1
$\begingroup$

Does the integral $$\int (1-x^a)^\frac{1}{a} dx$$ have a closed form solution? The parameter $a$ is a constant and both $a$ and $x$ are positive real numbers less than or equal to $1$.

I tried plugging it in to Wolfram and got a result I don't understand. What e.g. does "$x_2 F_1(\ldots)$" mean? Or "$B_{x^a}(\ldots)$"?

$\endgroup$
  • 3
    $\begingroup$ Through the substitution $x^a=z$ the integral $\int_{0}^{1}(1-x^a)^{1/a}\,dx$ is a value for Euler Beta function, namely $\Gamma(1+1/a)^2 / \Gamma(1+2/a)$.$\;$ $\phantom{}_p F_q$ is the standard notation for hypergeometric functions. $\endgroup$ – Jack D'Aurizio Aug 14 '18 at 22:15
  • 2
    $\begingroup$ The $B_z(a,b)$ is called the incomplete beta function $\endgroup$ – Holo Aug 14 '18 at 22:23
  • $\begingroup$ @Jack D'Aurizio: Thanks. But what is solution to the indefinite integral? $\endgroup$ – Jens Aug 14 '18 at 22:46
  • $\begingroup$ @Jens: an incomplete Beta function. $\endgroup$ – Jack D'Aurizio Aug 14 '18 at 22:51
  • $\begingroup$ Maybe try expanding it with binomial theorem and then integrate? $\endgroup$ – Anurag B. Aug 15 '18 at 6:00
1
$\begingroup$

Yes, you can use binomial theorem then integrate.

$$ (1-x^a)^{1/a} = \sum_{k=0}^\infty \binom{1/a}{k}(-1)^k x^{ak} \\ \int(1-x^a)^{1/a} dx = \sum_{k=0}^\infty \binom{1/a}{k}(-1)^k \frac{x^{ak+1}}{ak+1} $$ Then do the usual manipulation to recognize this as a hypergeometric function \begin{align} \sum_{k=0}^\infty & \binom{1/a}{k}(-1)^k \frac{x^{ak+1}}{ak+1} = x \sum_{k=0}^\infty \frac{(1/a)(1/a-1)\cdots(1/a-k+1)}{k!}\;(-1)^k\frac{1}{ak+1}\;x^{ak} \\ &= x \sum_{k=0}^\infty \frac{(-1/a)(-1/a+1)\cdots(-1/a+k-1)}{k!}\frac{1}{a} \frac{1}{(1/a+k)}\;x^{ak} \\ &= x \sum_{k=0}^\infty \frac{(-1/a)(-1/a+1)\cdots(-1/a+k-1)}{k!}\; \frac{(1/a)(1/a+1)\cdots(1/a+k-1)}{(1/a+1)(1/a+2)\cdots(1/a+k)}\;x^{ak} \\&= x\;{}_2F_1\left(-\frac{1}{a},\frac{1}{a};\frac{1}{a}+1;x^a\right) \end{align} We used Gauss's definition $$ {}_2F_1(a,b;c;z) = \sum_{k=0}^\infty \frac{a(a+1)(a+2)\cdots (a+k-1)\;b(b+1)(b+2)\cdots(b+k-1)}{c(c+1)(c+2)\cdots(c+k-1)\; k!}\;z^k $$

$\endgroup$
1
$\begingroup$

Just rephrazing what was already in comments.

$$I=\int (1-x^a)^\frac{1}{a} dx=x \, _2F_1\left(-\frac{1}{a},\frac{1}{a};1+\frac{1}{a};x^a\right)$$

Now, as suggested, make $$x^a=z \implies x=z^\frac{1}{a} \implies dx=\frac{1}{a}{z^{\frac{1}{a}-1}}\,dz$$ to get $$I=\frac{1}{a} \int (1-z)^{\frac{1}{a}} z^{\frac{1}{a}-1}\,dz=\frac{1}{a}\,B_z\left(\frac{1}{a},\frac{1}{a}+1\right)$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.