3
$\begingroup$

We have to find $\lim_{n\rightarrow \infty}\int_0^1 e^{x^n}$ dx and also justify the existence.

Answer is 1. Writing it rigorously (to justify existence) is a little problem for me. I have tried through uniform convergence and attached.

My problem is that the proof which I have written may not be right as the N is dependent on 'b' which should not be the case in uniform convergence. Kindly help suggesting a better proof.

enter image description here

$\endgroup$
  • 1
    $\begingroup$ Can you use dominated convergence? $\endgroup$ – user223391 Dec 24 '16 at 19:40
  • $\begingroup$ Can you move the limit inside the integral sign? $\endgroup$ – Jacob Wakem Dec 24 '16 at 19:40
  • $\begingroup$ I want to use the definition to get the existence. $\endgroup$ – manhattan Dec 24 '16 at 19:41
  • $\begingroup$ @ alephnull that needs a justification which I am looking for, and which I am trying to prove $\endgroup$ – manhattan Dec 24 '16 at 19:42
  • $\begingroup$ @manhattan The integral is a limit of reimann sums. Maybe you can exchange limits that way. $\endgroup$ – Jacob Wakem Dec 24 '16 at 19:43
6
$\begingroup$

To show $\lim\limits_{n\rightarrow\infty}{\int\limits_{0}^{1}{e^{x^n}\,dx}} = 1$, it suffices to show that $\lim\limits_{n\rightarrow\infty}{\int\limits_{0}^{1}{(e^{x^n}-1)\,dx}} = 0$. Clearly, we have $\int\limits_{0}^{1}{(e^{x^n}-1)\,dx}\ge 0$. On the other hand, for any $0\le y\le 1$, we have $$e^y-1 = \int\limits_{0}^{y}{e^t\,dt}\le \int\limits_{0}^{y}{e\,dt} = ey$$ since, for $0\le t\le y\le 1$, we have $e^t\le e$. It follows that $e^{x^n}-1\le ex^n$ for all $0\le x\le 1$ for any $n$, and so $$ \int\limits_{0}^{1}{(e^{x^n}-1)\,dx} \le \int\limits_{0}^{1}{ex^n\,dx} = \frac{e}{n+1}.$$ It follows that $$ 0\le \int\limits_{0}^{1}{(e^{x^n}-1)\,dx} \le \frac{e}{n+1} $$ and so by the Squeeze theorem we have $\lim\limits_{n\rightarrow\infty}{\int\limits_{0}^{1}{(e^{x^n}-1)\,dx}} = 0$, as desired.

$\endgroup$
  • $\begingroup$ That was a great proof indeed. If Possibe could you share any proof using uniform convergence as I was trying? $\endgroup$ – manhattan Dec 24 '16 at 20:28
  • $\begingroup$ @manhattan since $e^{x^n}$ does not converge uniformly to $1$ on $(0,1)$, I'm not sure what you mean by "uniform convergence". For what it's worth, I think the proof in your picture is also correct. $\endgroup$ – Joey Zou Dec 24 '16 at 21:04
5
$\begingroup$

Clearly $e^{x^n} \ge 1$, hence the integral is $\ge 1$ for all $n$. On the other hand, we have for each $\delta>0$ $$ \int_{0}^1e^{x^n}\mathrm{d}x\le \int_{0}^{1-\delta}e^{x^n}\mathrm{d}x+\delta e \le e^{(1-\delta)^n}+\delta e. $$

Now, $(1-\delta)^n<\delta$ if $n$ is sufficiently large, then $$ \limsup_{n\to \infty}\int_{0}^1e^{x^n}\mathrm{d}x\le \inf_{\delta>0}\limsup_{n\to \infty}\left(e^{(1-\delta)^n}+\delta e\right)\le \inf_{\delta>0}(e^{\delta}+3\delta)\le 1. $$

$\endgroup$
  • $\begingroup$ the n which you choose to get $e^{ \delta ^n}$ close to 1 is dependent on $\delta$ which I did, hence the convergence is not uniform. $\endgroup$ – manhattan Dec 24 '16 at 19:56
  • $\begingroup$ I didn't read what you wrote. But it is enough to note that $e^x=1+O(x)$ as $x \to 0$. Hence $e^{\delta^n}=1+O(\delta^n)=1+O(\delta)$. This does not depend on $n$. $\endgroup$ – Paolo Leonetti Dec 24 '16 at 19:57
  • $\begingroup$ Sir, may be $e^{\delta ^n}-1 = O(\delta ^n)$ and for $ O(\delta ^n) \le \epsilon$ we require dependence of n on $\delta$. Sorry, but I am not looking for a proof in this context but without using the expansion. Thanks...please suggest a better idea if you come across. $\endgroup$ – manhattan Dec 24 '16 at 20:02
  • $\begingroup$ How do you get for $x\in[0,1-δ]$ that $x^n<δ^n$? I would expect $x^n\le(1-δ)^n$ and if $n$ is large enough, you can get $(1-δ)^n\leδ$. $\endgroup$ – LutzL Dec 24 '16 at 21:32
  • $\begingroup$ @LutzL you are right, thanks for the correction! $\endgroup$ – Paolo Leonetti Dec 24 '16 at 21:57
3
$\begingroup$

All we need is the inequality $e^u\le 1+3u, u \in [0,1].$ (To prove it, note they agree at $u=0$ and the derivative of the right side is $3,$ which is greater than the derivative of $e^u$ on this interval.)

Thus

$$1 \le \int_0^1e^{x^n}\, dx \le \int_0^1(1+3x^n)\, dx = 1 + 3/(n+1).$$

The limit is therefore $1$ by the squeeze theorem.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.