Curious Integral Proof

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.

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

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

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

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):

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