## Laplace Transform

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.

 PhysOrg.com science news on PhysOrg.com >> Hong Kong launches first electric taxis>> Morocco to harness the wind in energy hunt>> Galaxy's Ring of Fire

Blog Entries: 9
Recognitions:
Homework Help
 Quote by anv 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.
Let me make sure i've got it straight.I'll think about it later.So u wanna compute
$$I=\frac{k}{p^{2}}L(\frac{\sqrt{z^{2}+2pz}}{z+p})+\frac{1}{p^{2}}L[-\frac{\sqrt{z^{2}+2pz}}{(z+p)^{2}}]$$
??That's an ugly "animal"...

Did u try other version of Mathematica ??

Daniel.

PS.I'll work on it...I smell some Bessel functions...

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

Blog Entries: 9
Recognitions:
Homework Help

## Laplace Transform

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

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

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

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

Therefore
$$I=p e^{sp}\int_{0}^{+\infty} e^{-sp\cosh t}\frac{\sinh^{2} t}{\cosh t} dt$$ (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
$$J=e^{sp} \int_{0}^{+\infty} e^{-sp\cosh t}\frac{\sinh^{2} t}{\cosh^{2} t} dt$$ (8)

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

Daniel.

 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 )
 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.

Blog Entries: 9
Recognitions:
Homework Help
 Quote by anv 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.

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.