Unpredictable!

Some results

$\displaystyle \begin{array}{c}\\ I_1 & \int\limits_0^1 \ln^2 \left(\frac{1-x}{1+x}\right) \; dx = \frac{\pi^2}{3} \\ I_2= & \int\limits_0^\infty \frac{\sin x \cos ax}{x}\; dx =\frac{\pi}{2} \\ I_3 = & \int\limits_0^\infty \frac{\sin x \sin ax}{x}\; dx = \frac{1}{2}\ln \left(\frac{1-a}{1+a}\right)\end{array}\\$

Now consider , $\displaystyle F(a)=\int\limits_0^\infty \int\limits_0^\infty \int\limits_0^\infty \dfrac{\sin x\sin y\sin z\sin(a(x+y+z))}{xyz(x+y+z)}\; dx\;dy\; dz$

By differentiating once we have,

$\displaystyle F'(a) = \int\limits_0^\infty \int\limits_0^\infty \int\limits_0^\infty \dfrac{\sin x\sin y\sin z\cos(ax+ay+az)}{xyz(x+y+z)}\; dx\;dy\; dz$

Now expanding $\cos(ax+ay+az)$ and then expressing the integral as a sum integral we find that that there are no homogenous terms and the integrals are in variable separable form like,

$\displaystyle F'(a) = \left( \int_0^\infty \frac{\sin x \cos ax}{x}\; dx\right)\left( \int_0^\infty \frac{\sin y \cos ay}{y}\; dy\right)\left( \int_0^\infty \frac{\sin z \cos az}{z}\; dz\right) - \left( \int_0^\infty \frac{\sin x \cos ax}{x}\; dx\right)\left( \int_0^\infty \frac{\sin y \sin ay}{y}\; dy\right)\left( \int_0^\infty \frac{\sin z \cos az}{z}\; dz\right)- \left( \int_0^\infty \frac{\sin x \sin ax}{x}\; dx\right)\left( \int_0^\infty \frac{\sin y \cos ay}{y}\; dy\right)\left( \int_0^\infty \frac{\sin z \sin az}{z}\; dz\right)- \left( \int_0^\infty \frac{\sin x \sin ax}{x}\; dx\right)\left( \int_0^\infty \frac{\sin y \sin ay}{y}\; dy\right)\left( \int_0^\infty \frac{\sin z \cos az}{z}\; dz\right)$

Now the integrals in brackets are actually independent integrals so this thing can be rewritten as,

$\displaystyle F'(a)= (I_2)^3-3(I_2)(I_3)^2 = \frac{\pi^3}{8}-3\frac{\pi}{8}\left(\ln^2 \left(\frac{1-a}{1+a}\right)\right)$

Since $F(0)=0$ and we want $F(1)$ so,

$\displaystyle F(1)= \int_0^1 \frac{\pi^3}{8}\; da -3\frac{\pi}{8}I_1 = \frac{\pi^3}{8}-\frac{\pi^3}{8}=0$

Interested users having issues with the proofs of first three results may ask me for proofs.

If we define $\begin{aligned} F(\alpha,\beta,\gamma,\delta) & = \; \int_0^\infty \int_0^\infty \int_0^\infty \frac{\sin x \sin y \sin z \,e^{i(x+y+z)}e^{-\alpha x}e^{-\beta y}e^{-\gamma z}e^{-\delta(x+y+z)}}{x y z (x+y+z)}\,dx\,dy\,dz \\ & = \int_0^\infty \int_0^\infty \int_0^\infty \frac{\sin x \sin y \sin z \,e^{-(\alpha+\delta-i)x}e^{-(\beta+\delta-i)y}e^{-(\gamma+\delta-i)z}}{x y z (x+y+z)}\,dx\,dy\,dz \end{aligned}$ then $\begin{aligned} \frac{\partial^4F}{\partial\alpha\partial\beta\partial\gamma\partial\delta} & = \int_0^\infty \int_0^\infty \int_0^\infty \sin x \sin y \sin z \,e^{-(\alpha+\delta-i)x}e^{-(\beta+\delta-i)y}e^{-(\gamma+\delta-i)z}\,dx\,dy\,dz \\ & = \frac{1}{\big[(\alpha+\delta- i)^2+1\big]\big[(\beta+\delta-i)^2 + 1\big]\big[(\gamma+\delta-i)^2+1\big]} \\ & = \; \frac{1}{(\alpha+\delta)(\alpha+\delta-2i)(\beta+\delta)(\beta+\delta-2i)(\gamma+\delta)(\gamma+\delta-2i)} \end{aligned}$ and so $\frac{\partial F}{\partial \delta} \; = \; -g(\alpha+\delta)g(\beta+\delta)g(\gamma+\delta)$ where $g(u) \; = \; \int_u^\infty \frac{dv}{v(v-2i)} \; = \; \tfrac12i\ln\big(1 - \tfrac{2i}{u}\big) \hspace{1cm} u > 0$ Thus $F(\alpha,\beta,\gamma,\delta) \; = \; -\tfrac18i\int_\delta^\infty \ln\big(1 - \tfrac{2i}{\alpha+u}\big)\,\ln\big(1 - \tfrac{2i}{\beta+u}\big)\,\ln\big(1 - \tfrac{2i}{\gamma+u}\big)\,du$ and hence $F(0,0,0,\delta) \; = -\tfrac18i\int_\delta^\infty \left[\ln\big(1 - \tfrac{2i}{u}\big)\right]^3\,du \; = \; -\tfrac14i\int_0^{\frac{2}{\delta}} \frac{\ln^3(1 - vi)}{v^2}\,dv$ and so $F(0,0,0,0) \; = \; -\tfrac14i \int_0^\infty \frac{\ln^3(1 - vi)}{v^2}\,dv$ A simple piece of contour integration shows that $F(0,0,0,0) \; = \; -\tfrac14i\int_0^\infty \frac{\ln^3(1 - vi)}{v^2}\,dv \; = \; -\tfrac14\int_0^\infty \frac{\ln^3(1+x)}{x^2}\,dx$ and hence we deduce that $\int_0^\infty \int_0^\infty \int_0^\infty \frac{\sin x \sin y \sin z \sin(x+y+z)}{x y z (x+y+z)}\,dx\,dy\,dz \; = \; \boxed{0}$ We have also shown that $\int_0^\infty \int_0^\infty \int_0^\infty \frac{\sin x \sin y \sin z \cos(x+y+z)}{x y z (x+y+z)}\,dx\,dy\,dz \; = \; -\tfrac14\int_0^\infty \frac{\ln^3(1+x)}{x^2}\,dx \; = \; -\tfrac32\zeta(3)$ for free.