Prove that $\displaystyle\int^1_0\frac{\ln(x)}{\sqrt{x(1-x)^3}}dx$ converges.

Asked Jul 11 '21 at 20:21

Active Jul 11 '21 at 21:18

Viewed 77 times

I'm trying to prove that $\displaystyle\int^1_0\frac{\ln(x)}{\sqrt{x(1-x)^3}}dx$ converges. When I plug the integral into a calculator, it gives me that the indefinite integral is equal to $$\frac{2\sqrt{x}\ln(x)}{\sqrt{1-x}}-4\arcsin(\sqrt{x})$$

and then when evaluated on $[0,1]$, it is equal to $-2\pi$. How is it evaluated if the first term of the above expression is undefined since $\sqrt{x-1}=0$ for $x=1$?

Is this a correct/good way to determine convergence for this specific integral?

edited Jul 11 '21 at 21:18

metamorphy

39,111

asked Jul 11 '21 at 20:21

2

at $x\to1 \ln x=\ln(1-\epsilon)\sim -\epsilon$ ; the integrand behaves like $\sim -\frac{\epsilon}{\epsilon^{3/2}}=-\frac{1}{\sqrt\epsilon}$ and is, therefore, integrable – Svyatoslav Jul 11 '21 at 20:38
1

As $x\to 1^-$, $\log(x)=O(x-1)$. – Mark Viola Jul 11 '21 at 20:38
If $x\to 0^+$, the integrand is$\sim \frac{\log (x)}{\sqrt{x}}$ and $\int \frac{\log (x)}{\sqrt{x}},dx=2 \sqrt{x} (\log (x)-2)$ – Claude Leibovici Jul 12 '21 at 04:39

Prove that $\displaystyle\int^1_0\frac{\ln(x)}{\sqrt{x(1-x)^3}}dx$ converges.

0 Answers0