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Laplace Transform

  1. Jan 4, 2005 #1


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    I have a proplem to analitic calculate this :


    Im(k)=0, k>0
    Im(p)=0, p>0

    The Mathematica 5 doesn't calculate this.

    Very glad to help.
  2. jcsd
  3. Jan 4, 2005 #2


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    Let me make sure i've got it straight.I'll think about it later.So u wanna compute
    [tex] I=\frac{k}{p^{2}}L(\frac{\sqrt{z^{2}+2pz}}{z+p})+\frac{1}{p^{2}}L[-\frac{\sqrt{z^{2}+2pz}}{(z+p)^{2}}] [/tex]
    ??That's an ugly "animal"... :tongue2:

    Did u try other version of Mathematica ?? :tongue2:


    PS.I'll work on it...I smell some Bessel functions...
  4. Jan 5, 2005 #3


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    You undestand me right!

    I didn't use other version, but I found the new release 5.1 at http://www.wolfram.com
  5. Jan 5, 2005 #4


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    [tex] I=L(\frac{\sqrt{z^{2}+2pz}}{z+p})=\int_{0}^{+\infty} \frac{\sqrt{z^{2}+2pz}}{z+p}e^{-sz} dz [/tex](1)

    [tex]J=L[-\frac{\sqrt{z^{2}+2pz}}{(z+p)^{2}}]=-\int_{0}^{+\infty} \frac{\sqrt{z^{2}+2pz}}{(z+p)^{2}}e^{-sz} dz [/tex](2)

    The first integral,i complete the square under the sq.root and make a substitution:
    [tex] I=\int_{0}^{+\infty} \frac{\sqrt{(\frac{z+p}{p})^{2}-1}}{\frac{z+p}{p}} e^{-sz} dz [/tex](3)

    And now i make the substitution
    [tex] \frac{z+p}{p}\rightarrow \cosh t [/tex](4)
    [tex]dz=p\sinh t dt [/tex](5)
    The limits of integration are the same ([itex] \arg\cosh 1 =0 [/itex]).
    The exponential becomes:
    [tex] e^{-sz}=e^{sp}e^{-sp\cosh t} [/tex](6)

    [tex] I=p e^{sp}\int_{0}^{+\infty} e^{-sp\cosh t}\frac{\sinh^{2} t}{\cosh t} dt[/tex] (7)

    Can u convince 'Mathematica' to evaluate this integral??Maybe numerically...

    Using the same kind of substitution,for evaluating the second integral (L.transform),u get
    [tex] J=e^{sp} \int_{0}^{+\infty} e^{-sp\cosh t}\frac{\sinh^{2} t}{\cosh^{2} t} dt [/tex] (8)

    Again,i don't know what to do to it.

    Last edited: Jan 5, 2005
  6. Jan 5, 2005 #5


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    Thx for help, but I don't simplify your answer.

    I'm going use the answer (if it possible) in math model.
    Numerical answer is suit, but if it in the form:

    I try to apply the teory residues for a solve.
    What do you think about?

    P.S. I should note that the root of initial problem is

    \int {0} {\inf} { exp(-rk)/r*(1/r+k) }dt

    r = sqrt ( p^2 + (vt)^2 )
  7. Jan 5, 2005 #6


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    without ambiguity:

    \int {0} {\inf} { (exp(-rk)/r)*(1/r+k) }dt

    r = sqrt ( p^2 + (vt)^2 )

    p.s. my calculation is big, that is why I don't insert its.
  8. Jan 5, 2005 #7


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    I'm sorry,but it doesn't work with residues,because the functions doesn't have poles."Im(p)=0, p>0",and the integration is not on entire R,but only on its positive semiaxis.,where the denominator is never zero.

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