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Evaluate $$\lim_{n \to \infty}\int_0^1 n \log(1+ {(\frac xn)}^\alpha) dx$$ for some $0< \alpha \leq 1$.

Initially, I was thinking to prove the uniform convergence of $n \log(1+ {(\frac xn)}^\alpha)$ so that I can interchange integration and limit but later I observed that for $0< \alpha < 1$ $n \log(1+ {(\frac xn)}^\alpha) \to \infty$ if $2\alpha>1$ and $n \log(1+ {(\frac xn)}) \to x$ just by expanding the series so it will be complicated in that sense. Is there any way out to solve it easily?

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    $\begingroup$ it would help noticing that $y-y^2/2\le\ln(1+y)\le y$ for $|y|<1$ $\endgroup$ – Masacroso Jun 13 '19 at 23:25
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First show that $$ h\ge\log(1+h)\ge \frac{h}{1+h} \tag{1} $$ for $h\ge 0$, by integrating $$ 1\ge\frac{1}{1+t}\ge \frac{1}{(1+t)^2} $$ in $[0,h]$.

Case 1. $a=1$.

Then, (1) implies that $$ x=n\cdot\frac{x}{n}\ge n\log\left(1+\frac{x}{n}\right)\ge n\cdot \frac{\frac{x}{n}}{1+\frac{x}{n}}=\frac{x}{1+\frac{x}{n}}=x-\frac{x^2}{n+x}\ge x-\frac{1}{n} $$ and thus $$ \frac{1}{2}=\int_0^1x\,dx\ge\int_0^1 n\log\left(1+\frac{x}{n}\right)\,dx\ge \int_0^1x\,dx-\frac{1}{n}=\frac{1}{2}-\frac{1}{n}\to \frac{1}{2}. $$

Case 2. $0<a<1$.

Hence $$ n\log\left(1+\frac{x^a}{n^a}\right)\ge n\cdot\frac{\frac{x^a}{n^a}}{1+\frac{x^a}{n^a}}=\frac{nx^a}{n^a+x^a}\ge \frac{nx^a}{n^a}=n^{1-a}x^a, $$ and thus $$ \int_0^1 n\log\left(1+\frac{x^a}{n^a}\right)\,dx\ge n^{1-a}\int_0^1 x^a\,dx=\frac{n^{1-a}}{a+1}\to \infty, $$ as $n\to\infty$.

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  • $\begingroup$ For $\alpha =1$ the limit is $\frac 1 2$ $\endgroup$ – Kavi Rama Murthy Jun 13 '19 at 23:38
  • $\begingroup$ @KaviRamaMurthy See my modified answer. $\endgroup$ – Yiorgos S. Smyrlis Jun 13 '19 at 23:41
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Hint: denote $\epsilon=\frac{1}{n}$, $t=\epsilon x$, then $$ \int_0^1n\log(1+(x/n)^\alpha)\,dx=\frac{\int_0^\epsilon\log(1+t^\alpha)\,dt}{\epsilon^2}. $$ Now apply L'Hôpital's rule ($\epsilon\to 0$).

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The limit is $\frac 1 2$ for $\alpha =1$ and $\infty$ for $\alpha <1$. Both of these follows from the fact that $\frac {\log(1+t)} t \to 1$ as $ t \to 0$. [Note that $(\frac x n)^{\alpha} \to 0$ uniformly for $x \in [0,1]$].

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  • $\begingroup$ Let me know if you want more details. $\endgroup$ – Kavi Rama Murthy Jun 13 '19 at 23:36

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