
#1
Jan405, 06:48 PM

P: 4

I have a proplem to analitic calculate this :
(k/p^2)*LaplaceTransform[Sqrt[z^2+2p*z]/(z+p)],z,k]+ (1/p^2)*LaplaceTransform[Sqrt[z^2+2p*z]/(z+p)^2],z,k] Im(k)=0, k>0 Im(p)=0, p>0 The Mathematica 5 doesn't calculate this. Very glad to help. 



#2
Jan405, 08:24 PM

Sci Advisor
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P: 11,863

[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"... Did u try other version of Mathematica ?? Daniel. PS.I'll work on it...I smell some Bessel functions... 



#3
Jan505, 03:52 AM

P: 4

You undestand me right!
I didn't use other version, but I found the new release 5.1 at http://www.wolfram.com 



#4
Jan505, 07:54 AM

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P: 11,863

Laplace Transform
[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) Therefore [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. Daniel. 



#5
Jan505, 12:50 PM

P: 4

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: f(k,p)*numerical_answer 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 ) 



#6
Jan505, 12:54 PM

P: 4

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. 



#7
Jan505, 01:07 PM

Sci Advisor
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P: 11,863

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. Daniel. 


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