1. A limit

I tried to calculate this limit
$\lim_{n \to +\infty} \int_0^1 n\left(-1\right)^{[nx]}e^{-x^2}dx$ where $[]$ represents the floor.
I wrote $I_n = \int_0^1 n\left(-1\right)^{[nx]}e^{-x^2}dx
=n\sum_{k=0}^{n-1}\int_{\frac jn}^{\frac{j+1}n}\left(-1\right)^je^{-x^2}dx$
.
If $n$ is even $\exists p \in \mathbf{N} \, n=2p$ and
$I_{2p} = 2p\sum_{j=0}^{p-1}\int_{\frac {2j}{2p}}^{\frac{2j+1}{2p}}e^{-x^2}dx -
2p\sum_{j=0}^{p-1}\int_{\frac {2j+1}{2p}}^{\frac{2j+2}{2p}}e^{-x^2}dx$

$f:x \mapsto e^{-x^2}$ is continuous so
$\exists a_{j,p} \in \left[ \frac{2j}{2p},\frac{2j+1}{2p}\right] \, \int _{\frac {2j}{2p}}^{\frac{2j+1}{2p}}e^{-x^2}dx
=e^{-a_{j,p}^2}$

and $\exists b_{j,p} \in \left[ \frac{2j+1}{2p},\frac{2j+2}{2p}\right] \, \int _{\frac {2j+1}{2p}}^{\frac{2j+2}{2p}}e^{-x^2}dx
=e^{-b_{j,p}^2}$
so
$I_{2p} = \sum_{j=0}^{p-1}e^{-a_{j,p}^2}-e^{-b_{j,p}^2}$
but I haven't information enough about $a_{j,p}$ and $b_{j,p}$ to conclude.
If $n$ is odd there is one more term ( $-\int_{\frac{2p+1}{2p}}^1e^{-x^2}dx$) and its limit is $0$.
But it doesn't show that the sequence $\left(I_n\right)$ converges.

2. $\int_0^1 ne^{i\pi\lfloor nx\rfloor}e^{-x^2}dx.$ Substitute nx = y, dx=dy/n. Now we have this integral: $\int_0^n e^{i\pi\lfloor y\rfloor}e^{-(y/n)^2}dy$. Since in the interval a-1<y<a for some positive integer a $e^{i\pi\lfloor y\rfloor}=e^{i\pi(a-1)}$, we can consider the equivalent integrals

$\int_0^n e^{i\pi\lfloor y\rfloor}e^{-(y/n)^2}dy=\sum_{k=0}^{n}e^{i\pi k}\int_{k}^{k+1}e^{-(y/n)^2}dy$ $=\frac{n\sqrt{\pi}}{2}\sum_{k=0}^{n-1}(-1)^k\left[\mathrm{erf}\left(\frac{k+1}{n}\right)-\mathrm{erf}(k/n)\right]$ where erf is the error function. As n goes to infinity this converges to some constant around 0.683938, but I don't know how to tackle that sum analytically. Hopefully this can help you out though =).

3. Originally Posted by Texxy
$\int_0^1 ne^{i\pi\lfloor nx\rfloor}e^{-x^2}dx.$ Substitute nx = y, dx=dy/n. Now we have this integral: $\int_0^n e^{i\pi\lfloor y\rfloor}e^{-(y/n)^2}dy$. Since in the interval a-1<y<a for some positive integer a $e^{i\pi\lfloor y\rfloor}=e^{i\pi(a-1)}$, we can consider the equivalent integrals

$\int_0^n e^{i\pi\lfloor y\rfloor}e^{-(y/n)^2}dy=\sum_{k=0}^{n}e^{i\pi k}\int_{k}^{k+1}e^{-(y/n)^2}dy$ $=\frac{n\sqrt{\pi}}{2}\sum_{k=0}^{n-1}(-1)^k\left[\mathrm{erf}\left(\frac{k+1}{n}\right)-\mathrm{erf}(k/n)\right]$ where erf is the error function. As n goes to infinity this converges to some constant around 0.683938, but I don't know how to tackle that sum analytically. Hopefully this can help you out though =).
And how do you show that it converges to this constant ?

girdav : l'île des maths ?

4. Originally Posted by Texxy
$\int_0^1 ne^{i\pi\lfloor nx\rfloor}e^{-x^2}dx.$ Substitute nx = y, dx=dy/n. Now we have this integral: $\int_0^n e^{i\pi\lfloor y\rfloor}e^{-(y/n)^2}dy$. Since in the interval a-1<y<a for some positive integer a $e^{i\pi\lfloor y\rfloor}=e^{i\pi(a-1)}$, we can consider the equivalent integrals

$\int_0^n e^{i\pi\lfloor y\rfloor}e^{-(y/n)^2}dy=\sum_{k=0}^{n}e^{i\pi k}\int_{k}^{k+1}e^{-(y/n)^2}dy$ $=\frac{n\sqrt{\pi}}{2}\sum_{k=0}^{n-1}(-1)^k\left[\mathrm{erf}\left(\frac{k+1}{n}\right)-\mathrm{erf}(k/n)\right]$ where erf is the error function. As n goes to infinity this converges to some constant around 0.683938, but I don't know how to tackle that sum analytically. Hopefully this can help you out though =).
An other attemp led me to this result but I can prove that sum convergs.
I tried to separate the case $n$ odd or even.
girdav : l'île des maths ?
And "Forum mathématique", "Mathématex".
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