TITLE: Evaluate $\int_0^\infty\!\!\int_0^\infty\!\!\int_0^\infty\!\frac{(xyz)^{-1/7}(yz)^{-1/7}z^{-1/7}}{(x+1)(y+1)(z+1)}dx\,dy\,dz$
QUESTION [17 upvotes]: 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.

REPLY [2 votes]: The integral can be split into three parts which form 
$$
\int_0^\infty\dfrac{x^{\large -c}}{1+x}\ dx.
$$
Let's generalize the problem. We will evaluate
$$
\int_0^\infty\dfrac{x^{\large a-1}}{1+x^b}\ dx.
$$
Let $$y=\dfrac{1}{1+x^b}\quad\Rightarrow\quad x=\left(\dfrac{1-y}{y}\right)^{\large\frac1b}\quad\Rightarrow\quad dx=-\left(\dfrac{1-y}{y}\right)^{\large\frac1b-1}\ \dfrac{dy}{by^2}\ ,$$ then
\begin{align}
\int_0^\infty\dfrac{x^{\large a-1}}{1+x^b}\ dx&=\int_0^1 y\left(\dfrac{1-y}{y}\right)^{\large\frac{a-1}b}\left(\dfrac{1-y}{y}\right)^{\large\frac1b-1}\ \dfrac{dy}{by^2}\\&=\frac1b\int_0^1y^{\large1-\frac{a}{b}-1}(1-y)^{\large\frac{a}{b}-1}\ dy,
\end{align}
where the last integral in RHS is a Beta function.
$$
\text{B}(x,y)=\int_0^1t^{\ \large x-1}\ (1-t)^{\ \large y-1}\ dt=\frac{\Gamma(x)\cdot\Gamma(y)}{\Gamma(x+y)}.
$$
Hence
\begin{align}
\int_0^\infty\dfrac{x^{\large a-1}}{1+x^b}\ dx&=\frac1b\int_0^1y^{\large1-\frac{a}{b}-1}(1-y)^{\large\frac{a}{b}-1}\ dy\\&=\frac1b\cdot\Gamma\left(1-\frac{a}{b}\right)\cdot\Gamma\left(\frac{a}{b}\right)\\&=\large{\color{blue}{\frac{\pi}{b\sin\left(\frac{a\pi}{b}\right)}}}.
\end{align}
The last part uses Euler's reflection formula for Gamma function provided $\color{red}{0<a<b}$. Now, the given integral can easily be solved.