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

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 ?

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  • $\begingroup$ $\ln x<0$ in $(0,1)$. $\endgroup$ – A.Γ. Dec 17 '18 at 13:41
  • $\begingroup$ @A.Γ. but that just implies that the L-1 norm is zero. not the limit itself $\endgroup$ – rapidracim Dec 17 '18 at 13:47
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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 $$ \lim_{n\to\infty} (-1)^n I_n = \infty. $$

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A shorter proof uses $\int_0^1\frac{(-\ln x)^n dx}{\sqrt{1-x^2}}\ge\int_0^1(-\ln x)^n dx=\int_0^\infty y^n e^{-y}dy=n!$.

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Graphing the initial few functions for integer-valued n, it is clear that the integral will change signs repeatedly, and be negative for odd n and positive for even n. As a result, I would suggest that the limit does not exist unless it converges at 0, which seems unlikely when plotting the graph for high n-values

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  • $\begingroup$ 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. $\endgroup$ – nicomezi Dec 17 '18 at 13:49
  • $\begingroup$ @nicomezi That's an entirely valid point. I'll have a think about how to phrase a proper proof, but I think a much better written proof than I could come up with has been supplied in another answer. $\endgroup$ – M.M. Dec 17 '18 at 13:55

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