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I want to show that

$$\lim_{n \rightarrow \infty} n^2 \int_0^{2n} e^{-n \vert x-n \vert} \log \left[ 1+ \frac{1}{x+1} \right]dx=2.$$

The reason I think this is the limit is because if we make the substitution $y = n(x-n)$, then we obtain

$$ \int_{-n^2}^{n^2} e^{-\vert y \vert} \log \left[ \left( 1 + \frac{n}{n^2+n+y} \right)^n \right] dx. $$

Then if I can show $ \lim_{n \rightarrow \infty} \left( 1 + \frac{n}{n^2+n+y} \right)^n = e$, dominated convergence should give the rest. Certainly we have

$$\limsup_{n \rightarrow \infty} \left( 1 + \frac{n}{n^2+n+y} \right)^n \leq e$$

Is there a way to get the reverse inequality for the $\liminf$?

Philip Hoskins
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  • Why isn't that as easy as I believe? You are computing the limit of an elementary sequence, since $y$ is fixed. – Siminore Jul 19 '15 at 16:54
  • Probably because I am not clever enough to come up with a trick to get a useful lower bound. – Philip Hoskins Jul 19 '15 at 17:02
  • What I meant is that it looks like a limit from elementary calculus (take a log, discard lower-order terms, reduce to $e$) – Siminore Jul 19 '15 at 17:17
  • Since $\left(1+\frac n{n^2+n+y}\right)^n=2^n$ when $y=-n^2$, you pretty much need to break up your integral to use Dominated Convergence. – robjohn Jul 19 '15 at 17:51
  • But then $y$ would not be fixed... – Philip Hoskins Jul 19 '15 at 17:55
  • Ah, I see, you're talking about the dominating function. – Philip Hoskins Jul 19 '15 at 18:01
  • @PhilipHoskins: To apply Dominated Convergence, you need the family of functions whose limit you are taking to be dominated by one single integrable function. In this case no single multiple of $e^{-|y|}$ will do – robjohn Jul 19 '15 at 18:02

2 Answers2

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$$ \begin{align} &n^2\int_0^{2n}e^{-n|x-n|}\log\left(1+\frac1{x+1}\right)\,\mathrm{d}x\\ &=n^2\int_{-n}^ne^{-n|x|}\log\left(1+\frac1{x+n+1}\right)\,\mathrm{d}x\tag{1}\\ &=n\int_{-n^2}^{n^2}e^{-|x|}\log\left(1+\frac1{x/n+n+1}\right)\,\mathrm{d}x\tag{2}\\ &=n\int_{-n}^{n^2}e^{-|x|}\log\left(1+\frac1{x/n+n+1}\right)\,\mathrm{d}x\tag{3a}\\ &+n\int_{-n^2}^{-n}e^{-|x|}\log\left(1+\frac1{x/n+n+1}\right)\,\mathrm{d}x\tag{3b}\\ \end{align} $$ Explanation:
$(1)$: substitute $x\mapsto x+n$
$(2)$: substitute $x\mapsto x/n$
$(3)$: split integral at $x=-n$
$\hphantom{(3)\text{:}}$ in $\text{(3a)}$, $n\log\left(1+\frac1{x/n+n+1}\right)\le \log\left(1+\frac1n\right)^n\le\log(e)=1$
$\hphantom{(3)\text{:}}$ in $\text{(3b)}$, $n\log\left(1+\frac1{x/n+n+1}\right)\le n\log(2)$ and $\int_{-n^2}^{-n}e^{-|x|}\,\mathrm{d}x\le e^{-n}$

Thus, by Dominated Convergence, taking the limit of $\text{(3a)}$, $$ \begin{align} \lim_{n\to\infty}n\int_{-n}^{n^2}e^{-|x|}\log\left(1+\frac1{x/n+n+1}\right)\,\mathrm{d}x &=\int_{-\infty}^\infty e^{-|x|}\,\mathrm{d}x\\[6pt] &=2\tag{4} \end{align} $$ since $\lim\limits_{n\to\infty}\left(1+\frac1{x/n+n+1}\right)^n=e$.

Furthermore, by the estimates given in the Explanation above, $$ \begin{align} \lim_{n\to\infty}n\int_{-n^2}^{-n}e^{-|x|}\log\left(1+\frac1{x/n+n+1}\right)\,\mathrm{d}x &\le\lim_{n\to\infty}ne^{-n}\log(2)\\ &=0\tag{5} \end{align} $$ Adding $(4)$ and $(5)$, we get $$ \bbox[5px,border:2px solid #C0A000]{\lim_{n\to\infty}n^2\int_0^{2n}e^{-n|x-n|}\log\left(1+\frac1{x+1}\right)\,\mathrm{d}x=2}\tag{6} $$

robjohn
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  • To apply Dominated Convergence, we need to have a family of functions bounded by a single integrable function. Considering the integral in $(2)$ above, it is not bounded by any constant multiple of $e^{-|x|}$ since $n\log\left(1+\frac1{x/n+n+1}\right)$ gets to be $n\log(2)$ when $x=-n^2$. – robjohn Jul 19 '15 at 17:58
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To answer your last question : $$n\ln\left(\frac{n}{n^2+n+y}\right)=n\frac{n}{n^2+n+y}+O\left(n^{-1}\right)$$ when $n\gg 1$, and $$\lim_{n\to+\infty}\frac{n^2}{n^2+n+y}=1.$$ Now take the exponential to obtain $\mathrm{e}$ as the limit.

Nicolas
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