<s>[INST] Evaluate $\int_0^\infty\!\!\int_0^\infty\!\!\int_0^\infty\!\frac{(xyz)^{-1/7}(yz)^{-1/7}z^{-1/7}}{(x+1)(y+1)(z+1)}dx\,dy\,dz$ I am looking for guidance evaluating the following integral. $$\int_0^\infty\!\!\!\int_0^\infty\!\!\int_0^\infty\!\!\dfrac{(xyz)^{-1/7}(yz)^{-1/7}z^{-1/7}}{(x+1)(y+1)(z+1)}\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z$$
Its value can be found easily with a run through Mathematica (I will not post the answer) but there is apparently a common method to computing integrals of this type.
This problem be reduced to evaluating integrals of the type $$\int_0^\infty \dfrac{x^{-a}}{x+1}\mathrm{d}x.$$
I imagine there is some sort of differentiation under the integral sign that can be used but I am at loss. 
 [/INST] $$\int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{\infty}\frac{(xyz)^{-1/7}(yz)^{-1/7}(z)^{-1/7}}{(x+1)(y+1)(z+1)}dxdydz$$
$$
I = \int\limits_0^{ + \infty } {\int\limits_0^{ + \infty } {\int\limits_0^{ + \infty } {\frac{{\left( {xyz} \right)^{ - \frac{1}
{7}} \left( {yz} \right)^{ - \frac{1}
{7}} z^{ - \frac{1}
{7}} }}
{{\left( {x + 1} \right)\left( {y + 1} \right)\left( {z + 1} \right)}}dxdydz} } }  = \left( {\int\limits_0^{ + \infty } {\frac{{x^{1 - \frac{1}
{7} - 1} }}
{{\left( {x + 1} \right)^{1 - \frac{1}
{7} + \frac{1}
{7}} }}dx} } \right)\left( {\int\limits_0^{ + \infty } {\frac{{y^{1 - \frac{2}
{7} - 1} }}
{{\left( {y + 1} \right)^{1 - \frac{2}
{7} + \frac{2}
{7}} }}dy} } \right)\left( {\int\limits_0^{ + \infty } {\frac{{z^{1 - \frac{3}
{7} - 1} }}
{{\left( {z + 1} \right)^{1 - \frac{3}
{7} + \frac{3}
{7}} }}dz} } \right)
$$
now use
$$
{\rm B}\left( {x,y} \right) = \int\limits_0^{ + \infty } {\frac{{t^{x - 1} }}
{{\left( {1 + t} \right)^{x + y} }}dt} 
$$
then
$$
 = {\rm B}\left( {1 - \frac{1}
{7},\frac{1}
{7}} \right){\rm B}\left( {1 - \frac{2}
{7},\frac{2}
{7}} \right){\rm B}\left( {1 - \frac{3}
{7},\frac{3}
{7}} \right) = {\rm B}\left( {\frac{6}
{7},\frac{1}
{7}} \right){\rm B}\left( {\frac{5}
{7},\frac{2}
{7}} \right){\rm B}\left( {\frac{4}
{7},\frac{3}
{7}} \right)
$$
$$
 = \frac{{\Gamma \left( {\frac{6}
{7}} \right)\Gamma \left( {\frac{1}
{7}} \right)}}
{{\Gamma \left( {\frac{6}
{7} + \frac{1}
{7}} \right)}} \cdot \frac{{\Gamma \left( {\frac{5}
{7}} \right)\Gamma \left( {\frac{2}
{7}} \right)}}
{{\Gamma \left( {\frac{5}
{7} + \frac{2}
{7}} \right)}} \cdot \frac{{\Gamma \left( {\frac{4}
{7}} \right)\Gamma \left( {\frac{3}
{7}} \right)}}
{{\Gamma \left( {\frac{4}
{7} + \frac{3}
{7}} \right)}} = \Gamma \left( {\frac{6}
{7}} \right)\Gamma \left( {\frac{1}
{7}} \right) \cdot \Gamma \left( {\frac{5}
{7}} \right)\Gamma \left( {\frac{2}
{7}} \right) \cdot \Gamma \left( {\frac{4}
{7}} \right)\Gamma \left( {\frac{3}
{7}} \right)
$$
$$
 = \Gamma \left( {1 - \frac{1}
{7}} \right)\Gamma \left( {\frac{1}
{7}} \right) \cdot \Gamma \left( {1 - \frac{2}
{7}} \right)\Gamma \left( {\frac{2}
{7}} \right) \cdot \Gamma \left( {1 - \frac{3}
{7}} \right)\Gamma \left( {\frac{3}
{7}} \right)
$$
$$
 = \frac{\pi }
{{\sin \left( {\frac{\pi }
{7}} \right)}} \cdot \frac{\pi }
{{\sin \left( {\frac{{2\pi }}
{7}} \right)}} \cdot \frac{\pi }
{{\sin \left( {\frac{{3\pi }}
{7}} \right)}} = \frac{{8\sqrt 7 }}
{7}\pi ^3 $$
</s>