Studying the convergence of $\int_0^1\frac{dx}{\sqrt{1-x^4}}$

Question

I am studying the convergence of improper integrals. In this case I am asked to calculate if the following integral converges.$$\int_0^1\frac{dx}{\sqrt{1-x^4}}$$

To do so my professor used the comparison test with the function $\frac{1}{\sqrt{1-x}}$ since the limit $$\lim_{x\to 1}\frac{\frac{1}{\sqrt{1-x^4}}}{\frac{1}{\sqrt{1-x}}}\neq0,\infty$$ And therefore the initial integral will converge if and only if $$\int_0^1\frac{dx}{\sqrt{1-x}}$$ converges. Indeed, this last integral is $2$, so the initial integral converges.

My question is about the limit, I don't know how to compute it so I don't understand why my professor chose to compare the limit to the function $\frac{dx}{\sqrt{1-x}}$.

Answer 1

$$\frac{\frac{1}{\sqrt{1-x^4}}}{\frac{1}{\sqrt{1-x}}}=\frac{\sqrt{1-x}}{\sqrt{1-x}\sqrt{1+x}\sqrt{1+x^2}}=\frac1{\sqrt{1+x}\sqrt{1+x^2}}\xrightarrow[x\to1]{}\frac1{\sqrt2\sqrt2}=\frac12$$

Answer 2

The comparison test does not really require that $\lim_{x\to 1^-} (1-x^4)^{-1/2}/(1-x)^{-1/2}$ exists. Let $g(r)=\int_0^r(1-x^4)^{-1/2}dx$ for $r\in [0,1).$

It suffices to find some $h(x)$ and some $K>0$ such that $(1-x^4)^{-1/2} <Kh(x)$ for $x\in [0,1)$ and such that $\sup_{r\in [0,1)}\int_0^rh(x)dx=M<\infty.$

Because this implies that $g(r)\le KM$ for all $r\in [0,1),$ and $g(r)$ is an increasing function, so $\lim_{r\to 1^-}g(r)$ exists.

In the Q, let $K=1$ and $h(x)=(1-x)^{-1/2},$ as detailed in the other A's.