Why $\lim_{n\rightarrow \infty}\int_0^1 e^{x^n} = 1$

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.

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

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.

• That was a great proof indeed. If Possibe could you share any proof using uniform convergence as I was trying? – manhattan Dec 24 '16 at 20:28
• @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. – Joey Zou Dec 24 '16 at 21:04

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 $$\limsup_{n\to \infty}\int_{0}^1e^{x^n}\mathrm{d}x\le \inf_{\delta>0}\limsup_{n\to \infty}\left(e^{(1-\delta)^n}+\delta e\right)\le \inf_{\delta>0}(e^{\delta}+3\delta)\le 1.$$

• 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. – manhattan Dec 24 '16 at 19:56
• 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 $n$. – Paolo Leonetti Dec 24 '16 at 19:57
• 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. – manhattan Dec 24 '16 at 20:02
• 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 $(1-δ)^n\leδ$. – LutzL Dec 24 '16 at 21:32
• @LutzL you are right, thanks for the correction! – Paolo Leonetti Dec 24 '16 at 21:57

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.