A devious little problem indeed! I am interested in where you found it. We in fact have a very nice formula:
$$\sum_{k=1}^n (-1)^{k-1} \binom{n}{k} (x-k)^2 =x^2$$
It follows from:
$$F(x)=\sum_{k=1}^n (-1)^{k-1} \binom{n}{k} (x-k)^2 \\ =\sum_{k=1}^n (-1)^{k-1}\binom{n}{k}\big(x^2-2kx+k^2\big) \\ =x^2\color{red}{\sum_{k=1}^n (-1)^{k-1}\binom{n}{k}}-2x\color{green}{\sum_{k=1}^n (-1)^{k-1}\binom{n}{k}k}+\color{blue}{\sum_{k=1}^n (-1)^{k-1}\binom{n}{k}k^2}$$
Now, we need to somehow show
$$\color{red}{\blacksquare}=1 \\ \color{green}{\blacksquare}=\color{blue}{\blacksquare}=0$$
These sums have been studied before:
$\color{red}{\blacksquare}$ : Alternating sum of binomial coefficients equal to $1$
Follows from doing a binomial expansion of $(1-1)^n$.
$\color{green}{\blacksquare}$ : Binomial coefficient series $\sum\limits_{k=1}^n (-1)^{k+1} k \binom nk=0$
Follows from the recurrence $k\binom{n}{k}=n\binom{n-1}{k-1}$.
$\color{blue}{\blacksquare}$ : This is the hard one. We proceed as follows:
$$\sum_{k=1}^n (-1)^{k-1}k^2\binom{n}{k}=n\sum_{k=1}^n (-1)^{k-1}k \binom{n-1}{k-1} \tag{1}$$ $$ =n\sum_{l=0}^{n-1}(-1)^l(l+1)\binom{n-1}{l} \\ =n\sum_{l=0}^{n-1}(-1)^l l\binom{n-1}{l}+n\underbrace{\sum_{l=0}^{n-1}(-1)^l \binom{n-1}{l}}_{=(1-1)^{n-1}=0} $$
Finally, since $\binom{n-1}{n}=0$, and since the $l=0$ summand is zero, we can remove the $l=0$ index and add a $l=n$ index, and then rename the index back to $k$:
$$n\sum_{l=0}^{n-1}(-1)^l l\binom{n-1}{l}=n\sum_{k=1}^n(-1)^k k\binom{n-1}{k}$$
Now, using the recurrence relation for the binomials,
$$n\sum_{k=1}^n(-1)^k k\binom{n-1}{k}=n\underbrace{\sum_{k=1}^n k(-1)^k \binom{n}{k}}_{=\color{green}{\blacksquare}=0}-n\sum_{k=1}^n (-1)^kk \binom{n-1}{k-1}$$
Again, $k\binom{n-1}{k-1}=k^2\binom{n}{k}$ and hence
$$-n\sum_{k=1}^n (-1)^kk \binom{n-1}{k-1}=-n\sum_{k=1}^n (-1)^kk^2 \binom{n}{k}=n\sum_{k=1}^n (-1)^{k-1}k^2 \binom{n}{k}\tag{2}$$
But, retracing our steps from $(1)$ to $(2)$, we have just proved that
$$\sum_{k=1}^n (-1)^{k-1}k^2\binom{n}{k}=~\boldsymbol{n}~\sum_{k=1}^n (-1)^{k-1}k^2\binom{n}{k}$$
This can only be true for general $n$ if the sum is zero. Hence,
$$\boxed{\sum_{k=1}^n (-1)^{k-1} \binom{n}{k} (x-k)^2=x^2\color{red}{\blacksquare}-2x\color{green}{\blacksquare}+\color{blue}{\blacksquare} \\ =x^2}$$
QED!!
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.
Best Answer
$$\int_0^2f=\int_0^1f+\int_1^2f=\int_0^1f+\int_0^1g=\int_0^1(f+g)$$
Where $g(x)=f(x+1)$.
Can you continue?
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