How to evaluate this definite integral $\int_0^2(1-x^2)^\frac{1}{3}~dx$

A student asked me to help him calculate this definite integral $$\int_0^2(1-x^2)^\frac{1}{3}~dx$$ Although I have tried almost all the methods I have learned, I can not still do with it. I have tried the change of variable $x=\sec t$ , and the method of integral by parts. Can anyone help me?

• Apparently it's kind of nasty: wolframalpha.com/input/… Hypergeometric functions, whatever those are, and you know you have a problem when your series expansion is full of gamma functions. – dfeuer Dec 15 '13 at 5:47
• Where did the student get the question? Because this looks beyond first-year calculus to me. (unless they're missing an x multiplied by the integrand? ie. $x(1-x^2)^{1/3}$ because then a simple substitution would have worked) – andraiamatrix Dec 15 '13 at 5:48
• @Alec, that should be wolframalpha.com/input/… – dfeuer Dec 15 '13 at 5:49
• My bad, yes. Thanks dfeuer. – Alec Dec 15 '13 at 5:50
• @Integrals and Series: It sounds great, but when we substitute $x=2$ into it we will get trouble. – doraemonpaul Dec 15 '13 at 11:18

$\int_0^2(1-x^2)^\frac{1}{3}~dx$

$=\int_1^{-\sqrt3}x~d\left((1-x^3)^\frac{1}{2}\right)$

$=\int_1^{-\sqrt3}\dfrac{3x^3(1-x^3)^{-\frac{1}{2}}}{2}dx$

$=\int_{-1}^\sqrt3\dfrac{3(-x)^3(1-(-x)^3)^{-\frac{1}{2}}}{2}d(-x)$

$=\int_{-1}^\sqrt3\dfrac{3x^3(1+x^3)^{-\frac{1}{2}}}{2}dx$

$=\int_{-1}^1\dfrac{3x^3(1+x^3)^{-\frac{1}{2}}}{2}dx+\int_1^\sqrt3\dfrac{3x^3(1+x^3)^{-\frac{1}{2}}}{2}dx$

$=\int_{-1}^1\dfrac{3x^3(1+x^3)^{-\frac{1}{2}}}{2}dx+\int_1^\sqrt3\dfrac{3x^\frac{3}{2}~(1+x^{-3})^{-\frac{1}{2}}}{2}dx$

For the binomial series of $(1+x)^{-\frac{1}{2}}$ , $(1+x)^{-\frac{1}{2}}=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^n}{4^n(n!)^2}$

$\therefore\int_{-1}^1\dfrac{3x^3(1+x^3)^{-\frac{1}{2}}}{2}dx+\int_1^\sqrt3\dfrac{3x^\frac{3}{2}~(1+x^{-3})^{-\frac{1}{2}}}{2}dx$

$=\int_{-1}^1\dfrac{3x^3}{2}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{3n}}{4^n(n!)^2}dx+\int_1^\sqrt3\dfrac{3x^\frac{3}{2}}{2}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{-3n}}{4^n(n!)^2}dx$

$=\int_{-1}^1\sum\limits_{n=0}^\infty\dfrac{3(-1)^n(2n)!x^{3n+3}}{2^{2n+1}(n!)^2}dx+\int_1^\sqrt3\sum\limits_{n=0}^\infty\dfrac{3(-1)^n(2n)!x^{\frac{3}{2}-3n}}{2^{2n+1}(n!)^2}dx$

$=\biggl[\sum\limits_{n=0}^\infty\dfrac{3(-1)^n(2n)!x^{3n+4}}{2^{2n+1}(n!)^2(3n+4)}\biggr]_{-1}^1+\biggl[\sum\limits_{n=0}^\infty\dfrac{3(-1)^n(2n)!x^{\frac{5}{2}-3n}}{4^n(n!)^2(5-6n)}\biggr]_1^\sqrt3$

$=\sum\limits_{n=0}^\infty\dfrac{3(-1)^n(2n)!}{2^{2n+1}(n!)^2(3n+4)}-\sum\limits_{n=0}^\infty\dfrac{3(-1)^n(2n)!(-1)^{3n+4}}{2^{2n+1}(n!)^2(3n+4)}-\sum\limits_{n=0}^\infty\dfrac{3(-1)^n(2n)!3^{\frac{5}{6}-n}}{4^n(n!)^2(6n-5)}+\sum\limits_{n=0}^\infty\dfrac{3(-1)^n(2n)!}{4^n(n!)^2(6n-5)}$

$=\sum\limits_{n=0}^\infty\dfrac{3(-1)^n(2n)!}{4^n(n!)^2(6n-5)}-\sum\limits_{n=0}^\infty\dfrac{3^\frac{11}{6}(-1)^n(2n)!}{12^n(n!)^2(6n-5)}-\sum\limits_{n=0}^\infty\dfrac{3(4n+2)!}{4^{2n+1}((2n+1)!)^2(6n+7)}$

No, no one can help your student calculate this definite integral. It is too horribly nasty by far. Tell him to wait till he gets to grad school and see if he's still interested in awful integrals then.

• Lol. Good answer. I'm assuming something must have been copied down incorrectly because there's no way that's a first-year integral question. – andraiamatrix Dec 15 '13 at 5:58
• Actually the student that asked me the above problem is one of my undergraduate student, who takes part in my Calculus course. – nuage Dec 15 '13 at 6:31

Let $x = 1 -u^3$, we can rewrite the integral $\mathcal{I}$ as

$$\mathcal{I} = \int_0^2\sqrt{1-x^2}dx = 3\sqrt{2}\int_{-1}^1 u^3 \left(1 - \frac{u^3}{2}\right)^{\frac13} du\\$$ Since the power series expansion of $\left(1 - \frac{u^3}{2}\right)^{\frac13}$ at $u = 0$ has radius of convergence $> 1$, we can expand it inside the integral sign and evaluate the expansion term by term. We have

\begin{align} \mathcal{I} \stackrel{[\color{blue}{1}]}{=}& 3\sqrt{2}\int_{-1}^1 u^3 \sum_{k=0}^{\infty} \frac{\left(-\frac13\right)_k}{k!}\left(\frac{u^3}{2}\right)^k du\\ =&3\sqrt{2}\sum_{k=0}^\infty\frac{(-\frac13)_k}{k!}\frac{1}{3k+4}\left[\left(\frac12\right)^k - \left(-\frac12\right)^k\right]\\ \stackrel{[\color{blue}{2}]}{=}&\frac{3\sqrt{2}}{4}\sum_{k=0}^\infty\frac{(-\frac13)_k (\frac43)_k}{k!(\frac73)_k} \left[\left(\frac12\right)^k - \left(-\frac12\right)^k\right]\\ =&\frac{3\sqrt{2}}{4}\left[\,_2F_1(-\frac13,\frac43;\,\frac73;\,\frac12) -\,_2F_1(-\frac13,\frac43;\,\frac73;\,-\frac12)\right] \end{align} Throwing the last expression to WA give us $$\mathcal{I} \sim -0.18490339160722117817295686175099263891533938048269736635284...$$ consistent with what will get if you ask WA to numerically evaluate the original integral.

Notes

$[\color{blue}{1}]$ $(\alpha)_k = \alpha(\alpha+1)\cdots(\alpha+k-1)$ is the rising Pochhammer symbol.
$[\color{blue}{2}]$ We are using the identity $\frac{(\gamma)_k}{(\gamma+1)_k} = \frac{\gamma}{\gamma+k}$ here.

• What's $\left(-\frac13\right)_k$ mean? – dfeuer Dec 16 '13 at 4:16
• @dfeuer, $(\alpha)_k = \alpha(\alpha+1)\cdots(\alpha+k-1)$, the rising Pochhammer symbol – achille hui Dec 16 '13 at 7:07