Integration of (x^2-x^3)^-(1/3) from 0 to 1

In summary, Daniel is trying to find the limit for the integral of (x^2-x^3)^-(1/3) around the singularities 0 and 1. He found a method to integrate around the contour around the singularities and found the limit to be -2 pi i e^(pi i / 3).
  • #1
DE7
8
0
Hey all,

I need some help with a multivalued complex integration problem.

Evaluate the integral of (x^2-x^3)^-(1/3) from 0 to 1.

I know you need to pick branches of the cube root function, and this will give you multiples of the integral which you can then equate to integrals along contours around the singularities 0 and 1. However, I'm having trouble finding those integrals around the contours. How can I use residues when fractional powers are involved? Any help is appreciated.
 
Physics news on Phys.org
  • #2
Here's the indefinite integral

[tex] \int \frac{dx}{\sqrt[3]{x^{2}-x^{3}}} =\frac{1}{\sqrt[3]{x^{2}-x^{3}}}\left[\left(3x\sqrt[3]{1-x}\right) \ _{2}F_{1}\left(\frac{1}{3},\frac{1}{3},\frac{4}{3},x\right)\right] +\mathcal{C} [/tex]

Good luck with those limits (in case u use the FTC).

Daniel.
 
Last edited:
  • #3
Do you actually have a methodology to get solutions in the form of hypergeometric polynomials or did you just plug that into Mathematica or Maple?

I mean either of those programs will tell you what the limit is too (at least maple does). But if there's a method to get that answer I'd be curious how you go about it. Always nice to have a new method of integration in your arsenal.

Thanks
Steven
 
  • #4
Well, I solved this in time to turn it in for class. If anyone's interested, the method is to integrate the function on a barbell shaped contour around the singularities 0 and 1. Then the curved paths around the singularities go to 0 by limiting arguments, and the straight paths give (1-e^(2 pi i / 3)) I, where I is the desired integral, by branch arguments. Deforming the contour to infinity and using limiting arguments gives the integral around the contour also equals -2 pi i e^(pi i / 3). Equating the two and solving for I gives I = pi / sin (pi / 3), or 2 pi / sqrt(3).

By the way, this is the screenname i usually use...
 

Related to Integration of (x^2-x^3)^-(1/3) from 0 to 1

1. What is the formula for integrating (x^2-x^3)^-(1/3) from 0 to 1?

The formula for integrating (x^2-x^3)^-(1/3) from 0 to 1 is ∫(x^2-x^3)^-(1/3) dx = -3((x^2-x^3)^2/3)/(2x^3-3x^4) + C.

2. How do you solve for the indefinite integral of (x^2-x^3)^-(1/3)?

To solve for the indefinite integral of (x^2-x^3)^-(1/3), use the substitution u = x^2-x^3. Then, du = (2x-3x^2)dx, and the integral becomes ∫u^(-1/3) du. Using the power rule for integration, the solution is u^(2/3)/ (2/3) + C = 3(u^(2/3))/2 + C = 3((x^2-x^3)^(2/3))/2 + C.

3. Is it possible to use a different substitution to solve for the integral of (x^2-x^3)^-(1/3)?

Yes, it is possible to use a different substitution to solve for the integral of (x^2-x^3)^-(1/3). Another option is to use the substitution u = x^3-x^2, which results in the integral becoming ∫u^(-1/3) du. The solution using this substitution is (u^(2/3))/(2/3) + C = 3(u^(2/3))/2 + C = 3((x^3-x^2)^(2/3))/2 + C.

4. Can the integral of (x^2-x^3)^-(1/3) from 0 to 1 be solved using partial fractions?

No, it is not possible to solve the integral of (x^2-x^3)^-(1/3) from 0 to 1 using partial fractions. This is because the integrand is not a rational function, and therefore cannot be expressed as a sum of simpler fractions.

5. What is the significance of solving for the integral of (x^2-x^3)^-(1/3) from 0 to 1?

The significance of solving for the integral of (x^2-x^3)^-(1/3) from 0 to 1 lies in its application in various fields such as physics, engineering, and economics. This integral can be used to calculate the volume of a solid of revolution, the work done in a thermodynamic process, or the average value of a function. It is also an important concept in calculus and helps in understanding the relationship between the derivative and the integral.

Similar threads

Replies
5
Views
2K
  • Calculus
Replies
8
Views
2K
Replies
4
Views
2K
Replies
8
Views
1K
Replies
12
Views
1K
  • Calculus
Replies
6
Views
2K
Replies
3
Views
1K
Replies
4
Views
3K
  • Calculus
Replies
29
Views
1K
Replies
3
Views
1K
Back
Top