Find the value of $\lim_{n\rightarrow\infty}\int_{-\infty}^{\infty}e^{-x^{2n}}\:\:dx$ 
Find the value of $$\lim_{n\rightarrow\infty}\int_{-\infty}^{\infty}e^{-x^{2n}}\:\:dx$$

I just searched on the internet and learned that the given integral given is a special integral.
If we consider, $f(x)=e^{-x^{2n}}$ then we seen that $f(x)$ is an even function. Maybe it can help$?$
When I put $n=\infty$ in the $2n$ I got $e^{-x^{\infty}}$ How do I evaluate it$?$ As it's value depends on the valuenof $x$.
Any help is greatly appreciated.
Apparently this is a MIT Integration Bee problem. But I ain't sure.
 A: Assuming that $n$ is an integer, symmetry gives that the integral is $$2 \int_0^\infty e^{-x^{2n}} dx$$
Then applying the substitution $$u = x^{2n}, \qquad du = 2n x^{2 n - 1}\,dx$$ yields
$$\frac{1}{n} \int_0^{\infty} u^{\frac{1}{2 n} - 1} e^{-u} du = \frac1n \Gamma\left(\frac{1}{2n}\right) ,$$
where $\Gamma$ is the Gamma function. Substituting in the well-known Laurent expansion
$$\Gamma(z) = \frac{1}{z} + R(z)$$
for $\Gamma$ about $z = 0$, where $R(z) \in O(1)$, thus gives
$$\int_{-\infty}^\infty e^{-x^{2n}} dx = 2 + S(n), $$ where $S(n) \in O\left(\frac{1}{n}\right)$. Thus, the limit as $n \to \infty$ is $$\boxed{\lim_{n \to \infty} \int_{-\infty}^\infty e^{-x^{2n}} dx = 2}$$
as claimed.
Remark Maybe interestingly, keeping one more term of the Laurent expansion gives the refined asymptotic $\int_{-\infty}^\infty e^{-x^{2n}} dx = 2 - \frac{\gamma}{n} + T(n)$, where $\gamma$ is the Euler-Mascheroni constant and $T(n) \in O\left(\frac{1}{n^2}\right)$.
A: I thought it might be instructive to proceed without appealing to the Dominated Convergence Theorem.  To that end, we now proceed.

First, enforcing the substitution $x\mapsto x^{1/2n}$ we have
$$\int_{-\infty}^\infty e^{-x^{2n}}\,dx=\frac1{n}\int_0^\infty \frac{e^{-x}}{x^{1-1/2n}}\,dx\tag1$$
Next, integrating by parts the integral on the right-hand side of $(1)$ with $u=e^{-x}$ and $v=2nx^{1/2n}$ we find that
$$\int_{-\infty}^\infty e^{-x^{2n}}\,dx=2\int_0^\infty x^{1/2n}e^{-x}\,dx\tag2$$
The integral on the right-hand side of $(2)$ is $\Gamma(1+1/2n)$, which has limit $\Gamma(1)=1$.
If one is unfamiliar with the Gamma function, then one can exploit the uniform convergence of the improper Riemann integral on the right-hand side of $(2)$, pass the limit under the integral, and obtain the coveted result
$$\begin{align}
\lim_{n\to \infty}\int_{-\infty}^\infty e^{-x^{2n}}\,dx&=2\lim_{n\to \infty}\int_0^\infty x^{1/2n}e^{-x}\,dx\\\\
&=2\int_0^\infty \lim_{n\to \infty} x^{1/2n}e^{-x}\,dx\\\\
&=2\int_0^\infty e^{-x}\,dx\\\\
&=2
\end{align}$$

If one is unfamiliar with the concept of uniform convergence of an improper Riemann integral, then one can proceed as follows.
Choose any number $\varepsilon>0$.  Then, fix a number $L>0$ such that $\int_L^\infty (x^{1/2n}-1)e^{-x}\,dx<\frac\varepsilon2$ for any $n$.  Note that this is possible since
$$\begin{align}
\left|\int_L^\infty (x^{1/2n}-1)e^{-x}\,dx\right|\le Le^{-L}
\end{align}$$
Then, with $L$ fixed write
$$\begin{align}
\left|\int_0^\infty x^{1/2n}e^{-x}\,dx-1\right|&=\int_0^\infty (x^{1/2n}-1)e^{-x}\,dx\\\\
&\le \int_0^L (x^{1/2n}-1)e^{-x}\,dx+\int_L^\infty (x^{1/2n}-1)e^{-x}\,dx
\end{align}$$
We know that for the chosen $\varepsilon>0$ there exists a number $N$ such that whenever $n>N$, $|x^{1/2n}-1|<\frac{\varepsilon}{2(1-e^{-L})}$.
Hence, we see that for any chosen $\varepsilon>0$ there exists a number $N$ such that $n>N$ implies
$$\left|\int_0^\infty x^{1/2n}e^{-x}\,dx-1\right|<\frac\varepsilon2+\frac\varepsilon2=\varepsilon$$
And we are done!
A: Consider the limit piecewise
$$\lim_{n\to\infty}x^{2n}=\begin{cases}0 & |x| < 1 \\ +\infty & |x| > 1\end{cases}$$
which by the continuity of $e^x$ gives the following limits
$$\lim_{n\to\infty}\exp(-x^{2n}) = \begin{cases}e^0 = 1 & |x| < 1 \\ e^{-\infty} = 0 & |x| > 1\end{cases}$$
so the integral converges to
$$\lim_{n\to\infty}\int_{-\infty}^\infty e^{-x^{2n}}\:dx = \int_{-1}^1 dx = 2$$
by dominated convergence.
