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Beta function and changing variables 
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#1
Feb1409, 06:28 PM

P: 126

1. The problem statement, all variables and given/known data
[tex]\int_{0}^{3}\sqrt[3]{\frac{3x}{x^2}}\: \mathrm{d}x[/tex] 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: [tex]\int_{0}^{3}\sqrt[3]{\frac{3x}{x^2}}\: \mathrm{d}x=\int_{0}^{1}\sqrt[3]{\frac{33u}{(3u)^2}}\: 3\cdot\mathrm{d}u=3^{\frac{2}{3}}\cdot\beta\left(\frac{1}{3},\frac{4}{3 }\right)[/tex] Thanks a lot for your help. 


#2
Feb1509, 04:13 AM

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P: 4,300

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
Feb1509, 08:48 AM

P: 126

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


#4
Feb1509, 10:49 AM

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Beta function and changing variables
But when I evaluate both numerically, I get the same answer up to 4*$MachineEpsilon.
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
Feb1509, 11:43 AM

P: 126

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
Feb1509, 01:06 PM

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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. 


#7
Feb1509, 02:39 PM

P: 126

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). [tex]\int_{0}^{\infty}\frac{\sqrt[3]{x}\sqrt{x}}{x^ba^b}\: \mathrm{d}x[/tex] I'm thinking that it's going to be a beta too (duh, it's the gammabeta 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. 


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