Beautiful or Difficult?

first note $\sin(x+y) = \sin(x)\cos(y)+\cos(x)\sin(y)$ .by symmetry of x,y, we would have the integral be $\int_0^\infty \int_0^\infty \dfrac{\sin(x)\sin(y) \sin(x+y)}{xy(x+y)} dx dy =2 \int_0^\infty \int_0^\infty \dfrac{\sin^2(x)\sin(y)\cos(y) }{xy(x+y)} dx dy = \int_0^\infty \int_0^\infty \dfrac{\sin^2(x)\sin(2y) }{xy(x+y)} dx dy$ now, we make the double integral a triple integral: $\int_0^\infty \int_0^\infty \int_0^\infty \dfrac{\sin^2(x)\sin(2y) }{xy}e^{-t(x+y)} dt dx dy= \int_0^\infty \left( \int_0^\infty \dfrac{\sin^2 (x)}{x} e^{-tx} dx \right)\left( \int_0^\infty \dfrac{\sin (2y)}{y} e^{-ty} dy \right) dt$ we can solve each by differentiating under the integral sign(shown at the end of solution). this integral becomes so we are left evaluating $\dfrac{1}{4} \int_0^\infty \ln\left(1+\dfrac{4}{t^2} \right)\arctan\left(\frac{2}{t}\right) dt=\int_0^{\pi/2} x \ln(\sec(x)) \csc^2(x) dx$ since we had $1+\left(\dfrac{2}{t}\right)^2$ inside the log, and $\frac{2}{t}$ inside the arctan, a tangent substitution was motivated, i.e $\frac{2}{t} = \tan(x)$ . using integration by parts this turns into $\int_0^{\pi/2} \left[x \ln(\sec(x))\right] \left(\csc^2(x) dx\right) = -\left[x \ln(\sec(x))\right] \cot(x)\big|_0^{\pi/2}+\int_0^{\pi/2} \left[ \ln(\sec(x))+x\tan(x)\right]\cot(x) dx\\ = \int_0^{\pi/2} x dx+\int_0^{\pi/2} \ln(\sec(x)) \cot(x) dx = \dfrac{\pi^2}{8} +\int_0^{\pi/2} \dfrac{\ln(\sec(x))}{\sec^2(x)-1} \tan(x) dx$ making the obvious substitution in the last integral: $\dfrac{\pi^2}{8} +\int_0^{\infty} \dfrac{x}{e^{2x}-1} dx = \dfrac{\pi^2}{8} +\dfrac{1}{4}\int_0^{\infty} \dfrac{x}{e^{x}-1} dx = \dfrac{\pi^2}{8}+\dfrac{1}{4}\dfrac{\pi^2}{6} = \boxed{\dfrac{\pi^2}{6}}$ note in the last integral the integral defination of the zeta function was used.

the two integrals were $I_1 (t) = \int_0^\infty \dfrac{\sin^2 (x)}{x} e^{-tx} dx \to I_1 '(t) = -\int_0^\infty \sin^2 (x) e^{-tx} dx =-\dfrac{1}{2}\int_0^\infty (1-\cos(2x))e^{-tx} dx = -\dfrac{1}{2} \left(\dfrac{1}{t} -\dfrac{t}{t^2+4} \right)$ integrating this w.r.t to t and using the fact that $I_1(\infty) = 0$ , we have $I_1(t) = -\dfrac{1}{2} \left(\ln(t)-\dfrac{1}{2} \ln(t^2+4)\right) = \dfrac{1}{4}\ln\left(1+\dfrac{4}{t^2} \right)$ similarly: $I_2(t) = \int_0^\infty \dfrac{\sin (2y)}{y} e^{-ty} dy \to I_2'(t) = -\int_0^\infty \sin (2y) e^{-ty} dy = -\dfrac{2}{t^2+4}$ using $I_2(0) = \dfrac{\pi}{2}$ , we have $I_2(t) = \dfrac{\pi}{2} -\arctan\left(\frac{t}{2}\right) = \arctan\left(\frac{2}{t}\right)$