Beta function and changing variables

[springo]

Homework Statement

\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.

 

[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.

 

[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)

 

[CompuChip]

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

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.

 

[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.

 

[CompuChip]

Just a general remark...

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

I suggest always FullSimplify'ing the difference betweeen your and Mathematica's answer
MyExpression - MathematicaResult // FullSimplify
and checking it gives zero.

 

[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.