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A Convergence properties of integrals

  1. Oct 6, 2016 #1
    It is perfectly fine to do the following:

    ##\displaystyle{\int_{-\infty}^{\infty}\ d\phi\ e^{-\phi^{2}/2}e^{-\lambda \phi^{4}/4!} = \int_{-\infty}^{\infty}e^{-\phi^{2}/2}\sum\limits_{n=0}^{\infty}}\frac{(-\lambda\phi^{4})^{n}}{(4!)^{n}\ n!}=\sum\limits_{n=0}^{\infty}\frac{(-1)^{n}\lambda^{n}}{(4!)^{n}n!}\int_{-\infty}^{\infty}e^{-\phi^{2}/2}\phi^{4n}##

    and then to continue with the integration, but the following is not valid:

    ##\displaystyle{\int_{-\infty}^{\infty}\ d\phi\ e^{-\phi^{2}/2}\ \phi^{4n}=\int^{\infty}_{-\infty}d\phi\ \phi^{4n}\ \sum\limits_{m=0}^{\infty} \frac{(-1)^{m}\ \phi^{2m}}{2^{m}\ m!} = \sum\limits_{m=0}^{\infty} \frac{(-1)^{m}}{2^{m}\ m!} \int^{\infty}_{-\infty}d\phi\ \phi^{4n+2m}}##

    The reason is that the integral after the Taylor expansion is not convergent, but it would be helpful if you could provide details.
     
  2. jcsd
  3. Oct 7, 2016 #2
    if you are interested in a way out to solve it one can proceed using Gamma functions
    which has a form exp(-t) t^n.dt integrated from 0 to infinity..
     
  4. Oct 7, 2016 #3
    Thanks for the help. I'll try it out.

    But can you help me understand why my first integral is convergent but the second integral is not convergent?
     
  5. Oct 7, 2016 #4

    mathman

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    It is not absolutely convergent, although it probably still converges.
     
  6. Oct 7, 2016 #5
    Oh, so you mean that it's the first series (and not the integral) that is convergent, and that the second series is not absolutely convergent?
     
  7. Oct 8, 2016 #6

    mathman

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    The final series is not convergent. The individual integrals are each infinite. Expanding [itex]e^{-\phi^2/2}[/itex] is a bad idea.
     
    Last edited: Oct 8, 2016
  8. Oct 9, 2016 #7

    vanhees71

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    Well, also the series in #1 is not convergent. You get
    $$I_n=\int_{-\infty}^{\infty} \mathrm{d} \phi \phi^{4n} \exp(-\phi^2/2)=2^{2n+1/2} \Gamma(2n+1/2).$$
    The coefficients of the power series (with expansion parameter ##\lambda/24##) is
    $$c_n=\frac{I_n}{n!}=\frac{(-1)^n 2^{2n+1/2}}{n!} \Gamma(2n+1/2).$$
    The radius of convergence is
    $$\lim_{n \rightarrow \infty} \left |\frac{c_n}{c_{n+1}} \right|=\lim_{n \rightarrow \infty} \frac{n+1}{16n^2+16n+1}=0.$$
    Thas the series is divergent everywhere (except for ##\lambda=0## of course).
     
  9. Oct 11, 2016 #8
    Thanks! But if the original series ##\displaystyle{\sum\limits_{n=0}^{\infty}\frac{(-1)^{n}\lambda^{n}}{(4!)^{n}n!}\int_{-\infty}^{\infty}e^{-\phi^{2}/2}\phi^{4n}}## is divergent (except for ##\lambda=0##), does it make still make sense of talk about the integral ##\displaystyle{\int_{-\infty}^{\infty}\ d\phi\ e^{-\phi^{2}/2}e^{-\lambda \phi^{4}/4!}}##?

    My hunch is that it still makes sense to talk about this integral, because the Taylor expansion of the integral may not be correct, but the integral could still be a finite quantity.

    But, I don't see why the final series is not convergent - the individual integrals do appear to be finite. Consider the second line of my original post:

    ##\displaystyle{\int_{-\infty}^{\infty}\ d\phi\ e^{-\phi^{2}/2}\ \phi^{4n}=\int^{\infty}_{-\infty}d\phi\ \phi^{4n}\ \sum\limits_{m=0}^{\infty} \frac{(-1)^{m}\ \phi^{2m}}{2^{m}\ m!} = \sum\limits_{m=0}^{\infty} \frac{(-1)^{m}}{2^{m}\ m!} \int^{\infty}_{-\infty}d\phi\ \phi^{4n+2m}}##

    Now,

    ##\displaystyle{\int^{\infty}_{-\infty}d\phi\ \phi^{4n+2m}=\frac{1}{4n+2m+1}{[\phi^{4n+2m+1}]}^{\infty}_{-\infty}}=0##,

    because the fields vanish at infinity!!!

    Where's my mistake here?
     
  10. Oct 12, 2016 #9

    vanhees71

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    This integral is of course simply divergent. The original integral is of course convergent and can be easily solved by numerical integration.
     
  11. Oct 12, 2016 #10
    I don't see why the integral is divergent.

    After all, I get ##0## for each of the integrals.
     
  12. Oct 12, 2016 #11

    vanhees71

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    What's ##\infty^n##?
     
  13. Oct 12, 2016 #12
    Right! My bad!
     
  14. Oct 12, 2016 #13
    So, this means that every series is divergent, except for the original integral, which is convergent.

    But the first series is an asymptotic series expansion of the original integral, is it not?

    You have clearly shown to me that the first series is not convergent, but how do you prove that the series is asymptotic?
     
  15. Oct 13, 2016 #14

    vanhees71

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    I don't know, why you are surprised. Why should the series be convergent only because the integral is convergent? Note that you tacitly change the order of integration and series summation, which is not always allowed!
     
  16. Oct 13, 2016 #15
    Ah, I see!


    I think I have a way to prove that the series is asymptotic.

    From before,

    ##Z(\lambda)=\displaystyle{\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\ d\phi\ e^{-\phi^{2}/2}e^{-\lambda \phi^{4}/4!} = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}d\phi\ e^{-\phi^{2}/2}\sum\limits_{n=0}^{\infty}}\frac{(-\lambda\phi^{4})^{n}}{(4!)^{n}\ n!}=\frac{1}{\sqrt{2\pi}}\sum\limits_{n=0}^{\infty}\frac{(-1)^{n}\lambda^{n}}{(4!)^{n}n!}\int_{-\infty}^{\infty}d\phi\ e^{-\phi^{2}/2}\phi^{4n}##

    so that, after integration (which you have already done), we have

    ##Z(\lambda)=\displaystyle{\frac{1}{\sqrt{2\pi}}\sum\limits_{n=0}^{\infty}\frac{(-1)^{n}\lambda^{n}}{(4!)^{n}n!}(2)^{n+\frac{1}{2}}\Gamma(2n+\frac{1}{2})=\sum\limits_{n=0}^{\infty}\frac{(-1)^{n}(4n-1)!!}{(4!)^{n}n!}\lambda^{n}=\sum\limits_{n=0}^{\infty}\frac{(-1)^{n}(4n)!}{(4!)^{n}n!4^{n}(2n)!}\lambda^{n}}##

    (This is all taken from the previous posts except the last two expressions which are my own simplifcation.)

    So, the asymptotic series is ##\displaystyle{Z(\lambda)=\sum\limits_{n=0}^{\infty}c_{n}\lambda^{n}}##, where ##\displaystyle{c_{n}=\frac{(-1)^{n}(4n)!}{(4!)^{n}n!4^{n}(2n)!}}##.


    My idea is to use the method of stationary points to show that, for a fixed value of ##\lambda##, the size of the successive terms decrease until they increase again. To accomplish this task, I will need to differentiate ##c_n## with respect to ##n##, set it to ##0## and show that the equation has a solution for a finite value of ##n##.

    To make the differentiation easy, I plan to use Stirling's approximation to find the asymptotic form of the ##c_n## for large ##n##. My hunch is that the stationary value of ##n## is large (in which case Stirling's approximation is valid).

    What do you think?
     
    Last edited: Oct 13, 2016
  17. Oct 14, 2016 #16
    I would find the original post much clearer if it used parentheses so that I know exactly what is supposed to be under the integral sign and what is not. In the same vein, it's a good idea to always use dϕ when writing an integral with respect to ϕ (or whatever the dummy variable is) — this seems to have been omitted in several terms. Finally, modern notation places the dϕ after the integrand (the expression being integrated), not before the integrand as in the past.
     
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