# Definite integral with fractional power

Prompted by this question, and the answer which suggests a bug in Wolfram Alpha's shell-integration algorithm, I tried to evaluate the integral:

$$I = \int_0^2 u(1-(u-1)^{1/3})du.$$

But doing this by parts I run into problems with the last step:

$$I = \int_0^2 udu - \int_0^2 u(u-1)^{1/3} du$$ $$= 2 - \left[u\frac{3}{4}(u-1)^{4/3}\right]_0^2 + \frac{3}{4}\int_0^2(u-1)^{4/3}du$$

$$= 2 - \frac{3}{2} + \frac{3}{4}\left[\frac{3}{7}(u-1)^{7/3}\right]_0^2$$

... and now I'm stuck with raising a negative number to the power $\frac{7}{3}$ and don't know how to proceed. What is the correct way to tackle this?

• Raising a negative number to a rational power with odd denominator shouldn't be a problem -- you just get $x^{7/3} = (\sqrt[3] x)^7$. And negative numbers do have cube roots. Mar 18, 2015 at 14:58
• In general it would probably be simpler to start with substituting $t=u-1$. Then the entire integrand simplifies to a generalized (fractional-powers) polynomial, and you don't need to integrate by parts. Mar 18, 2015 at 14:59
• Thanks: I was being a bit dim not to recognise the cube root... I now get the same answer ($8/7$) as I do with disk-integration. There's definitely something up with Wolfram Alpha here.
– Tom
Mar 18, 2015 at 15:08
• @Tom: you have to tell WA which branch of the cube root you are taking, WA cannot guess from nothing that for $0<u<1$ we are taking $(u-1)^{1/3}$ as $-(1-u)^{1/3}$. Mar 18, 2015 at 15:30
• @Jack D'Aurizio: the problem isn't with the evaluation of this integral but with the integral that WA sets up in order to find the volume of revolution by the shell method: see the original question linked in my post.
– Tom
Mar 18, 2015 at 15:58

There's nothing wrong with fractional powers of negative numbers, as long as the denominator of the fraction (in lowest terms) is odd. For instance, $(-8)^{1/3} = -2$, since $(-2)^3 = -8$. So your evaluation (assuming everything else is correct) can be carried out like this: $$(-1)^{7/3} = ((-1)^{1/3})^7 = (-1)^7 = -1.$$

Just to avoid any ambiguity, assuming that for $\alpha<0$ we are defining $\alpha^{1/3}$ as $-(-\alpha)^{1/3}$, our integral equals:

$$\begin{eqnarray*} I &=& \int_{0}^{1}u\left(1+(1-u)^{1/3}\right)\,du+\int_{1}^{2}u\left(1-(u-1)^{1/3}\right)\,du\\&=&\int_{0}^{1}(1-u)(1+u^{1/3})\,du+\int_{0}^{1}(1+u)(1-u^{1/3})\,du\end{eqnarray*}$$ and now, by replacing $u$ with $v^3$, $$I = \int_{0}^{1}\left(3v^2(1-v^3)(1+v)+3v^2(1+v^3)(1-v)\right)\,dv =\int_{0}^{1}6v^2(1-v^4)\,dv$$ so: $$I = 6\left(\frac{1}{3}-\frac{1}{7}\right)=\color{red}{\frac{8}{7}}.$$

• Now I'm confused again... can you explain where the first term in your first line comes from (the $+$ sign), please?
– Tom
Mar 18, 2015 at 15:27
• @Tom: as I said, for $\alpha < 0$ we take $\alpha^{1/3}$ as $-(-\alpha)^{1/3}$. Mar 18, 2015 at 15:28
• Got it-- I think we agree on the final answer now.
– Tom
Mar 18, 2015 at 15:57