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Find the following limit, if it exists, and justify your calculation: $$\lim_{n\to \infty}\int_{0}^{n}(1+\frac{x}{n})^{-n}\log(2+\cos(\frac{x}{n}))\mathrm{d}x$$

where $\log$ denotes the natural logarithm.

This is a homework problem, and we have proved several theorems that may help with this - namely the Monotone and Dominated Convergence Theorems, as well as the fact that Riemann and Lebesgue integrals agree (provided that the integrand is both Riemann and Lebesgue integrable).

I have been thinking about the MCT and DCT. I thought there would be a problem with the fact that the limit as $n\to \infty$ is affecting the upper limit of the integral, but looking again at the theorem statements, they really do not say much about the particular set that one integrates over. However, if I take the limit inside of the integral and then integrate the result, I would be left with an answer which depends on $n$, and that clearly wouldn't happen if this was done as a Riemann integral.

Any hints would be appreciated, but since this is a homework problem, I would like only hints to be given, no solutions. Thanks!

EDIT:

Let $f_n(x)=(1+\frac{x}{n})^{-n}\log(2+\cos(\frac{x}{n}))$. Then $f_n(x)\rightarrow log(3)e^{-x}:=f(x)$ for all $x\in(0,\infty)$.

Since $1<\log(3)<2$, $|f_n(x)|=f_n(x)\leq f(x) \leq 2e^{-x}$ for all $x\in (0,\infty)$. Also, $2e^{-x}$ is Lebesgue integrable over $(0,\infty)$, since $\int_{(0,\infty)}2e^{-x}\mathrm{d}m=2$ (where $m$ is the Lebesgue measure).

Thus, by DCT, $$\lim_{n\to \infty}\int_{0}^{n}(1+\frac{x}{n})^{-n}\log(2+\cos(\frac{x}{n}))\mathrm{d}x=\lim_{n\to \infty}\int_{(0,\infty)}\chi_{(0,n)}(1+\frac{x}{n})^{-n}\log(2+\cos(\frac{x}{n}))\mathrm{d}m=\log(3)\int_{(0,\infty)}\chi_{(0,n)}e^{-x}\mathrm{d}m$$

Am I on the right track with this? Is this now simply a Riemann integral, and how does the characteristic function change the calculation?

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    $\begingroup$ Are you sure the exponent is $-1$ and not $-n$? (With $-1$ it is actually not that hard to present an explicit lower bound that implies your integral is $+\infty$. But with $-n$ you have a problem related to the exponential function which is more interesting.) $\endgroup$ – Ian Oct 12 '15 at 14:25
  • $\begingroup$ Yes, it is a {-n}, thank you! I see how it relates to the exponential, and in fact I evaluated the limit inside, but I can't justify interchanging the limit and integral yet $\endgroup$ – Ducky Oct 12 '15 at 14:52
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First I should mention that there is already a complete solution to a very similar problem on this site and it can be found here.(it contains more than just a hint). Since the request here is different here is a hint.

You can replace the upper bound of integration with $\infty$ multiplying the integrand by the bounded function $\chi_{(0,n)}$. Understanding the behavior of the $\log$ term should not be a problem once you look closely at what is inside. As Ian noticed in the comments section above, $$\Big(1 + \frac{x}{n}\Big)^n \uparrow e^x.$$ Luckily the exponential term in your question is raised to the $-1$.


EDIT (to answer the edited question): you are almost there, but it is still not quite correct. First, the major flaw in your reasoning is you don't keep track of the dependence on $n$ of the characteristic function when you send $n$ to infinity. Once you take the limit on $n$ that variable is gone and it should not appear again.

The second problem is more subtle: you say that $f_n(x) \le 2e^{-x}$ since $\log(3) < 2$, but notice that

$$\Big(1 + \frac{x}{n}\Big)^{-n} \downarrow e^{-x},$$ so what you can say is that that estimate is satisfied for $n$ sufficiently large (you can make this precise if you wish so).

I'd suggest to start over again considering $$f_n(x) = \Big(1 + \frac{x}{n}\Big)^{-n}\log(2 + \cos(x/n))\chi_{(0,n)}(x).$$

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  • $\begingroup$ Thank you for the answer! I'm not sure if I understand your first point; if $f_n(x)$ is the integrand, we can rewrite the integral as such? $\int_{0}^{\infty}\Chi_(0,n)(x)f_n(x)$? $\endgroup$ – Ducky Oct 12 '15 at 16:55
  • $\begingroup$ Exactly: as you mentioned, limit theorems don't explicitly deal with the case in which the domain of integration changes. The reason for this is that you can always integrate over the whole space, but the price you pay for that is you need to introduce a characteristic function that multiplies the integrand. In your case you can define $g_n(x) = \chi_{(0,n)}(x)f_n(x)$ and use the appropriate limit theorem. I hope this clarifies. :) $\endgroup$ – Giovanni Oct 12 '15 at 17:01
  • $\begingroup$ Thanks again for the clarification. I edited my original post and I think I'm on the right track, although it seems my answer will still have an $n$ in it, due to the characteristic function, so I'm skeptical about my work $\endgroup$ – Ducky Oct 12 '15 at 20:50
  • $\begingroup$ Sorry but I guess I am feeling dense tonight. $\log(3)e^{-x}$ won't work for a dominating function, since the DCT requires the sequence to be dominated for each $n$. Or is it sufficient for the sequence to be dominated for "most" $n$, i.e. for $n$ sufficiently large? $\endgroup$ – Ducky Oct 12 '15 at 23:48
  • $\begingroup$ It is sufficient to have the estimate for $n$ large: you don't care about the first terms of the sequence when computing the limit and you can always pretend your sequence started from the large value of $n$ as above (and perhaps relabel the sequence, but it is not needed). @Ducky $\endgroup$ – Giovanni Oct 12 '15 at 23:54

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