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I am currently working on the following expression:

$$\int\left(1-\left(1-x^{\frac{1}{a}}\right)^a\right)\mathrm dx$$

I am particularly interested in finding the closed-form solution for this integral. During my research, I came across the idea of utilizing a definition of the hypergeometric function for $(1-x)^{-a}$. However, I am unsure about the precise steps involved in applying this approach to my integral. Perhaps there is a simpler method or a more useful change of variable to solve this problem.

Travis Willse
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JuanmaGun
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    Why not just ask about $\int_{ }^{ }\left(1-x^{\frac{1}{a}}\right)^{a}dx$? And what is $a$? – Accelerator Jul 05 '23 at 20:39
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    There are these related problems on $[0,1]$ https://math.stackexchange.com/q/113250/399263 a,d also https://math.stackexchange.com/q/3962326/399263, do you really need the incomplete one ? – zwim Jul 05 '23 at 20:51

1 Answers1

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We may as well just compute the integral of the nonconstant term, $(1 - x^\frac1a)^a$. Substituting $x = u^a$, $dx = a u^{a - 1} \,du$ transforms the integral to the so-called Chebyshev Integral: $$\int (1 - x^\frac1a)^a \,dx = a \int u^{a - 1} (1 - u)^a \,du = a \mathrm{B}(u; a, a + 1) + C = a \mathrm{B}\left(x^\frac1a; a, a + 1\right) + C.$$ Here $\mathrm{B}$ is the incomplete beta function. It can be expressed in terms of the ordinary hypergeometric function ${}_2 F_1$ (see the link) as $$\int (1 - x^\frac1a)^a \,dx = x \cdot {}_2 F_1\left(a, -a; a + 1; x^\frac1a\right) + C .$$ In general this antiderivative will be expressible in terms of elementary functions if $a$ is a (nonzero) half-integer.

For $a > 0$ a particular definite integral (cf. this question about the superellipse) is: $$\int_0^1 (1 - x^\frac1a)^a \,dx = a \mathrm{B}(a, a + 1) = \frac12 a \mathrm{B}(a, a) = \frac{a \Gamma(a)^2}{2 \Gamma(2a)} .$$ Here $\mathrm{B}$ is the (complete) beta function.

Travis Willse
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