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I'm having some trouble evaluating the following integrals: $$ I=\int_{-\infty}^0\frac{dx}{\sqrt{1-x^3}},\quad J=\int_0^1\frac{dx}{\sqrt{1-x^3}}. $$ Any help will be appreciated.

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By the way, these are improper integrals, not indefinite. –  L. F. Jan 31 '13 at 15:52
    
Thanks for the precision! –  albmiz-mth Jan 31 '13 at 15:56

1 Answer 1

up vote 3 down vote accepted

The $J$ integral is reduced to the Euler integral of the first kind using $u$-substitution $u=x^3$: $$ J = \frac{1}{3} \int_0^1 \left(1-u\right)^{1/2-1} u^{1/3-1} \mathrm{d}u = \frac{1}{3} \operatorname{Beta}\left(\frac{1}{3}, \frac{1}{2} \right) $$ The $I$ integral can be reduced to the beta integral using $x^3 = \frac{u}{1-u}$, i.e. $u=\frac{x^3}{1+x^3}$: $$ I = \int_0^\infty \frac{\mathrm{d}x}{\sqrt{1+x^3}} =\frac{1}{3} \int_0^1 u^{1/3-1} \left(1-u\right)^{1/6-1} \mathrm{d}u = \frac{1}{3} \operatorname{Beta}\left(\frac{1}{3}, \frac{1}{6} \right) $$

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