# Beta function and changing variables

by springo
Tags: beta, function, variables
 P: 126 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.
 P: 126 Well, Mathematica says: $$\int_{0}^{3}\sqrt[3]{\frac{3-x}{x^2}}\: \mathrm{d}x=\left(\frac{3}{2}\right)^{\frac{2}{3}}\cdot\beta\left(\frac {1}{3},\frac{1}{2}\right)$$
 Sci Advisor HW Helper P: 4,300 Beta function and changing variables 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.
 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). $$\int_{0}^{\infty}\frac{\sqrt[3]{x}-\sqrt{x}}{x^b-a^b}\: \mathrm{d}x$$ 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.