# Beta function and changing variables

1. Feb 14, 2009

### springo

1. The problem statement, all variables and given/known data
$$\int_{0}^{3}\sqrt[3]{\frac{3-x}{x^2}}\: \mathrm{d}x$$

2. The attempt at a solution
I'm pretty sure I have to use Euler's Beta function, so I tried to change the limits to 0 and 1 by setting x = 3·u (so dx = 3·du). However there must be some mistake when I did it because I checked and I did it wrong:
$$\int_{0}^{3}\sqrt[3]{\frac{3-x}{x^2}}\: \mathrm{d}x=\int_{0}^{1}\sqrt[3]{\frac{3-3u}{(3u)^2}}\: 3\cdot\mathrm{d}u=3^{\frac{2}{3}}\cdot\beta\left(\frac{1}{3},\frac{4}{3}\right)$$

Thanks a lot for your help.

Last edited: Feb 14, 2009
2. Feb 15, 2009

### CompuChip

Hmm then what should be the correct answer. Because, unless I am still asleep, both your change of variables and your identifying the arguments of the beta function are quite all right. I even got the same prefactor with a 2/3 exponent.

3. Feb 15, 2009

### springo

Well, Mathematica says:
$$\int_{0}^{3}\sqrt[3]{\frac{3-x}{x^2}}\: \mathrm{d}x=\left(\frac{3}{2}\right)^{\frac{2}{3}}\cdot\beta\left(\frac{1}{3},\frac{1}{2}\right)$$

4. Feb 15, 2009

### CompuChip

But when I evaluate both numerically, I get the same answer up to 4*\$MachineEpsilon.

Code (Text):
Beta[1/3, 4/3] == (1/2)^(2/3) Beta[1/3, 1/2] // FullSimplify
gives True.

So probably there is some property of the Beta function (which can most likely be derived from the Gamma function) which you need to show.

5. Feb 15, 2009

### springo

OK thanks I think Mathematica isn't using beta function though because in all my integrals when I try to check them I get a different output, which I can't get to even when I try FullSimplify[Beta[whatever, whatever]]. However after try the comparison I get true.

6. Feb 15, 2009

### CompuChip

Just a general remark...

In my experience Mathematica sometimes does strange things with comparisons, and seen instances of
SomeExpression == DifferentExpression
giving True.

MyExpression - MathematicaResult // FullSimplify
and checking it gives zero.

7. Feb 15, 2009

### springo

Thanks, I'm just learning how to use Mathematica and I didn't know about that feature.

Now I'm stuck with the last integral in my problem sheet... this one I can't solve (b > 2 and a > 0).
$$\int_{0}^{\infty}\frac{\sqrt[3]{x}-\sqrt{x}}{x^b-a^b}\: \mathrm{d}x$$
I'm thinking that it's going to be a beta too (duh, it's the gamma-beta functions section of the problem set), so I should change that infinity to either 1 or pi/2.
Since changing it to pi/2 would imply having arctan in my integral (wouldn't it?) it doesn't look good. So I guess I need a 1... but I can't think of anything good there.

Last edited: Feb 15, 2009