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

I've been able to show that the integral is bounded above by 1 (several ways). One of the simplest is just letting $u=x/n$ and doing some bounding above. It must converge to something (if it is monotone increasing.. don't think so), but all my efforts to use Lebesgue Dominated Convergence have failed. I can't seem to dominate this function...

However, if I could dominate it the integrand converges pointwise to 0 and so then the integral would be 0, which I think is incorrect.

I have found it sufficient to deal with $$\lim_{n\to\infty}\int_{0}^{n}\frac{\sin(x/n)}{(1+x/n)^{n}}\, dx$$

I tried to turn it into a series and use the converse of the integral test, but that hasn't went well.

I've used Egoroff's theorem to try to use uniform convergence but that failed.

Some of the inequalities and ideas I've tried to use in a solution include:

$sin(x/n)\leq x/n$

after a $u$ sub I've tried to turn it into a riemann sum... failed

$(1+x/n)^{n}\geq 1+x$ and the binomial theorem

$(1+x/n)^{n}$ converges to $e^x$ and it is increasing in $n$ so $e^{-x}\leq \frac{1}{(1+x/n)^{n}}$

I've tried to use that the integrand is measurable and subtract out different measurable sets from the integral to make it more convenient..

I've been playing with this on and off for a month or so and no luck. I appreciate any help here.

Update: From mixedmath's suggestion

splitting it into $[0,10]$ and $[10,\infty]$ it is clear that the integrand is dominated on $[0,10]$ (say by 1) and so lebesgue sends that portion to 0.

On the larger interval I am still running into a problem making it bounded nicely. If I try a $u$ sub the bounds get messy. I would like to choose an $n$ so that

$2e^{-x} \leq (1+x/n)^n$ for all $x\in [10,\infty]$ .. maybe:

there exists an $N$ such that for all $n\geq N$

$e^{10}/2 < (1+10/n)^{n}$

but for $x>10$..

$e^{10}/2<(1+10/n)^{n}<(1+x/n)^{n}$

but that's not quite right..

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Just an idea: Perhaps you could break the interval into two parts, say $[0,10]$ and $[10,\infty]$. It's not so bad to show that the limit on $[0,10]$ is $0$. For the rest, maybe bound above by something like $\frac{2}{e^x}$ or something. –  mixedmath May 23 '12 at 2:34
    
I had tried something like that, only the bounds I used were smaller, like [0,2]. Maybe that's what caused a problem. I'll try it now. –  toypajme May 23 '12 at 2:35

2 Answers 2

up vote 3 down vote accepted

Let $$f_n(x) = \frac{\sin(x/n)}{(1+x/n)^n}.$$ Notice that $\lim_{n\to\infty} f_n(x) = 0$. But $$|f_n(x)| \leq \frac{1}{(1+x/n)^n} \leq \frac{1}{(1+x/2)^2}$$ for $n\ge 2.$ So, for $n\ge 2$, $f_n(x)$ is dominated by $1/(1+x/2)^2$, which is integrable, $$\int_0^\infty dx \frac{1}{(1+x/2)^2} = 2.$$ Therefore, the dominated convergence theorem applies, $$\lim_{n\to\infty} \int_0^\infty \frac{\sin(x/n)}{(1+x/n)^n} dx = 0.$$

In the comments, @RobertIsrael gives another option for a dominating function.

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For $n\geq 2$, unless I'm going crazy, $1+x/n\leq 1+x/2\leq (1+x/2)^{2}$ and so $\frac{1}{1+x/n}\geq \frac{1}{1+x/2}^{2}$ or $\frac{1}{1+x/n}^{n}\geq \frac{1}{1+x/2}^{n}$. I'm not sure where we'd fit in the square and reverse that inequality. Care to shed some light? –  toypajme May 23 '12 at 4:09
    
@toypajme: Notice that $(1+x/n)^n = 1+x + \frac{n-1}{2n}x^2 + h.o.$. But $\frac{n-1}{2n}\ge \frac{1}{4}$ for $n\ge2$. Therefore, $(1+x/n)^n \ge (1+x/2)^2$. –  user26872 May 23 '12 at 4:25

Just for an alternate solution that avoids the Dominated Convergence theorem you can do this.

First show:

$$\lim_{n \to \infty} \int_0^{n\pi} \frac{\sin(x/n)}{(1 + x/n)^n} dx = \lim_{n \to \infty} \int_0^{\infty} \frac{\sin(x/n)}{(1 + x/n)^n} dx.$$

You can do this by looking at the difference between their limit, and an obvious bound. Then use the variable substitution $x/n \mapsto x$ to get:

$$\int_0^{n\pi} \frac{\sin(x/n)}{(1 + x/n)^n} dx = n \int_0^{\pi} \frac{\sin(x)}{(1 + x)^n} dx.$$

Now this integral is set up nicely for integration by parts ($\sin$ is 0 at the end points) to get:

$$n \int_0^{\pi} \frac{\sin(x)}{(1 + x)^n} dx = \frac{n}{n-1} \int_0^{\pi} \frac{\cos(x)}{(1 + x)^{n-1}} dx.$$

Now you can use Bernoulli's inequality, which says $(1 + x)^{n-1} \geq 1 + (n-1)x$ to get:

$$\int_0^{\pi} \frac{\cos(x)}{(1 + x)^{n-1}} \leq \int_0^{\pi} \frac{1}{1 + (n-1)x} dx$$.

which is easily seen to go to 0.

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