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Inverse FT Integral

  1. Oct 14, 2011 #1
    I have a tricky integral of the form

    ∫∫[(L^4)((k_z)^2 + (k_p)^2)^2]/[(L^4)((k_z)^2 + (k_p)^2))^4 + (k_p)^2] * e^(i*k_z*z)*e^(i*k_p*cos(θ-θ_k))*dk_z*dk_p

    which is arising in an inverse Fourier transform. I know that integrals somewhat like this appear in smectic liquid crystals but I can't find a solution offhand. It's also one of those things where Mathematica churns and spits out some unreadable nonsense.

    Any ideas? No numerics please this is in relation to a Green's fn problem so there is guaranteed to be an analytic solution. Also if anyone thinks this could somehow be simplified by the convolution theorem I'm all ears.

    The above is kind of hard to read so I've attached the integral as .png

    -Ski
     

    Attached Files:

  2. jcsd
  3. Oct 15, 2011 #2
    I got an idea. Why not make the notation easier to read? Looks like you have:

    [tex]\iint \frac{(x^2+y^2)^2}{a(x^2+y^2)^4+y^2}e^{ixz}e^{iyk}dxdy[/tex]

    Is that correct?

    Ok, Mathematica does return a single integral expression for that in terms of RootSums. It's completely readable if you know how to interpret the RootSums although it is extremely complex. If that can help you, I could explain how to interpret them.
     
    Last edited: Oct 15, 2011
  4. Oct 16, 2011 #3
    Yes, I could use a different explanation for the RootSums, the one Mathematica gives doesn't help me much. The goal is to reduce this to as simple a form as possible because it still needs to be expressed as part of a larger vector field and differentiated, etc., before it will be useful in my research.
     
  5. Oct 16, 2011 #4
    Let's look at a simple one:


    [tex]
    \begin{align*}
    \int \frac{1}{x^5+11 x+1}dx &=
    -\text{RootSum}\left[1+11 \text{$\#$1}+\text{$\#$1}^5\&,\frac{\text{Log}[1-\text{$\#$1}]}{11+5 \text{$\#$1}^4}\&\right]\\ &+\text{RootSum}\left[1+11 \text{$\#$1}+\text{$\#$1}^5\&,\frac{\text{Log}[3-\text{$\#$1}]}{11+5 \text{$\#$1}^4}\&\right]
    \label{eq:}
    \end{align*}
    [/tex]


    The RootSum represents a sum of a function over the zeros of a polynomial. The syntax uses "pure functions" using the "#1" as place holders for the variable. So the construct:

    1+11#1+#1^5

    is to be interpreted as z^5+11z+1 and we want to compute the zeros of z^5+11z+1=0 and then substitute those zeros into the second argument of the RootSum which is also a pure function, and sum them. For example, the first one is:

    [tex]\frac{\text{Log}[1-\text{$\#$1}]}{11+5 \text{$\#$1}^4}[/tex]

    So say the first root is z_0, then the first term of the sum would be:

    [tex]\frac{\log(1-z_0)}{11+5 z_0^4}[/tex]

    We can write in general:

    [tex]\text{RootSum}\left[1+11 \text{$\#$1}+\text{$\#$1}^5\&,\frac{\text{Log}[1-\text{$\#$1}]}{11+5 \text{$\#$1}^4}\&\right]=\sum_{z^5+11z+1=0} \frac{\log(1-z_n)}{11+5 z_n^4}
    [/tex]


    Here's a Mathematica function you can use if you wish which accepts a RootSum and returns the sum of the RootSum terms:

    Code (Text):

    mySolution[rs_] :=
     Module[{mysystem, thepoly, theeqn, thevals, thesols},
      mysystem = rs /. RootSum[val___, ans_] :> {val, ans};
      thepoly = mysystem[[1]];
      theeqn = thepoly[y] == 0;
      thevals = y /. Solve[theeqn, y];
      Plus @@ mysystem[[2]] /@ thevals
      ]
     
     
  6. Oct 16, 2011 #5
    Ok, that makes sense. A ridiculous amount of terms, but that happens sometimes. Thanks!
     
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