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Math Help - A limit

  1. #1
    Super Member girdav's Avatar
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    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<br />
=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 -<br />
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 <br />
=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 <br />
=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.
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  2. #2
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    \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 =).
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  3. #3
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    Quote Originally Posted by Texxy View Post
    \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 ?
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  4. #4
    Super Member girdav's Avatar
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    Quote Originally Posted by Texxy View Post
    \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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