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 $\alpha$ ?

Extra Addition:

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 $$2\sqrt{2}\int_0^{\pi/4}\csc\theta d\theta\ge2\sqrt{2}\int_0^{\pi/4}\theta^{-1} d\theta=\infty.$$

  • $\begingroup$ Thank you. I find myself with your steps, but how to prove that the integral is bounded for all alpha<1, especially for theta tending to zero ? $\endgroup$ – C.C.12 Apr 26 at 10:27
  • 1
    $\begingroup$ @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. $\endgroup$ – J.G. Apr 26 at 10:30
  • 1
    $\begingroup$ Of course! sorry it is trivial you're right $\endgroup$ – C.C.12 Apr 26 at 10:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.