the convergence of$\int_{0}^1 \frac{(1-x^2)^{\alpha}}{\sqrt{x}} dx$

Question

I have to analyse the convergence of the improper integral, for different values of $\alpha$:
$\int_{0}^1 \frac{(1-x^2)^{\alpha}}{\sqrt{x}} dx$

---

I think that when $\alpha$ is bigger or equal to $0$, the improper integral is convergent, but when $\alpha$ is negative?

Answer 1

Suppose $p>0$. since $1\leq 1+x\leq2$ over $0\leq x\leq 1$,

$$ 2^{-p}\frac{1}{(1-x)^p\sqrt{x}}\leq \frac{1}{(1-x^2)^p\sqrt{x}}\leq \frac{1}{(1-x)^p\sqrt{x}} $$

In integrating over $[0,1]$, the difficulty arrises at the singularity at $x=1$ (around $x=0$ there is no problem since $\int^1_0x^{-1/2}\,dx<\infty$.

Suffices to consider the behavior of $$\int^1_{1/2}\frac{1}{(1-x)^p\sqrt{x}}\,dx= \int^{\frac12}_0\frac{1}{x^p\sqrt{1-x}}\,dx $$ For $0<p<1$ the integral above converges since $\frac{1}{x^p\sqrt{1-x}}\leq \frac{\sqrt{2}}{x^p}$ for $0\leq x\leq \frac12$. For $p\geq1$ the integral diverges since $\frac{1}{x^p\sqrt{1-x}}\geq \frac{1}{x^p}$ for $0\leq x\leq\frac12$.

In summary:

1. If $\alpha>-1$, the integral converges.
2. If $\alpha\leq-1$ the integral diverges.

Answer 2

HINT

If $\sqrt x = t$ then $x=t^2$ and $\operatorname d x = 2 t \operatorname d t$. Hence the integral becames

$$ \int_0^1 \frac{(1-t^4)^\alpha}{t}2t \operatorname d t = 2 \int_0^1 (1-t^4)^\alpha \operatorname d t $$

If $\alpha \ge 0$ then the integral is convergent (why?).

If $\alpha = -k$ and $k > 0$ then the integral becames

$$ \int_0^1 \frac{\operatorname d t}{(1-t^4)^k} =\int_0^1 \frac{\operatorname d t}{(1-t^2)^k(1+t^2)^k} =\int_0^1 \frac{\operatorname d t}{(1-t)^k(1+t)^k(1+t^2)^k} $$

Can you prooceed from here?

Answer 3

Starting from @Sewer Keeper's answer and aking the problem more general, sooner or later, you will learn about the antiderivative in terms of special function $$\int \left(1-t^a\right)^b \,dt=t \,\, _2F_1\left(\frac{1}{a},-b;1+\frac{1}{a};t^a\right)$$ where appears the gaussian hypergeomtric function.

For the definite integral $$\int_0^1 \left(1-t^a\right)^b \,dt=\frac{\Gamma \left(1+\frac{1}{a}\right) \Gamma (b+1)}{\Gamma \left(b+\frac{1}{a}+1\right)}\qquad \text{if}\qquad \Re(b)>-1\land \Re(a)>0$$ So, for your case $a=4$ and $b=\alpha$ the rsult is $$\int_0^1 (1-t^4)^\alpha \,dt=\frac{\Gamma \left(\frac{5}{4}\right) \Gamma (\alpha+1)}{\Gamma\left(\alpha+\frac{5}{4}\right)}\qquad \text{if}\qquad \Re(\alpha)>-1$$