Study this improper integral $\int_0^1 \frac{dt}{\sqrt{t}\,\sqrt{1-t}\,\sqrt{1-\alpha\,\sqrt{1-t}}}$

I'm trying to study the behavior of this improper integral

$$\int_0^1 \frac{dt}{\sqrt{t}\,\sqrt{1-t}\,\sqrt{1-\alpha\,\sqrt{1-t}}}$$

for:$$\alpha>0$$

While I just can't understand the behavior around $$\alpha=1$$, inasmuch for $$\alpha\rightarrow1^-$$ seems to converge, while for $$\alpha=1$$ obviously diverges.

Writing the integral as

$$\begin{equation} \int_0^1 \frac{t^{-\frac{1}{2}}\,(1-t)^{-\frac{1}{2}}}{\sqrt{1-\alpha\,\sqrt{1-t}}}\:dt \end{equation}\:(*)$$

it is observed that the numerator, and the extremes of integration are similar to Euler Beta function.It is possible to use this similarity to study the convergence or divergence of the integral as function of $$\alpha$$ ?

for $$\alpha\rightarrow0$$ the integral converges in that:

$$\int_0^1 {t^{-\frac{1}{2}}\,(1-t)^{-\frac{1}{2}}}dt=\beta(\,\frac{1}{2},\,\frac{1}{2})=\pi$$

This obviously implies that for $$\alpha<0$$ the integral (*) converges

Write $$t=\sin^2 2\theta$$ so your integral is $$\frac{4}{\sqrt{1-\alpha}}\int_0^{\pi/4}\frac{d\theta}{\sqrt{1-\frac{2\alpha}{\alpha-1}\sin^2\theta}}=\frac{4}{\sqrt{1-\alpha}}F\left(\frac{\pi}{4},\,\sqrt{\frac{2\alpha}{\alpha-1}}\right),$$ with $$F$$ an incomplete elliptic integral of the first kind. Unfortunately the second argument is imaginary, so for convergence interests let's start over. The integral is $$\int_0^{\pi/4}\frac{4d\theta}{\sqrt{1-\alpha\cos 2\theta}}.$$The integrand is bounded for any $$\alpha<1$$, but for $$\alpha=1$$ is equal to $$2\sqrt{2}\int_0^{\pi/4}\csc\theta d\theta\ge2\sqrt{2}\int_0^{\pi/4}\theta^{-1} d\theta=\infty.$$
• @Christian For $\theta\in [0,\,\pi/4]$, $\cos 2\theta\in [0,\,1]$ so the integrand has extrema $\{\frac{1}{\sqrt{1-\alpha}},\,1\}$. Multiplying this by $\pi/4$ givies bounds on the integral. – J.G. Apr 26 at 10:30