Curious Integral Proof

Question

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.

Answer 1

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

$$\int_{-\infty}^{\infty}e^{-x^{2k}}dx=2\int_0^{\infty}e^{-x^{2k}}dx=\frac{1}{k}\int_{0}^{\infty}t^{\frac{1}{2k}-1}e^{-t}dt=\frac{1}{k}\Gamma\left(\frac{1}{2k}\right)=2\Gamma\left(\frac{2k+1}{2k}\right)$$

So in the limit we have

$$\lim_{k\to\infty}\int_{-\infty}^{\infty}e^{-x^{2k}}dx=2\Gamma(1)=2$$

Answer 2

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

Answer 3

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

Answer 4

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):

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

Answer 5

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 : $$I = \int_{-\infty}^\infty\ e^{-x^n} = \int_{-\infty}^\infty\ e^{-x^{2k}} = 2\int_0^\infty\ e^{-x^{2k}}$$

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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 $|x| > 1$ .