# 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.

• The usual way to evaluate that last integral is by complex methods and residues. I couldn't find it through the MSE search box but I bet it's here somewhere. – David May 16 '14 at 14:04
• Do you think that there is a method of calculating the first integral without using residue theory? I found the problem in some papers from an old high school math competition where residue theory is unlikely but still possible. – Brad May 16 '14 at 14:09
• Can you please link the papers? I would definitely like to try those problems, I am myself a high school student. Thanks! :) – Pranav Arora May 16 '14 at 14:25
• @David, I dug up the answers. They are expecting contour integration, color me surprised. – Brad May 16 '14 at 14:31
• Don't know what school that was but I wish I had gone there ;-) – David May 16 '14 at 14:45

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

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

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

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

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

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

• If anybody is unfamiliar with the final evaluation. He/She uses Euler's Reflection Formula which is described here. – Ali Caglayan Jun 21 '14 at 14:07

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: $$\Gamma(1-a)\Gamma(a)=\frac{\pi}{\sin(\pi a)}$$

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.

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.

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?

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 $a^{n-m}\cdot\dfrac\pi m\cdot\csc\bigg(n\cdot\dfrac\pi m\bigg)$.