How do we prove that $$\int_1^\infty\int_0^\infty\dfrac{1}{(x^3+y^3)^3}\mathrm{d}x\ \mathrm{d}y=\dfrac{10\pi}{189\sqrt3}$$
I tried to expand and use partial fraction, but in vain. I don't have a clue what to do now. Please help me out. Thank you.
Please avoid using complex analysis, as I am not familiar with it.
First, $$\int_0^\infty \frac{dx}{(x^3+y^3)^3} = \frac{1}{y^9}\int_0^\infty \frac{dx}{((x/y)^3+1)^3}= \frac{1}{y^8}\int_0^\infty \frac{dt}{(t^3+1)^3}= \frac{1}{y^8}\int_0^\infty \frac{\frac13u^{-2/3}du}{(u+1)^3}=$$ $$= \frac{1}{3y^8}\mathrm{B}\left(\frac13,3-\frac13\right)= \frac{1}{3y^8}\frac{5}{3}\frac{2}{3}\frac{\pi}{\sin\frac{\pi}{3}}=\frac{10\pi}{27\sqrt3}\frac{1}{y^8}.$$ Since $$\int_1^\infty\frac{dy}{y^8}=\frac17$$ we get the desired result $$\frac{10\pi}{189\sqrt3}.$$
