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

Question

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 $2n=4$.

Answer 1

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.

Answer 2

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).

Answer 3

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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.

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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}$$

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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

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

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$$\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.

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