compute $\lim_{n \to \infty} \int_{[0,1]} \frac{(\ln x)^n}{\sqrt{1-x^2}} \, dx$

$$I = \lim_{n \to \infty}I_n = \lim_{n \to \infty} \int_{[0,1]} \frac{(\ln x)^n}{\sqrt{1-x^2}} \, dx$$

since we have $$(\ln x)^n \leq x \, \forall x \in [0,1], n\geq 1$$ then I could show that $$I_n \leq 1$$

I wanted then to use the DCT but $$\lim_{n \to \infty} (\ln x)^n$$ doesn't exist for positive $$x$$ less than $$1$$.

I'm a bit lost, anyone can give me hints or tell me what to do ?

• $\ln x<0$ in $(0,1)$. – A.Γ. Dec 17 '18 at 13:41
• @A.Γ. but that just implies that the L-1 norm is zero. not the limit itself – rapidracim Dec 17 '18 at 13:47

Note that $$-\log x\geq 1-x\geq0$$ on $$[0,1]$$. Thus your assertion that $$I_n \leq 1$$ is wrong. We may write $$(-1)^nI_n = \int_0^{\frac{1}{e}}\frac{(-\log x)^n}{\sqrt{1-x^2}}dx + \int_{\frac{1}{e}}^1\frac{(-\log x)^n}{\sqrt{1-x^2}}dx .$$ We can observe that for $$x\in (0,\frac{1}{e})$$, it holds $$1\leq (-\log x)^n \uparrow \infty,$$ and $$1\geq (-\log x)^n \downarrow 0$$ for $$x\in (\frac{1}{e},1]$$. By Lebesgue's dominated convergence theorem and monotone convergence theorem, it follows that $$\lim_{n\to\infty} (-1)^n I_n = \infty.$$
A shorter proof uses $$\int_0^1\frac{(-\ln x)^n dx}{\sqrt{1-x^2}}\ge\int_0^1(-\ln x)^n dx=\int_0^\infty y^n e^{-y}dy=n!$$.
• I agree with you that either the limit is $0$ or the limit does not exists, but unfortunately, plotting a graph is not a proof. – nicomezi Dec 17 '18 at 13:49