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I'd like to find

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

I am wondering if it is true that it converges to $\ln2$?

thanks in advance

user62498
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2 Answers2

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From L'Hospital's Rule we have

$$\begin{align} \lim_{n\to \infty}\frac1n\int_0^n\frac{x\log\left(1+\frac xn\right)}{1+x}\,dx&=\lim_{n\to \infty}\left(\frac{n\log(2)}{n+1}-\frac1n\int_0^n \frac{x^2}{(1+x)(n+x)}\right)\\\\ &=\log(2)-\lim_{n\to \infty}\frac1{n(n-1)}\int_0^n\left(\frac{nx}{n+x}-\frac{x}{1+x}\right)\,dx\\\\ &=\log(2)-\lim_{n\to \infty}\frac{1}{n(n-1)}\left.\left((n-1)x-n^2\log(x+n)+\log(x+1)\right)\right|_{x=0}^{x=n}\\\\ &=\log(2)-\lim_{n\to \infty}\frac{n(n-1)-n^2\log(2)+\log(n+1)}{n(n-1)}\\\\ &=2\log(2)-1 \end{align}$$

Mark Viola
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The involved integral equals $$ \int_{0}^{1}\log(1+x)\frac{nx}{nx+1}\,dx $$ that by the dominated convergence theorem converges to $$ \int_{0}^{1}\log(1+x)\,dx = \color{red}{\log\left(\frac{4}{e}\right)} $$ as $n\to +\infty$.

Jack D'Aurizio
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  • @dear Jack D'Aurizio thanks you answer, would you mind explaining your answer a little more – user62498 Feb 27 '17 at 20:53
  • Given that the tag is real analysis, dominated convergence theorem is a bit powerful – Brevan Ellefsen Feb 27 '17 at 20:53
  • @user62498: the first step is to perform the substitution $x\to nx$, the second step is to apply the dominated convergence theorem. The sequence of functions $f_n(x)=\frac{nx}{1+nx}$ is pointwise convergent to $1$ on the interval $(0,1)$. – Jack D'Aurizio Feb 27 '17 at 20:55
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    @BrevanEllefsen: true, but this is the most simple answer. Otherwise we may apply squeezing: the idea stays pretty the same, the proof gets longer. – Jack D'Aurizio Feb 27 '17 at 20:56
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    While the DCT is very powerful, it also has the advantage of... working, and being quite systematic. If one knows it, refraining from using it in favor of ad hoc, non-generalizable, and longer arguments, just for the sake of it, is a tad odd. (Now, if it's not known, or explicitly forbidden to rely on the DCT for an exercise, sure.) – Clement C. Feb 27 '17 at 20:59