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I would like to prove that this integral is convergent. Although it's easy to evaluate it, I would like more ellegant way to do it. Thanks $$\int\limits_{0}^{1}\frac{\sqrt{x}}{\sqrt{1-x^4}}\, dx $$

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2 Answers 2

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Close to $0$, there's no issue and close to $1$, the integrand is basically $\frac{1}{\sqrt{1-x^4}} \approx \frac{1}{\sqrt{1-x^2}}\frac{1}{\sqrt{1+x^2}} \approx \frac{1}{\sqrt{1-x^2}}\frac{1}{\sqrt{2}}$ but this is fine since $\frac{1}{\sqrt{1-x^2}}$ has antiderivative $\arcsin x$

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We only have problems at $x=1$

Close to $x=1$ we can see $$\int_0^1\frac{\sqrt{x} \ dx}{\sqrt{1 - x^4}} = \int_0^1 \left(\sqrt{\frac{{x}}{{(1+x^2)(1+x)}}} \frac{1}{\sqrt{1-x}}\right) dx \leq \int_0^1\frac{M}{\sqrt{1-x}} \ dx=2M < +\infty$$

where $M$ is an upper bound for $\sqrt{\frac{{x}}{{(1+x^2)(1+x)}}}$ on $[0,1]$.

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