Hot Integral - 25

$\displaystyle \int\limits_{0}^{\infty} \int\limits_{0}^{\infty} \int\limits_{0}^{\infty} \int\limits_{0}^{\infty} \frac{1}{wxyz(w+x+y+z+\frac1w + \frac1x + \frac1y + \frac1z)^2} dw dx dy dz$

First , we look at this 4-dimensional integral and starts thinking solving it internally one by one. But unfortunately, due to its complicated form we can't proceed further at one point.

Have a look on its limits. They are independent of $w,x,y,z$ . Make substitution $w=r , x=rx_0 , y=rx_1 , z=rx_2$

Solve, and one should get $dw dx dy dz = r^3 dr dx_0 dx_1 dx_2$

Hence , finally we get

$\displaystyle \int\limits_{0}^{\infty} \int\limits_{0}^{\infty} \int\limits_{0}^{\infty} \int\limits_{0}^{\infty} \frac{dr dx_0 dx_1 dx_2}{rx_0 x_1 x_2(r(1+x_0+x_1+x_2)+\frac1r(1+\frac{1}{x_0}+\frac{1}{x_1}+\frac{1}{x_2}))^2}$

Solve internal integrand, we will get easily, $\displaystyle \int\limits_{0}^{\infty} \int\limits_{0}^{\infty} \int\limits_{0}^{\infty} \frac{dx_0 dx_1 dx_2}{2x_0 x_1 x_2(1+x_0+x_1+x_2)(1+\frac{1}{x_0}+\frac{1}{x_1}+\frac{1}{x_2})}$

For solving the above integral we have used , this result

$\displaystyle \int\limits_{0}^{\infty} \frac{dr}{r(ar+\frac{b}{r})^2}=\frac{1}{2ab}$ Here , $a=1+x_0+x_1+x_2 , b=1+\frac{1}{x_0}+\frac{1}{x_1}+\frac{1}{x_2}$

Solve it again for the internal integral,

$\displaystyle =\frac12 \int\limits_{0}^{\infty} \int\limits_{0}^{\infty} \frac{\log((1+x_1+x_2)(1+\frac{1}{x_1}+\frac{1}{x_2}))}{(1+x_1+x_2)(1+\frac{1}{x_1}+\frac{1}{x_2})-1} \frac{dx_1}{x_1} \frac{dx_2}{x_2}$

Split the log term and in the second part,use transformation $\frac{1}{x_1} \to x_1 , \frac{1}{x_2} \to x_2$

and you will get

$\displaystyle \int\limits_{0}^{\infty} \int\limits_{0}^{\infty} \frac{\log(1+x_1+x_2)}{(1+x_1+x_2)(1+\frac{1}{x_1}+\frac{1}{x_2})-1} \frac{dx_1}{x_1} \frac{dx_2}{x_2}$

Again solve it internally , we get

$\displaystyle \int\limits_{0}^{\infty} \frac{\text{Li}_2(\frac{1}{x_2})-\text{Li}_2(x_2)}{x_2^2-1} dx_2$

Split this integral at $x_2=1$ , and use substitution $x_2 \to 1/x_2$ in the second integral , we get,

$\displaystyle I = 2\int\limits_{0}^{1} \frac{\text{Li}_2(\frac{1}{x_2})-\text{Li}_2(x_2)}{x_2^2-1} dx_2$ Integrate it using by parts, with first function as $\text{Li}_2(\frac{1}{x_2})-\text{Li}_2(x_2)$

On Further simplification we are only left with four integrals on the right hand side.

$\displaystyle I = 2\int\limits_{0}^{1}\frac{\log^2(t+1)}{t}dt -2\int\limits_{0}^{1}\frac{\log(1+t)\log(1-t)}{t} dt - \int\limits_{0}^{1}\frac{\log t \log(1+t)}{t} dt + \int\limits_{0}^{1}\frac{\log t \log(1-t)}{t} dt$

$I = \frac12\zeta(3)+\frac54\zeta(3)+\frac34\zeta(3)+\zeta(3)$

$\boxed{I=\frac72\zeta(3)}$