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fine the limit :

$$\lim_{ n \to \infty }\frac{1}{n}\int_{0}^{n}{ \frac{x\ln(1+\frac{x}{n})}{1+x}}=?$$

My Try: in the http://www.integral-calculator.com $$I=\int_{}^{}{ \frac{x\ln(1+\frac{x}{n})}{1+x}}=\left(x+n\right)\ln\left(\left|x+n\right|\right)+\left(\ln\left(n\right)-\ln\left(n-1\right)\right)\ln\left(\left|x+1\right|\right)+\operatorname{Li}_2\left(-\dfrac{x+1}{n-1}\right)+\left(-\ln\left(n\right)-1\right)+c$$

now?

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  • $\begingroup$ First idea: try the change of variable $u= \frac{x}{n}$, then apply the dominated convergence theorem? (If that's not enough, I can jot down a full answer -- my guess is that it works, and gives $2\ln 2-1$) $\endgroup$ – Clement C. Feb 24 '17 at 19:20
  • $\begingroup$ i have $$-1+2\ln(2)$$ as the searched answer $\endgroup$ – Dr. Sonnhard Graubner Feb 24 '17 at 19:27
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  • Step 1: make the change of variable $u=\frac{x}{n}$, so that $$ \frac{1}{n}\int_{0}^{n}{ \frac{x\ln(1+\frac{x}{n})}{1+x}} = \int_0^1 \frac{nu}{1+nu}\ln(1+u)du $$

  • Step 2: letting $(f_n)_n$ be defined on $[0,1]$ by $f_n(u) = \frac{nu}{1+nu}\ln(1+u)$, observe that (1) each $f_n$ is integrable on $[0,1]$, (2) $\lvert f_n(u)\rvert \leq g(u)\stackrel{\rm def}{=} \ln(1+u)$ for every $u\in[0,1]$ and $n\geq 1$, and (3) $(f_n)_n$ converges pointwise to $g$ on $(0,1)$.

    By the Dominated Convergence Theorem, it follows that $$ \int_0^1 f_n(u)du \xrightarrow[n\to\infty]{} \int_0^1 g(u)du $$

  • Step 3: compute $$ \int_0^1 g(u)du = 2\ln 2 -1. $$ observing that an antiderivative of $g$ is $G(x) = (1+x)\ln(1+x) - x$.

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  • $\begingroup$ $\int \ln(1+u) du=\frac{1}{u+1}$ but you$ G(x) = (1+x)\ln(1+x) - x$ why? $\endgroup$ – Almot1960 Feb 24 '17 at 20:54
  • $\begingroup$ @Almot1960 Err, no -- the first part of your comment is wrong. You want to integrate, not differentiate. $\endgroup$ – Clement C. Feb 24 '17 at 21:01
  • $\begingroup$ You're welcome! $\endgroup$ – Clement C. Feb 24 '17 at 21:20
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\lim_{n \to \infty}\bracks{% {1 \over n}\int_{0}^{n}{x\ln\pars{1 + x/n} \over 1 + x}\,\dd x} = \lim_{n \to \infty}\bracks{% {1 \over n}\int_{0}^{n}\ln\pars{1 + {x \over n}}\,\dd x - {1 \over n}\int_{0}^{n}{\ln\pars{1 + x/n} \over 1 + x}\,\dd x} \\[5mm] = &\ 2\ln\pars{2} - 1\ -\ \underbrace{\lim_{n \to \infty}\bracks{% {1 \over n}\int_{0}^{1}{\ln\pars{1 + x} \over 1/n + x}\,\dd x}}_{\ds{=\ 0}}\ =\ \bbx{\ds{2\ln\pars{2} - 1}} \end{align}

because

$$ 0 < \verts{{1 \over n}\int_{0}^{1}{\ln\pars{1 + x} \over 1/n + x}\,\dd x} < {1 \over \verts{n}}\verts{\int_{0}^{1}{\ln\pars{1 + x} \over \color{#f00}{0} + x}\,\dd x} \,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\to}\,\,\, {\large 0} $$

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  • $\begingroup$ Holy crap. This answer is gorgeous. +1!!!! $\endgroup$ – Brevan Ellefsen Feb 24 '17 at 21:11
  • $\begingroup$ @BrevanEllefsen Thanks. It's indeed... $\endgroup$ – Felix Marin Feb 24 '17 at 23:43
  • $\begingroup$ Pretty nice using elementary techniques +1 $\endgroup$ – Paramanand Singh Feb 25 '17 at 4:52
  • $\begingroup$ @ParamanandSingh Thanks... whenever is possible, simple is better. $\endgroup$ – Felix Marin Feb 25 '17 at 6:04
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Not a new idea, but avoiding dominated convergence:

$$\frac{1}{n} \int_0^n \frac{ x \ln (1+x/n) }{1+x} dx \underset{y=x/n}{=}\int_0^1 \ln (1+y)\frac{1+ny-1 }{1+ny} dy =\int_0^1 \ln (1+y) dy - \underset{(*)}{\underbrace{\int_0^1 \frac{\ln (1+y)}{1+ny}dy}}. $$

Now

$$0\le (*) \le \ln 2 \int_0^1 \frac{dy }{1+yn}=\ln 2 \frac{1}{n} \ln (1+n)\underset{n\to\infty}{\to} 0.$$

Therefore the answer is $\int_0^1\ln (1+y) dy=\ln 4 - 1$.

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  • $\begingroup$ why ?$0\le (*) \le \ln 2 \int_0^1 \frac{dy }{1+yn}$ $\endgroup$ – Almot1960 Feb 24 '17 at 22:03
  • $\begingroup$ $\ln 2 \ge \ln (1+y)$ for $y \in [0,1]$ $\endgroup$ – poweierstrass Feb 24 '17 at 22:48

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