# Limits at infinity involving integrals

$$\lim_{n\to\infty}\int_{n}^{n+1} e^{-x^2}dx$$

Trouble is, I have no real idea how to go about evaluating that integral -- u-substitution doesn't really seem to work, at least. I also considered doing something hacky involving the squeeze theorem, but I have no idea what the bounds on that might be.

• I understand what you mean intuitively, and see why it should converge to zero, but is there a more rigorous way to do it? Edit: Didn't see your edit. – Why-Seven-Six Oct 22 '15 at 6:03
• To add, the primitive of $e^{-x^2}$ cannot be computed or written in finite terms. The given integral is evaluated as $\frac{\sqrt{\pi}}{2}\lim_{n\to\infty} \left[\mathrm{erf}(n+1)-\mathrm{erf}(n)\right]=0$ where "erf" is the error function. Of course, to actually derive the answer, go by André's method. – Corellian Oct 22 '15 at 6:12
• Set $y=nx$. Then the result is immediately clear. – Urgje Oct 22 '15 at 11:05
• A far better question would have been proving that $\displaystyle\lim_{n\to\infty}n~e^{n^2}\int_n^{n+1}e^{-x^2}~dx~=~\frac12$ – Lucian Oct 22 '15 at 11:06

You don't need to evaluate, the function is very small for $n$ large, and the interval of integration has length $1$.

If you want to give an explicit argument, you could observe that if $x\gt 1$ we have $x^2\gt x\gt 0$, so $$0\lt \int_n^{n+1}e^{-x^2}\,dx\lt \int_n^{n+1}e^{-x}\,dx.$$ This last integral is $e^{-n}-e^{-(n+1)}$, which has limit $0$ as $n\to\infty$. So by Squeezing, $\lim_{n\to\infty}\int_n^{n+1}e^{-x^2}\,dx=0$.

• @Why-Seven-Six: I prefer the argument of sky90. – André Nicolas Oct 22 '15 at 6:32

$f(x)=e^{-x^2}\geq 0 \ \forall x\in \mathbb{R}$ and the integral of a positive function is also positive. So now you can try to estimate it using that $f$ is monotonic decreasing:

$\int_n^{n+1}{e^{-x^2}dx}\leq \max\{e^{-n^2},e^{-(n+1)^2}\}(n+1-n)=e^{-n^2}$

so: $0\leq\lim_{n\rightarrow \infty}{\int_n^{n+1}{e^{-x^2}dx}}\leq\lim_{n\rightarrow \infty}{e^{-n^2}}\leq 0$

And so all must be $0$.