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Can someone identify for me the value of this expression and prove it? $$\lim_{n\rightarrow\infty}{\int^{\infty}_{-\infty}{e^{-x^n}} dx}$$

where $n$ is an even positive integer.

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Visibly, i can see that it approaches 2 units but i require proof – Keith Afas Sep 3 '13 at 3:25
What's the domain of $n$? The integral doesn't even converge for many values of $n$. – aschepler Sep 3 '13 at 3:29
$$\{n|\exists k\in N\wedge n=2k \wedge n>0\}$$ – Keith Afas Sep 3 '13 at 3:40
@KeithAfas It would be tidier if you just considered $lim_{n\rightarrow\infty} {\int^{\infty}_{-\infty}{e^{-x^{2n}}} dx} $ instead. – David H Sep 3 '13 at 3:45
The condition that $n$ is even should be incorporated into the question. Comments can be deleted and are not really part of the question. – robjohn Sep 3 '13 at 9:22
up vote 22 down vote accepted

We suppose $n$ is even, else the integral does not converge.


So in the limit we have


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Assuming $n=2k$ is even, by Dominated Convergence, we have $$ \begin{align} \lim_{k\to\infty}\int_{-\infty}^\infty e^{-x^{2k}}\,\mathrm{d}x &=\lim_{k\to\infty}\overbrace{\int_{-\infty}^{-1}e^{-x^{2k}}\,\mathrm{d}x}^{\text{dominated by $e^{-x^2}$}} \hspace{-8mm}&&+\lim_{k\to\infty}\overbrace{\int_{-1}^1e^{-x^{2k}}\,\mathrm{d}x}^{\text{dominated by $1$}} \hspace{-8mm}&&+\lim_{k\to\infty}\overbrace{\int_{1}^\infty e^{-x^{2k}}\,\mathrm{d}x}^{\text{dominated by $e^{-x^2}$}}\\ &=\hphantom{\lim_{k\to\infty}}\int_{-\infty}^{-1}0\,\mathrm{d}x &&+\hphantom{\lim_{k\to\infty}}\int_{-1}^11\,\mathrm{d}x &&+\hphantom{\lim_{k\to\infty}}\int_{1}^\infty 0\,\mathrm{d}x\\[6pt] &=\hphantom{\lim_{k\to\infty}}0 &&+\hphantom{\lim_{k\to\infty}}2 &&+\hphantom{\lim_{k\to\infty}}0 \end{align} $$

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excellent simplification – Steven Kramer Sep 3 '13 at 13:31

I like robjohn's answer, as it does not require to use the gamma function. If you are wondering how you could come up with the idea of dividing the integral at points –1 and 1, plot the functions $f(x) = \exp(-x^n)$ for increasing values of $n$ (even):

enter image description here

You see that it tends to zero for $x<-1$ and $x>1$, and to $1$ in the interval $[-1,1]$.

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This does not really address the main question, it is more of a comment. However, you cannot include images in comments, so to prevent flags saying that this is not an answer, I usually preface these comments-as-answers with something like "This is really a comment on ... but images cannot be included in comments", or "this was too long for a comment", etc. – robjohn Sep 3 '13 at 14:50
This also gives a geometric intuition on why the integral has a value of $2$: it looks like a two unit by 1 unit rectangle... :) – apnorton Oct 15 '13 at 0:30

Hint: Make the change of variable $x^{2m}=t$ and then use the gamma function. Then to find the limit use Stirling approximation.

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According to Jared's answer, all you need is $\Gamma(1)$ - there is no need for Stirling's approximation. – marty cohen Sep 3 '13 at 4:05
@martycohen: there is no need for $\Gamma$ at all :-) – robjohn Sep 3 '13 at 6:05
@martycohen: As you see it is a hint for the OP and it is the same technique like Jared's one. No need for Stirling approximation, you can see that when you workout the problem but as I said it is a hint. Thanks for the comment. – Mhenni Benghorbal Sep 3 '13 at 17:46
@robjohn: It is a different approach to the problem! I can see that the OP accepted this approach maybe because of their background. By the way, you could've used uniform convergence which I think that the OP has been exposed to it. – Mhenni Benghorbal Sep 3 '13 at 17:52
@robjohn: Here are some answers (I), (II). – Mhenni Benghorbal Sep 3 '13 at 18:01

We could simply apply the famous Gaussian identity : $$\boxed{\displaystyle\quad\int_0^\infty e^{-x^n} = \frac1n ! \qquad\iff\qquad I = 2\,\left(\frac1\infty\right) ! = 2\cdot 0\,! = 2\cdot1 = 2\quad}$$ since for an even n, the graphic is symmetrical with regards to the vertical axis Oy, $(-t)^{2k} = t^{2k}$ , and our integral thus becomes : $$I = \int_{-\infty}^\infty\ e^{-x^n} = \int_{-\infty}^\infty\ e^{-x^{2k}} = 2\int_0^\infty\ e^{-x^{2k}}$$

Otherwise, we might directly compute the limit at $\infty$ as follows : $$I = 2\int_0^\infty\ e^{-x^{2\infty}} = 2\,\left(\,\int_0^1\ e^{-x^\infty} + \int_1^\infty\ e^{-x^\infty}\right) = 2\,\left(\,\int_0^1\ e^{-0} + \int_1^\infty\ e^{-\infty}\right)$$ $$\quad\qquad = 2\,\left(\,\int_0^1 1\ dx + \int_1^\infty 0\ dx\right)\quad = 2\,(1\cdot1 + 0) = 2$$ since $x^\infty = 0$ , for $|x| < 1$ , and $x^\infty = \infty$ , for $|x| > 1$ .

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$\frac{1}{\infty} \ne 0$, really... (first boxed result) – apnorton Oct 15 '13 at 0:31

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