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

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.

Answer 1

$$\int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{\infty}\frac{(xyz)^{-1/7}(yz)^{-1/7}(z)^{-1/7}}{(x+1)(y+1)(z+1)}dxdydz$$ $$ I = \int\limits_0^{ + \infty } {\int\limits_0^{ + \infty } {\int\limits_0^{ + \infty } {\frac{{\left( {xyz} \right)^{ - \frac{1} {7}} \left( {yz} \right)^{ - \frac{1} {7}} z^{ - \frac{1} {7}} }} {{\left( {x + 1} \right)\left( {y + 1} \right)\left( {z + 1} \right)}}dxdydz} } } = \left( {\int\limits_0^{ + \infty } {\frac{{x^{1 - \frac{1} {7} - 1} }} {{\left( {x + 1} \right)^{1 - \frac{1} {7} + \frac{1} {7}} }}dx} } \right)\left( {\int\limits_0^{ + \infty } {\frac{{y^{1 - \frac{2} {7} - 1} }} {{\left( {y + 1} \right)^{1 - \frac{2} {7} + \frac{2} {7}} }}dy} } \right)\left( {\int\limits_0^{ + \infty } {\frac{{z^{1 - \frac{3} {7} - 1} }} {{\left( {z + 1} \right)^{1 - \frac{3} {7} + \frac{3} {7}} }}dz} } \right) $$ now use

$$ {\rm B}\left( {x,y} \right) = \int\limits_0^{ + \infty } {\frac{{t^{x - 1} }} {{\left( {1 + t} \right)^{x + y} }}dt} $$ then

$$ = {\rm B}\left( {1 - \frac{1} {7},\frac{1} {7}} \right){\rm B}\left( {1 - \frac{2} {7},\frac{2} {7}} \right){\rm B}\left( {1 - \frac{3} {7},\frac{3} {7}} \right) = {\rm B}\left( {\frac{6} {7},\frac{1} {7}} \right){\rm B}\left( {\frac{5} {7},\frac{2} {7}} \right){\rm B}\left( {\frac{4} {7},\frac{3} {7}} \right) $$

$$ = \frac{{\Gamma \left( {\frac{6} {7}} \right)\Gamma \left( {\frac{1} {7}} \right)}} {{\Gamma \left( {\frac{6} {7} + \frac{1} {7}} \right)}} \cdot \frac{{\Gamma \left( {\frac{5} {7}} \right)\Gamma \left( {\frac{2} {7}} \right)}} {{\Gamma \left( {\frac{5} {7} + \frac{2} {7}} \right)}} \cdot \frac{{\Gamma \left( {\frac{4} {7}} \right)\Gamma \left( {\frac{3} {7}} \right)}} {{\Gamma \left( {\frac{4} {7} + \frac{3} {7}} \right)}} = \Gamma \left( {\frac{6} {7}} \right)\Gamma \left( {\frac{1} {7}} \right) \cdot \Gamma \left( {\frac{5} {7}} \right)\Gamma \left( {\frac{2} {7}} \right) \cdot \Gamma \left( {\frac{4} {7}} \right)\Gamma \left( {\frac{3} {7}} \right) $$

$$ = \Gamma \left( {1 - \frac{1} {7}} \right)\Gamma \left( {\frac{1} {7}} \right) \cdot \Gamma \left( {1 - \frac{2} {7}} \right)\Gamma \left( {\frac{2} {7}} \right) \cdot \Gamma \left( {1 - \frac{3} {7}} \right)\Gamma \left( {\frac{3} {7}} \right) $$

$$ = \frac{\pi } {{\sin \left( {\frac{\pi } {7}} \right)}} \cdot \frac{\pi } {{\sin \left( {\frac{{2\pi }} {7}} \right)}} \cdot \frac{\pi } {{\sin \left( {\frac{{3\pi }} {7}} \right)}} = \frac{{8\sqrt 7 }} {7}\pi ^3 $$

Answer 2

Write the integral as: $$\int_0^{\infty} \int_0^{\infty} x^{-a} e^{-(x+1)t}\,dt\,dx=\int_0^{\infty} \left(\int_0^{\infty} x^{-a}e^{-xt}\,dx\right) e^{-t}\,dt$$ First I evaluate: $$\int_0^{\infty} x^{-a}e^{-xt}\,dx$$ Use the substitution $xt=y$ to obtain: $$\frac{1}{t^{-a+1}} \int_0^{\infty} y^{-a} e^{-y}\,dy=\frac{1}{t^{1-a}}\Gamma(1-a)$$ Hence, $$\int_0^{\infty} \left(\int_0^{\infty} x^{-a}e^{-xt}\,dx\right) e^{-t}\,dt=\Gamma(1-a)\int_0^{\infty} t^{a-1}e^{-t}\,dt=\Gamma(1-a)\Gamma(a)$$ ....which is by Euler's reflection formula: $$\Gamma(1-a)\Gamma(a)=\frac{\pi}{\sin(\pi a)}$$

Answer 3

Using residues you can show that $$\int_0^\infty \frac{x^{-a}}{1+x}\,dx=\frac{\pi}{\sin a\pi}$$ if $a$ is real and $0<a<1$. Details omitted ;-)

The integrals can be done in principle by real methods because the exponents in the numerator are all rational. We have for example $$\int_0^\infty \frac{x^{-1/7}}{1+x}\,dx=\int_0^\infty \frac{u^{-1}}{1+u^7}\,7u^6\,du$$ and it is now possible to factorise the denominator, obtain partial fractions and integrate to get a logarithm and three inverse tangents. But it will be hard going, to say the least.

Answer 4

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.

Answer 5

You can separate the $x,y,z$-terms and finally you will find the integral as product of three Beta function of the second kind, do you get it?

Answer 6

This problem be reduced to evaluating integrals of the type $\displaystyle\int_0^\infty \dfrac{x^{-a}}{x+1}\mathrm{d}x.$

The substitution you're looking for is $t=\dfrac1{x+1}$ , which automatically transforms its expression into that of the famous beta function. Then, with the help of Euler's reflection formula for the $\Gamma$ function, this in its turn becomes $\dfrac\pi{\sin\pi a}$ . The exact same trick works for all integrals of the form $\displaystyle\int_0^\infty \dfrac{x^{n-1}}{x^m+a^m}\mathrm{d}x,$ yielding the general result $a^{n-m}\cdot\dfrac\pi m\cdot\csc\bigg(n\cdot\dfrac\pi m\bigg)$.