Integral of sqrt(x^3 -1) dx

1. Mar 21, 2005

One-D

integral of sqrt(x^3 -1) dx.
I have tries it with trigonometri subs, but I only found till
integral of 2/3(sin^5/3 (tetha) / cos^2 (tetha) ) d(tetha)

2. Mar 21, 2005

Data

I don't believe this has a simple antiderivative.

3. Mar 21, 2005

One-D

hm you may be right

4. Mar 21, 2005

Crosson

Yes, this requires the use of elliptical integrals (non-elementary functions).

5. Mar 21, 2005

One-D

what's that? I don't know about it...

6. Mar 21, 2005

mathwonk

that is sort of a tautological stetament, i.e. elliptic functions are defined in terms of integrals, of (usually reciprocals of) square roots of cubics.

i.e. just as sin can be defiend as the inmverse of the integral of dx/sqrt(1-x^2), so also one can define some interesting fucntions as the inverse of the integral of

things like dx/sqrt(1-x^3).

7. Mar 21, 2005

Tom McCurdy

wow... can someone walk me through how this would work out

I just typed it in and got

$$\int \sqrt{x^3-1}dx=\frac{2x\sqrt{x^3-1}}{5}-\frac{3\int{\frac{1}{\sqrt{x^3-1}}}}{5}$$

8. Mar 21, 2005

dextercioby

Okay,here's the result for the curious.

Daniel.

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9. Mar 21, 2005

Tom McCurdy

ummm I don't really follow what ur attachment is showing.

10. Mar 21, 2005

dextercioby

You mean the elliptic integral...?Go to wolfram's site (mathworld) and search for Legendre elliptic integrals.

Daniel.

11. Mar 22, 2005

saltydog

The formula Daniel reported is the anti-derivative. That is:

$$\int\frac{1}{\sqrt{x^3-1}}=\frac{2i\sqrt{(-1)^{5/6}(x-1)}\sqrt{1+x+x^2}EllipticF[\arcsin(\frac{\sqrt{-(-1)^{5/6}-ix}}{3^{1/4}}),(-1)^{\frac{1}{3}}]}{3^{\frac{1}{3}}\sqrt{x^3-1}}$$

It's just not in terms of elementary functions.

Great. Now suppose I want to integrate it from 2 to 4 using the formula and for the moment I want to concentrate on the elliptical integral portion, the EllipticF part. A definition first. EllipticF is the eliptical integral of the first (F) kind defined as:

$$F(x,k)=\int_0^x\frac{dx}{\sqrt{(1-x^2)(1-k^2x^2)}}$$

Alright, it's already getting messy. Let's just concentrate on the ArcSin part first then: Let x=2 so we have:

$$\arcsin[\frac{\sqrt{-(-1)^{5/6}-2i}}{3^{1/4}}]$$

Jesus. That's a problem in itself (for me anyway). Will need to spend time on it first. Anyway, I want to see how the imaginary parts of the anti-derivative are eliminated since this is a definite integral. I've attached a plot of the function.

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Last edited: Mar 22, 2005
12. Mar 22, 2005

saltydog

Alright, I'm stuck. Let's just concentrate on the radical:

$$\sqrt{-(-1)^{5/6}-2i}$$

Sinking deeper still . . . how about just the -1 to the 1/6 power:

$$\sqrt[6]{-1}$$

Wouldn't I get six roots from this? Which one do I use? Going to bed . . . will work on it latter. Any comments would be appreciated though.

13. Mar 22, 2005

Crosson

If we define one new function, and it allows to solve a whole new class of integrals, are these solutions tautological?

Answer: Yes, but that is the way math is. Mathematical satisfaction comes from getting so used to the tautologies that they talk on a life of there own:

Von Neumann: "You don't understandthings in mathematics, you just get used to them".

14. Mar 23, 2005

saltydog

Well, I'm still working on the elliptical integral expression and finding it interesting. You know, even the part:

$$EllipticF[\arcsin(\frac{\sqrt{-(-1)^{5/6}-ix}}{3^{1/4}}),(-1)^{\frac{1}{3}}]$$

is a challenge for me (well, I don't mean just plug it into Mathematica and turn the crank you know but I am using it to study the expression). Seems the definite integral ends up being (a+bi)-(c+bi) for any limits! That's how the i is eliminated. How's that? I'll work with it more.

15. Mar 25, 2005

One-D

I got this problem from someone by forum too. I thought it wasn't so difficult.. thanx for reply. but is the function really can be integrated?

16. Mar 26, 2005

dextercioby

Of course it can.It's an elliptic integral.

Daniel.

17. Apr 2, 2005

One-D

thanx everyone

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