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Complex integral.

  1. Aug 22, 2006 #1
    It's quite a "strange" thing..why people have so many difficulties with dealing with integrals of the form:

    [tex] \int _{c-i\infty}^{c+i\infty}dsf(s)e^{st}=I(t) [/tex] ?

    You can always make the change of variable s=c+ix so you get:


    [tex] \int _{-\infty}^{+\infty}dxf(c+ix)e^{ixt}=I(t)e^{-ict} [/tex] (2)

    And (2) is just simply an improper integral that can be evaluated approximately by some quadrature formula.....:rolleyes: :approve :
     
    Last edited: Aug 22, 2006
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  3. Aug 22, 2006 #2

    shmoe

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    No one ever thought to try to numerically approximate an integral like that before :rolleyes:

    How about this, go find someone who had "difficulties" with an integral like that and show them how your numeric approximation will solve their problem. Go on, actually do an example (have you heard me say this before?)
     
  4. Aug 22, 2006 #3

    saltydog

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    I think you're missing an i: If [itex]s=c+ix[/itex] then [itex]ds=idx[/itex]. Making this substitution, I get:

    [tex]I(t)=ie^{ct}\int_{-\infty}^{\infty}e^{ixt}f(c+ix)dx[/tex]

    Evaluating these integrals numerically can, I think, "help" you arrive at the solution. Take for example, the expression:

    [tex]\mathcal{L}^{-1}\left\{\frac{1}{s-k}\right\}=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{e^{st}}{s-k}ds[/tex]

    This could be evaluated via Residue Integration about an expanding half-circle contour by splitting up the contour into two integrals.

    I suspect the integral about the curved portion of the contour tends to zero as the radius increases without bound. Therefore, any calculations I do to justify this suspicion, unless I'm absolutely sure (almost never), I would estimate numerically. If the numerical approximation is not tending to zero, there is a good chance I made a mistake in the analysis and I would go back and check it. Works for me.:smile:
     
  5. Aug 25, 2006 #4
    But what's the problem doing the integral:

    [tex] \int_{-\infty}^{\infty}dxe^{ixt}f(c+ix) [/tex] ??

    This is just the "Fourier transform" plus the function exp(ct) of the function f(c+ix) "Shmoe" i have "Mathematica 5.0 " (cracked) if you tell me what i must introduce to evaluate the integral, I don't mind performing it Numerically......or if you indicates me a webpage to evaluate integrals by Numerical Methods
     
  6. Aug 25, 2006 #5

    TD

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    It puzzles me why you would want to explicitly mention that it is a cracked version.
    It's not really something to be proud of (:grumpy:), nor is it of any relevance :confused:

    Sorry for the off topic :blushing:
     
  7. Aug 25, 2006 #6

    shmoe

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    I wouldn't know how to do anything in mathematica. I wasn't willing to either pay for or steal it.

    Any of these integrals can be approximated as accurately as you like given enough time and resources, assuming you can determine f with the required accuracy and you have some way to estimate the rate of convergence. This doesn't solve a problem like determining the asymptotic behavior of psi(x) for example as you are only able to approximate these integrals for one "x" at a time.

    You've read the Odlyzko/Lagarias paper on calculating pi(x) via analytic techniques by now right? It takes some work to get around the slow convergence of the relevant integral, they have to worry about approximating zeta quickly, etc. I'd suggest you read their paper very carefully, it will be able to teach you much more about this sort of thing than I ever will (Odlyzko=king of computations, shmoe=boob).
     
  8. Aug 25, 2006 #7
    I really can't believe there's NO method to evaluate for big t (t-->oo) integrals of the form:

    [tex] \int_{-\infty}^{\infty}dxf(c+ix)e^{ixt} [/tex] when the argument of the t is big t>>1, i think this has to deal with "Cauchy's Stationary phase method" but i don't know how to apply this to the Fourier transform.

    - by the way isn't a reliable and fast method to compute Fourier transforms? (FFT Algorithm)
     
  9. Aug 25, 2006 #8

    shmoe

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    Of course there are ways to work with an integral like that as t->infinity, I did not say there wasn't. I said calculating one of these integrals for a given value of t (or as many values of t as you have time to do) says nothing at all about the behavior as t->infinity. How you will deal with an integral like this depends very much on f.

    What can I say, I'm assuming that you are hoping to apply this in number theory land and I was just seeing if you had an example of where someone had 'difficulty' with an integral like this which could be dealt with by a numerical approximation and they had overlooked this possibility. I guess it's too much to ask you to justify any of your claims in your OP?

    The FFT is for the discrete fourier transform, no? I don't see how this will help?
     
  10. Aug 25, 2006 #9
    Of course my intention..was (and i'm not the first thinking this..) is to apply that to the Mertens function given by the integral:(omitting some constants and the exp(ct) )

    [tex] \int_{-\infty}^{\infty} dx \frac{e^{ixt}}{\zeta(c+ix)} [/tex]

    to evaluate an asymptotic behaviour of [tex] M(e^t ) [/tex] (and its relation to RH) and also although is proved before to give an asymptotic behaviour of [tex] \psi (e^t) [/tex] to obtain a fast proof of PNT
     
  11. Aug 25, 2006 #10

    saltydog

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    I suspect Mathematica won't handle the integral:

    [tex] \int_{-\infty}^{\infty} dx \frac{e^{ixt}}{\zeta(c+ix)} [/tex]

    easily because of its highly oscillatory nature as t gets bigger due to the [itex]e^{itx}[/itex] component. You can change the various parameters to the NIntegrate function such as MaxRecursion, WorkingPrecision, but that slows the calculations down significantly. For example, the code:


    Code (Text):
    NIntegrate[e^{ixt}/Zeta[c+ix]},{x,-10,10},
    WorkingPrecision->50,MaxRecursion->50]
     
    with t=50 and c=2 took about 3 minutes at 2.2 GHz. Omitting those parameters and Mathematica reports "integration converging too slowly"
    Better way I'd like to know.
     
    Last edited: Aug 25, 2006
  12. Aug 26, 2006 #11

    saltydog

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    Thanks to Hurkly, I can now ask my question: The integral above speaks of this:

    [tex]
    \mathcal{L}^{-1}\left\{\frac{1}{\zeta(s)}\right\}=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{e^{st}}{\zeta(s)}ds
    [/tex]

    I think this could be solved via residue integration by taking an expanding square contour (avoiding the zeros of zeta). I wonder if the integral over the remainder of the contour goes to zero, that is:

    [tex]
    \lim_{R\to\infty}\;\int \! \! \! \! \! \! \square
    \frac{e^{st}}{\zeta(s)}ds=\;
    2\pi i \sum_{n=1}^{\infty}
    \mathop\text{Res}\limits_{z=z_i}
    \left\{\frac{e^{st}}{\zeta(s)}\right\}
    \substack{\displaystyle{?}\\ \displaystyle{=}}\;
    \int_{c-i\infty}^{c+i\infty}\frac{e^{st}}{\zeta(s)}ds\;


    [/tex]

    Be interesting to work on.:smile:
     
    Last edited: Aug 26, 2006
  13. Aug 26, 2006 #12

    shmoe

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    Have you stopped to ask if this integral is even convergent?

    It's doesn't give the summatory function for the mobius function in any case, there's a 1/s missing, look up Perron's formula.
     
  14. Aug 26, 2006 #13
    Residue integration is known however you can't use it to compute Mertens function...by residue theorem (and assuming you only have roots with multiplicity one) then:

    [tex] M(x)= \sum_{\rho} \frac{x^{\rho}}{\zeta( \rho)}-2+\sum_{n>0} A(n)(-1)^{n-1}x^{-2n} [/tex] i don't remember what A(n) was although it included the functions n!, n and [tex] \zeta (1+2n) [/tex] and a constant involving pi=3.14159.....

    EDIT: Sorry "Shmoe" i forgot the s factor inside the integral ..i apologize..:frown: :frown: for it.

    By the way..is not a similar to "Euler's sum formula so...

    [tex] \int_{-\infty}^{\infty}dx f(x) = \sum_{n=-\infty}^{\infty}f(n)+ R [/tex]

    Where "R" is a rest that can be exact or approximated..in a similar form ot the "Euler-Mc Laurin" sum formula...
     
    Last edited: Aug 26, 2006
  15. Aug 26, 2006 #14

    saltydog

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  16. Aug 26, 2006 #15

    saltydog

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    Ok Lokofer. Thanks. I'm not familiar with Mertens functions and not well explained in Mathworld. Will be something for me additionally to look into this fall in class.
     
  17. Aug 26, 2006 #16

    saltydog

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    Alright guys, think I have a basic outline of it:

    [tex]\frac{1}{\zeta(s)}=\sum_{n=1}^{\infty}\frac{\mu(n)}{n^s}=\sum_{n=1}^{\infty}\mu(n)e^{-ln(n)s

    }[/tex]

    Therefore, by Perron's formula:

    [tex]\sum_{ln(n)\leq x}^{'}\mu(n)=\frac{1}{2\pi

    i}\int_{c-i\infty}^{c+i\infty}\frac{1}{\zeta(s)}\frac{e^{sx}}{s}ds[/tex]

    Letting [itex]x=ln(u)[/itex] this then becomes:

    [tex]\sum_{n\leq u}^{'}\mu(n)=\frac{1}{2\pi
    i}\int_{c-i\infty}^{c+i\infty}\frac{1}{\zeta(s)}\frac{u^s}{s}ds[/tex]

    And Mertens function is defined as:

    [tex]M(x)=\sum_{n\leq x} \mu(n)[/tex]

    I assume the prime above the summation acts similarly as with Chebyshev's [itex]\psi(x)[/itex]

    Sorry Lokofer if I slightly hogged your thread.:rolleyes:
     
    Last edited: Aug 26, 2006
  18. Aug 26, 2006 #17

    shmoe

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    That doesn't show the limit is not zero, remember O is just an upper bound. You do know [tex]\zeta(s)=O(1)[/tex] on this line (and any right half plane to the right of s=1), which does tell you you don't converge to zero.

    However! Not converging to zero is not enough to conclude your intergral is divergent! Consider the limit of the Fresnel integrals:

    [tex]\int_0^\infty \cos(x^2)dx=\sqrt{\pi/8}[/tex]

    it's oscillations grow faster and faster, and it does converge.

    We don't have that here. The exponential is oscillating at a constant speed and the real part of 1/zeta is bounded away from zero on our line (take c=4 say, this will be easy to prove, the real part of 1/zeta will be 1+ the rest of the terms in the sum, which is bounded trivially by zeta(4)-1=0.082...). This means you can find intervals as high up as you like where the integral will be larger than some fixed constant.

    That's the basics in your last post saltydog. M(x) is kinda like a random walk, but not random of course. A good exercise, if you haven't already seen it, is to assume a bouund like M(x)=O(x^(1/2+e)) is true for any e>0 and use this to extend the region of convergence of the dirichlet series for 1/zeta into the critical strip. The other direction is also possible, but less straightforward.

    The expression lokofer gave for M(x) has some problems, this lives on the mathworld page I believe if anyone is interested.
     
    Last edited: Aug 26, 2006
  19. Aug 27, 2006 #18

    saltydog

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    You mean:

    [tex]M_0(x)=\sum_{\rho}\frac{x^{\rho}}
    {\rho\zeta'(\rho)}-2+\sum_{n=1}^{\infty}\frac{2\pi}{x}^{2n}
    \frac{(-1)^{n-1}}{(2n)!n\zeta(2n+1)}[/tex]

    I assume that's from residue integration. First term is from the complex zeros, second term is residue at 0, (residue at s=1 is zero), and the remaining term comes from the trivial zeros (need to check that last one though). I suspect the sum over the complex zeros is done symmetrically like the expression for [itex]\psi_0(x)[/tex].

    Therefore let me hypothesize (haven't checked it yet):

    [tex]
    \lim_{R\to\infty}\;\frac{1}{2\pi i}\int \! \! \! \! \! \! \square
    \frac{1}{\zeta(s)}\frac{e^{sx}}{s}ds=\;
    \sum_{n=1}^{\infty}
    \mathop\text{Res}\limits_{s=s_n}
    \left\{\frac{1}{\zeta(s)}\frac{e^{st}}{s}\right\}\;
    \substack{\displaystyle{?}\\ \displaystyle{=}}\;
    \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty}\frac{1}{\zeta(s)}\frac{e^{st}}{s}ds\;
    =M_0(x)
    [/tex]

    In this case, I would asssume the expansion of the contour is done in discrete "jumps" so as to assure an upper bound on the integrand as it passes across the critial line as is done for [itex]\psi_0(x)[/tex].

    The contribution from the remaining sides of the contour approach zero as R grows without bounds (need to verify that though).

    You know Lokofer . . . . you missed an s up there dude.:rolleyes:

    Anyway guys, I need some projects to work on this fall because I'm taking Complex Analysis. This looks very interesting to me as well. Thanks.:smile:
     
    Last edited: Aug 27, 2006
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