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Solve an integral $$\int\left(\frac{x^{1/3}-1}{x}\right)^{5/3}dx$$

I am using Chebyshev’s criterion (integrals of differential binomials), but can't figure out how to use it properly.

Integrand function is of the form $x^m(a+bx^n)^p$ where $a,b$ are coefficients and $m,n,p$ are rational numbers.

$$\left(\frac{x^{1/3}-1}{x}\right)^{5/3}=x^{-5/3}\left(-1+x^{1/3}\right)^{5/3}$$

Here $a=-1,b=1,m=-5/3,n=1/3$.

Since $$\frac{m+1}{n}$$ is an integer, we use substitution $$a+bx^n=t$$

Using this substitution gives an integral $$3\int t^{5/3}(1+t)^{-3}dt$$

Now $p=-3$ is an integer, so we use the first case in Chebyshev’s criterion. I don't understand how to use this case.

It says: Expanding $(a+bt^n)^p$ by the binomial formula we write the integrand as a rational function of the simple radicals $t^{j/k}$. Then the substitution $t=u^r$, where $r$ is the largest of all denominators $k$ will remove the radicals entirely.

How to use this case?

Reference: Chebyshe

  • 0
    this integral leads to a hypergeometric series – 2017-02-10
  • 1
    Personally, I would take $\frac{x^{\frac{1}{3}} - 1}{x} = t^3$ and see what happens... – 2017-02-10
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    @nikola, do you know how to use the first case of Chebyshe's criterion? – 2017-02-10
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    @Dr. Sonnhard Graubner, Does that mean it is not integrable over $\mathbb R$? – 2017-02-10
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    I think in the paper you have referred, **integer** means $\mathbb{N}$. – 2017-02-10

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If you set $u^3=x^\frac{1}{3}-1$ or equivalently $u^3=t$ you get $\int \frac{9u^7}{(u^3+1)^3}du$.

$\frac{9u^7}{(u^3+1)^3}=\frac{2-u}{(u^2-u+1)^3}+\frac{u-17}{3(u^2-u+1)^2}+\frac{5u+13}{3(u^2-u+1)}-\frac{1}{3(u+1)^3}+\frac{4}{3(u+1)^2}-\frac{5}{3(u+1)}$

Happy integration...

Note: maybe there is a better method than partial fraction expansion at this stage, but you'll have to read "Manuel Bronstein - Symbolic Integration I".

Edit: I've now read the interesting part of this book, let's apply the Horowitz-Ostrogradsky algorithm !

$A=9u^7$

$D=(u^3+1)^3$

$D\,'=9u^2(u^3+1)^2$

$D^-=gcd(D,D\,')=(u^3+1)^2$

$D^*=D/D^-=(u^3+1)$

$B=\sum_{i=0}^{deg(D^-)-1}b_iu^i=b_0+b_1u+b_2u^2+b_3u^3+b_4u^4+b_5u^5$

$C=\sum_{i=0}^{deg(D^*)-1}c_iu^i=c_0+c_1u+c_2u^2$

And let's identify to the null polynom $\forall u, H(u)=A-B\,'D^*+BD^*{D^{-}}'/D^--CD^-=0$

$H(u)=-c_2u^8+(b_5-c_1+9)u^7+(-c_0+2b_4)u^6+(3b_3-2c_2)u^5+(-5b_5+4b_2-2c_1)u^4+(-4b_4-2c_0+5b_1)u^3+(-3b_3-c_2+6b_0)u^2+(-2b_2-c_1)u-b_1-c_0=0$

This system solves to $(b_i)_i=(0,0,\frac{-5}{2},0,0,-4)$ and $(c_i)_i=(0,5,0)$

$$\int \frac{9u^7}{(u^3+1)^3}du=\frac{B}{D^-}+\int\frac{C}{D^*}du=\frac{-\frac52u^2-4u^5}{(u^3+1)^2}+\int\frac{5u}{u^3+1}du$$

The last part is solved classically:

$\int=-\frac53\ln(u+1)+\frac56\ln(u^2-u+1)+\frac{5}{\sqrt 3}\arctan(\frac{2u-1}{\sqrt 3})$