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

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?

• it would help noticing that $y-y^2/2\le\ln(1+y)\le y$ for $|y|<1$ – Masacroso Jun 13 '19 at 23:25

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.

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 $$n\to\infty$$.

• For $\alpha =1$ the limit is $\frac 1 2$ – Kavi Rama Murthy Jun 13 '19 at 23:38
• @KaviRamaMurthy See my modified answer. – Yiorgos S. Smyrlis Jun 13 '19 at 23:41

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 ($$\epsilon\to 0$$).

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 $$x \in [0,1]$$].

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