Quadruple Integral $\int\limits_0^1\!\!\int\limits_0^1\!\!\int\limits_0^1\!\!\int\limits_0^1\frac1{1-xyzw}\,dw\,dz\,dy\,dx$

Regarding clarification of the question, Tom Apostol's Method of proving $\zeta(2)=\pi^2/6$ involves evaluating the double integral \begin{align*}\int_{0}^{1} \int_{0}^{1} \frac{1}{1-xy} \ dy \ dx,\end{align*} in two ways: converting the integrand into a geometric series and exchanging summation and integration to obtain $\zeta(2)$ and letting $x=u+v,y=u-v$ to get $\pi^2/6.$

We use Apostol's Double Integral Method to prove the result $\zeta(4)=\pi^4/90,$ using the quadruple integral \begin{align} \label{Apostol} \tag{1} \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{1}{1-xyzw} \ dw \ dz \ dy \ dx, \end{align} a similar geometric series argument shows \eqref{Apostol} is $\zeta(4).$ We now hope to evaluate \eqref{Apostol}.

Apostol's Method on $\zeta(4)$

We let $x=u+v,y=u-v, z=t/w.$ Then \eqref{Apostol} becomes \begin{align} \label{COV} \tag{2} \int_{\substack{0<u+v,u-v<1} } \int_{0}^{1} \int_{t}^{1} \frac{\frac{2}{w}}{1+(v^2-u^2)t} \ dw \ dt \ dv \ du. \end{align}

We split \eqref{COV} and consider the integral over the region $-u<v<u,0<u<1/2,$ that is,

\begin{align} \label{Part} \tag{3} \int_{0}^{\frac{1}{2}} \int_{-u}^{u} \int_{0}^{1} \int_{t}^{1} \frac{\frac{2}{w}}{1+(v^2-u^2)t} \ dw \ dt \ dv \ du= \int_{0}^{\frac{1}{2}} \int_{0}^{1} \int_{t}^{1} \int_{-u}^{u} \frac{\frac{2}{w}}{1+(v^2-u^2)t} \ dv \ dw \ dt \ du. \end{align} Integrating this, we find \eqref{Part} becomes \begin{align*} \int_{0}^{\frac{1}{2}} \int_{0}^{1} \int_{t}^{1} \frac{4\tan^{-1} \left(\frac{u\sqrt{t}}{\sqrt{1-tu^2}} \right)}{w\sqrt{t}\sqrt{1-tu^2}} \ dw \ dt \ du &=-\int_{0}^{\frac{1}{2}} \int_{0}^{1} \frac{4\log(t)\tan^{-1} \left(\frac{u\sqrt{t}}{\sqrt{1-tu^2}} \right)}{\sqrt{t}\sqrt{1-tu^2}} \ dt \ du \\ &=- \int_{0}^{1} \frac{2 \log(t) \left(\tan^{-1} \left(\sqrt{\frac{t}{4-t}}\right) \right)^2}{t} \ dt, \end{align*} after reversing the order of integration. Integrating by parts with $u=\left(\tan^{-1} \left(\sqrt{\frac{t}{4-t}}\right) \right)^2$ and $dv= \frac{-2 \log(t)}{t} \ dt$ and subsequently doing a change of a variables $t \mapsto x^2$ yields \eqref{Part} is equal to \begin{align} \label{Arctan} \tag{4} \int_{0}^{1} \frac{8\log^2(x)\tan^{-1} \left(\frac{x}{\sqrt{4-x^2}} \right)}{\sqrt{4-x^2}} \ dx. \end{align} It can be shown that \eqref{Arctan} upon expanding the integrand into a binomial series, is equal to \begin{align*} C(4)=\sum_{n=1}^{\infty} \frac{1}{n^4 \binom{2n}n}.\end{align*}

We wish to show $C(4)=17 \pi^4/3240.$ Expanding \eqref{Arctan} into the double integral \begin{align} \label{Double Integral} \tag{5} \int_{0}^{1} \int_{0}^{x} \frac{8\log^2(x)}{4-x^2+y^2} \ dy \ dx, \end{align} and applying the change of variables $x=u+v, y=u-v$ and $u \mapsto x, v \mapsto y,$ we get that \eqref{Double Integral} is \begin{align} \label{DI} \tag{6} \iint_{0<x-y<x+y<1} \frac{4\log^2(x+y)}{1-xy} \ dy \ dx. \end{align}

Integrating \eqref{DI} in the order presented, we split the region of integration into two triangular regions: $0<x<1/2, 0<y<x$ and $1/2<x<1,0<y<1-x$ and then use Mathematica and the integral additivity rule to get

\begin{align*}-\int_{0}^{1} \frac{4 \log^2(x)\log(x^2+1)}{x} \ dx + \int_{0}^{1} \frac{8 \log(x) Li_2 \left(\frac{x^2}{x^2+1} \right)}{x} \ dx - \int_{0}^{1} \frac{8 Li_3 \left(\frac{x^2}{x^2+1} \right)}{x} \ dx + \int_{\frac{1}{2}}^{1} \frac{8 Li_3 \left(\frac{x}{x^2+1} \right)}{x} \ dx + J, \end{align*} where $J$ is the sum of miscellaneous integrals. On the other hand, reversing the order of integration gives \eqref{DI} is \begin{align*}-J- \int_{0}^{\frac{1}{2}} \frac{8 Li_3 \left(\frac{y}{y^2+1} \right)}{y} \ dy \end{align*} Thus, equating these formulas, we see $C(4)$ is the sum of four integrals: \begin{align*} I_1 &=\int_{0}^{1} \frac{-2 \log^2(x)\log(x^2+1)}{x} \ dx \\ I_2 & = \int_{0}^{1} \frac{4 \log(x) Li_2 \left(\frac{x^2}{x^2+1} \right)}{x} \ dx \\ I_3 & = \int_{0}^{1} \frac{-4 Li_3 \left(\frac{x^2}{x^2+1} \right)}{x} \ dx \\ I_4 & =\int_{0}^{1} \frac{4 Li_3 \left(\frac{x}{x^2+1} \right)}{x} \ dx. \end{align*} Now we evaluate each integral.

To evaluate $I_1,$ we recall the series \begin{align*}\log(x^2+1)=\sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^{2n}}{n}, \quad |x|<1.\end{align*} Interchanging summation and integration, integrating term by term, and finally recalling the Eta function \begin{align*}\eta(4)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^4}=\frac{7}{8} \zeta(4),\end{align*} we see that \begin{align*}I_1=-\frac{7}{16} \zeta(4).\end{align*}

Using the substitution $u \mapsto \frac{x^2}{x^2+1}$, we get \begin{align} I_2 &=\int_{0}^{\frac{1}{2}} \frac{\log(\frac{u}{1-u}) Li_2 \left(u\right)}{u-u^2} \ du \nonumber \\ &= \int_{0}^{\frac{1}{2}} \frac{\log(1-u) \log^2\left( \frac{u}{1-u} \right)}{2u} \ du \label{I2IBP} \tag{7} \\ &= \int_{0}^{\frac{1}{2}} \frac{\log(1-u) \log^2(u)}{2u} \ du -\int_{0}^{\frac{1}{2}} \frac{\log(u) \log^2(1-u)}{u} \ du +\int_{0}^{\frac{1}{2}} \frac{ \log^3(1-u)}{u} \ du \label{splittedintegrals} \tag{8} \end{align} in which \eqref{I2IBP} follows from integrating by parts (with the function we differentiate being $Li_2(u)$) and \eqref{splittedintegrals} results from expanding the $log^2$ term. We will not evaluate the three integral in \eqref{splittedintegrals}; instead we will collect only the specific constants $Li_4(1), Li_4(-1),$ and constants that are rational multiples of $\pi^4$ The other constants are not necessary since they either will cancel out with one another or will cancel out with the constants resulting from the evaluation of $I_3.$ We recall
\begin{align*} Li_2(1) &= \zeta(2)= \frac{\pi^2}{6}, \\ Li_4(1) &=\zeta(4), \\ Li_4(-1) &=-\eta(4)=-\frac{7}{8}\zeta(4),\\ \left(Li_2 \left(\frac{1}{2} \right) \right)^2 &= \left(\frac{\zeta(2)}{2} - \frac{\log^2(2)}{2} \right)^2 = \frac{\pi^4}{144} - \frac{\pi^2}{12}\log^2(2) + \frac{\log^4(2)}{4}. \end{align*} Using Mathematica, the first integral in \eqref{splittedintegrals} does not consist of any constants we look for. The second integral, upon examining the antiderivative of the integrand yields the terms: \begin{align*}-2Li_4(1-u)+2Li_4(u)+2Li_4 \left( \frac{u}{-1+u} \right).\end{align*} Evaluating these terms at the endpoints yields the desired constant from the second integral in \eqref{splittedintegrals}, which is
$-\frac{7}{4}\zeta(4)+2\zeta(4)=\frac{1}{4}\zeta(4).$ The antiderivative of the integrand from the third integral in \eqref{splittedintegrals}, consisting of the term, \begin{align*}3 Li_4(1-u),\end{align*} yields the desired constant $-3\zeta(4)$ upon evaluation of the latter endpoint $u=0.$ Hence, for $I_2$ the desired constant is \begin{align*}-3\zeta(4)+\frac{1}{4}\zeta(4)=-\frac{11}{4}\zeta(4).\end{align*}

Using the substitution $u \mapsto \frac{x^2}{x^2+1}$, we get \begin{align} I_3 &= - \int_{0}^{\frac{1}{2}} \frac{2 Li_3(u)}{u-u^2} \ du \nonumber \\ &= - \int_{0}^{\frac{1}{2}} \frac{2Li_3(u)}{u} \ du - \int_{0}^{\frac{1}{2}} \frac{2Li_3(u)}{1-u} \ du \label{PF} \tag{9}, \end{align} in which \eqref{PF} is the result of using partial fractions. The first integral in \eqref{PF} does not consist of the aforementioned constants we seek. However, the integrand from the second integral in \eqref{PF} consists of the term \begin{align*} \left(Li_2(u) \right)^2,\end{align*} and upon evaluation of the endpoints, we get the constant $\pi^4/144.$

Lastly, for the fourth term, we apply $u \mapsto \frac{x}{x^2+1}$ to see \begin{align} I_4 & = \int_{0}^{\frac{1}{2}} \frac{4 Li_3(u)}{u\sqrt{1-4u^2}} \ du \nonumber \\ & = \sum_{n=1}^{\infty} \frac{\beta(n/2,n/2)}{n^3} \label{BS Conversion} \tag{10} \\ & = \sum_{n=1}^{\infty} \frac{\beta \left(\frac{2n}{2},\frac{2n}{2} \right)}{(2n)^3}+\sum_{n=1}^{\infty} \frac{\beta \left(\frac{2n-1}{2},\frac{2n-1}{2} \right)}{(2n-1)^3} \label{evenodd} \tag{11} \\ & = \frac{C(4)}{4} + \int_{0}^{1} \frac{\pi \log^2(u)}{\sqrt{4-u^2}} \ du \label{logsine} \tag{12} \\ & = \frac{C(4)}{4} + \frac{7 \pi^4}{216} \label{7pi^4/216} \tag{13}. \end{align} in which \eqref{BS Conversion} follows from recalling the definition of $Li_3(u)$, converting the integrand into a binomial series, and integrating term by term. Then the first term in \eqref{logsine} follows from simplifying the summand in the first term in \eqref{evenodd} with the properties of the beta function. The second term in \eqref{logsine} follows from converting the integrand into a binomial series, integrating term by term, and recognizing the expression corresponding to $\beta \left(\frac{2n-1}{2},\frac{2n-1}{2} \right).$ Lastly, the second term in \eqref{7pi^4/216} is the value of the integral in \eqref{logsine}, which we will not prove here (there are proofs available on StackExchange).

We already established $C(4)=I_1+I_2+I_3+I_4$ and asserted that only $\zeta(4)$ constants and constants that are rational multiples of $\pi^4$ remain from summing $I_2$ and $I_3.$ Hence, Rearranging terms yields the relation \begin{align} \label{sys1} \tag{14} \frac{3}{4} C(4)=-\frac{51}{16} \zeta(4) + \frac{17 \pi^4}{432} \end{align}

Now we obtain another system of equations relating $C(4)$ and $\zeta(4).$ This time we will reconsider $I_3$ and analyze the integral another way. Recall \begin{align*}I_3 = \int_{0}^{1} \frac{-4 Li_3 \left(\frac{x^2}{x^2+1} \right)}{x} \ dx.\end{align*} Integration by parts (with $dv=-4/x,$) yields

\begin{align} I_3 &= \int_{0}^{1} \frac{8 \log(x) Li_2 \left( \frac{x^2}{x^2+1} \right)}{x^3+x} \ dx \nonumber \\ & = \int_{0}^{1} \frac{8 \log(x) Li_2 \left( \frac{x^2}{x^2+1} \right)}{x} \ dx - \int_{0}^{1} \frac{8x \log(x) Li_2 \left( \frac{x^2}{x^2+1} \right)}{x^2+1} \ dx \label{PF2} \tag{15}\\ & = 2I_2- \int_{0}^{1} \frac{8x \log(x) Li_2 \left( \frac{x^2}{x^2+1} \right)}{x^2+1} \ dx \label{simplified} \tag{16}, \end{align}
in which \eqref{PF2} follows from partial fractions and \eqref{simplified} follows from the definition of $I_2.$ Now for the second term in \eqref{simplified}, we make the substitution $u \mapsto \frac{x}{x^2+1},$ to see it becomes \begin{align} \label{2nd term} \tag{17} - \int_{0}^{1} \frac{8x \log(x) Li_2 \left( \frac{x^2}{x^2+1} \right)}{x^2+1} \ dx & = - \int_{0}^{\frac{1}{2}} \frac{2 \log \left(\frac{u}{1-u} \right) Li_2(u)}{1-u} \ du \end{align} Integrating by parts on the right hand side of \eqref{2nd term} (with the differentiating function being $Li_2(u)$) yields \eqref{2nd term} is equal to \begin{multline} \label{split2} \tag{18} -\frac{7 \pi^4}{72} + \frac{7}{12} \pi^2 \log^2(2)+ \int_{0}^{\frac{1}{2}} \frac{\log(1-u) \log^2(u-1)}{u} \ du -\int_{0}^{\frac{1}{2}} \frac{2\log(u-1) \log^2(1-u)}{u} \ du \\ -\int_{0}^{\frac{1}{2}} \frac{2\log(1-u) Li_2(1-u)}{u} \ du \end{multline} Repeating the analysis of process of seeking the desired constants as done with \eqref{splittedintegrals}, we find the real part of the first integral in \eqref{split2} has a $\frac{3}{2} \zeta(4)$ constant, the second integral has a $12\zeta(4)$ constant, while the third has a $\frac{11}{8} \zeta(4)$ constant. Thus, \eqref{split2} has the desired constant \begin{align*}\frac{119}{8} \zeta(4)-\frac{7 \pi^4}{72},\end{align*} and from \eqref{simplified}, we see $I_3$ has the desired constant \begin{align*}\frac{119}{8} \zeta(4)- \frac{11}{2} \zeta(4)-\frac{7 \pi^4}{72}= \frac{75}{8} \zeta(4) -\frac{7 \pi^4}{72}. \end{align*} Hence, combining $I_1,\dots, I_4,$ we see that another relation arises: \begin{align} \label{sys2} \tag{19} \frac{3}{4}C(4)= \frac{99}{16} \zeta(4) - \frac{7 \pi^4}{108}. \end{align} Hence we have the system of equations \begin{align} \label{sys} \tag{20} \frac{3}{4} C(4)=-\frac{51}{16} \zeta(4) + \frac{17 \pi^4}{432}, \quad \frac{3}{4}C(4)= \frac{99}{16} \zeta(4) - \frac{7 \pi^4}{108}, \end{align} and solving this system yields $C(4)=17\pi^4/3240$ and $\zeta(4)=\pi^4/90.$

Remark To my knowledge, this method does not generalize any further to $\zeta(2n)$ for $n\geq 3.$ This is because the general version of the central binomial coefficient is not a rational multiple of $\pi^{2n}.$ The subtle point behind Apostol's method for $\zeta(2)$ is being able to evaluate the sum \begin{align*} \sum_{n=1}^{\infty} \frac{1}{\binom{2n}n n^2}= \frac{\pi^2}{18}, \end{align*} which is the result of evaluating $$\int_{0}^{1} \int_{0}^{1} \frac{1}{1-xy} \ dy \ dx$$ over the triangle $0<u<1/2, -u<v<u$ when making the change of variables $x=u+v, y=u-v.$

Note

$$\frac{1}{1-wxyz} = 1 + wxyz + (wxyz)^2+(wxyz)^3+\cdots$$

when $|wxyz|<1$, which is the case here. Integrating the series term by term gives the answer.