I'm trying to find the domain of this integral function: $$F(X)=\int_0^{x}\frac{1}{\ln({2-t^2})}dt$$

I know that the the integrand function's domain is $(-\sqrt2;-1) \cup (-1;1)\cup(1;+\sqrt2)$, so I have to discuss those points.

  • In $x\to\pm1$, the function is equal to $\frac1{\epsilon}$, where the infinitesimal is the one of the logarithm, so the order (standard: x, with $x\to0$) is less than $1$, so the integral converges.
  • In $x\to\pm\sqrt2$, the functions is equal to $\frac1{\infty}$, so the integral converges.

I don't really understand why I am wrong: the integral diverges in $-1$ and $+1$ (according to Desmos), so the domain isn't $(-\sqrt2;+\sqrt2)$, but $(-1;+1)$. Thanks so much for explanation.

  • $\begingroup$ Thanks, I just saw that I wrote $+\infty$ and not x. Fixed it. $\endgroup$ – L. Repetti Jan 28 '17 at 17:29
  • $\begingroup$ Yes, I wrote that, but isn't that log(1) an infinitesimal less than 1, so the integral converges? Like, $\int_0^1\frac1{x^{1/2}}$ for x->0, that converges? $\endgroup$ – L. Repetti Jan 28 '17 at 17:31

The domain of the integrand is indeed what you write. We can restrict to $x>0$, because of symmetry.

In order to see what happens for $1<x<\sqrt{2}$, we need to see whether both $$ \int_0^1\frac{1}{\ln(2-t^2)}\,dt \qquad\text{and}\qquad \int_1^x\frac{1}{\ln(2-t^2)}\,dt $$ converge.

Now $\ln a\le a-1$, so $\ln(2-t^2)\le 1-t^2$ and therefore, over $[0,1)$, $$ \frac{1}{\ln(2-t^2)}\ge\frac{1}{1-t^2} $$ and $$ \int_0^1\frac{1}{1-t^2}\,dt $$ diverges.

  • $\begingroup$ Is this one of the "case" that make us say that the improper integral, in a limited interval, converges always and if it is $+\infty$, it converges if the $+\infty$ is less than k less than 1? Is this k the thing here? We can't be sure if this $\ln{2-t^2}$ is or isn't in that range? $\endgroup$ – L. Repetti Jan 29 '17 at 10:03
  • $\begingroup$ @L.Repetti $\ln(1+x)$ at zero is like $x$ $\endgroup$ – egreg Jan 29 '17 at 10:13
  • $\begingroup$ Oh so I could do this taking $2$ and splitting log, and have pretty much same result!thanks, I really understood! $\endgroup$ – L. Repetti Jan 29 '17 at 10:23

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