right)}-2\ln{\left(z\right)}\operatorname{Li}_{2}{\left(1-z\right)},\\ \end{align}

and so we have

\begin{align} \mathcal{I}{\left(a\right)} &=-12S_{1,2}{\left(1-z\right)}-4\ln{\left(z\right)}\operatorname{Li}_{2}{\left(1-z\right)}\\ &=12\operatorname{Li}_{3}{\left(z\right)}-12\ln{\left(z\right)}\operatorname{Li}_{2}{\left(z\right)}-6\ln^{2}{\left(z\right)}\ln{\left(1-z\right)}-12\zeta{\left(3\right)}\\ &~~~~~+4\ln{\left(z\right)}\operatorname{Li}_{2}{\left(z\right)}+4\ln^{2}{\left(z\right)}\ln{\left(1-z\right)}-4\zeta{\left(2\right)}\ln{\left(z\right)}\\ &=12\operatorname{Li}_{3}{\left(z\right)}-8\ln{\left(z\right)}\operatorname{Li}_{2}{\left(z\right)}-2\ln^{2}{\left(z\right)}\ln{\left(1-z\right)}-4\,\zeta{\left(2\right)}\ln{\left(z\right)}-12\,\zeta{\left(3\right)},\tag{4}\\ \end{align}

where $$z=\frac{4a}{(1+a)^{2}}$$. As $$a\to\infty$$, this is asymptotically equivalent to

$$\mathcal{I}{\left(a\right)}\sim4\,\zeta{\left(2\right)}\ln{\left(\frac{(1+a)^{2}}{4a}\right)}\sim\frac23\pi^{2}\ln{\left(a\right)},$$

which is the expected result found in the other answers.