Generalization of Dirichlet integral

A lengthy method, but it works. First, we rewrite:

$$\sin(x+y) \sin(y+z) \sin(z+x)=\frac{1}{4} \left(\sin(2x)+\sin(2y)+\sin(2z)-\sin(2(x+y+z)) \right)$$

Now let's change the variables (keeping the same letters for simplicity):

$$x \to \frac{1}{2}x^2, \qquad y \to \frac{1}{2}y^2, \qquad z \to \frac{1}{2}z^2$$

$$I= 2 \int_0^\infty \int_0^\infty \int_0^\infty \frac{x y z dx dy dz}{(x^2+y^2)(y^2+z^2)(z^2+x^2)} \left(\sin(x^2)+\sin(y^2)+\sin(z^2)-\sin(x^2+y^2+z^2) \right)$$

Now we introduce spherical coordinates:

$$x= r \sin \theta \cos \phi, \qquad y = r \sin \theta \sin \phi, \qquad z= r \cos \theta$$

The limits are obvious:

$$I= 2 \int_0^\infty \int_0^{\pi/2} \int_0^{\pi/2} \frac{r^5 \sin^3 \theta \cos \theta \sin \phi \cos \phi dr d \theta d \phi}{r^6 \sin^2 \theta (\sin^2 \theta \sin^2 \phi+\cos^2 \theta)(\sin^2 \theta \cos^2 \phi+\cos^2 \theta)} \times \\ \times \left(\sin(r^2 \sin^2 \theta \cos^2 \phi)+\sin(r^2 \sin^2 \theta \sin^2 \phi)+\sin(r^2 \cos^2 \theta)-\sin(r^2) \right)$$

Simplifying and using:

$$\int_0^\infty \frac{\sin(a^2 r^2) dr}{r}=\frac{\pi}{4}$$

We obtain:

$$I= \pi \int_0^{\pi/2} \int_0^{\pi/2} \frac{\sin \theta \cos \theta \sin \phi \cos \phi d \theta d \phi}{(\sin^2 \theta \sin^2 \phi+\cos^2 \theta)(\sin^2 \theta \cos^2 \phi+\cos^2 \theta)}=\pi \int_0^{\pi/2} \int_0^{\pi/2} \frac{\sin 2 \theta \sin 2 \phi d \theta d \phi}{4 \cos^2 \theta+\sin^2 2 \phi \sin^4 \theta}$$

Changing the variables:

$$2 \theta \to \theta, \qquad 2 \phi \to \phi$$

$$I= \pi \int_0^{\pi} \int_0^{\pi} \frac{\sin \theta \sin \phi d \theta d \phi}{8 (1+\cos \theta)+\sin^2 \phi (1-\cos \theta)^2}$$

Substituting:

$$u= \cos \theta, \qquad v = \cos \phi$$

$$I= \pi \int_{-1}^1 \int_{-1}^1 \frac{d u d v}{8 (1+u)+(1-v^2) (1-u)^2}$$

Substituting:

$$1-u= t$$

$$I= \pi \int_0^2 \int_{-1}^1 \frac{d v d t}{(t-4)^2-t^2 v^2}=\pi \int_0^2 \int_{-1}^1 \frac{d v d t}{(t-4-tv)(t-4+tv)}$$

Taking the elementary integral w.r.t. $v$, we have:

$$I= \pi \int_0^2 \frac{dt}{t(t-4)} \ln \left(1-\frac{t}{2} \right)$$

Substituting:

$$t= 2s$$

$$I= \frac{\pi}{2} \int_0^1 \frac{ \ln (1-s) ds}{s(s-2)}=\frac{\pi}{4} \left( \int_0^1 \frac{ \ln (1-s) ds}{s-2} -\int_0^1 \frac{ \ln (1-s) ds}{s}\right)$$