Re: Problem with inverse laplace transform (FIX)

*To*: mathgroup at smc.vnet.net*Subject*: [mg81326] Re: Problem with inverse laplace transform (FIX)*From*: m.r at inbox.ru*Date*: Wed, 19 Sep 2007 05:31:47 -0400 (EDT)*References*: <fcb48a$fos$1@smc.vnet.net>

On Sep 13, 5:43 am, "Alexey Nikitin" <niki... at proc.ru> wrote: > Hello, > > I am very sorry that my function wasn't in my mail at first time. I > am a new user of Mathematica, and want to get the following inverse > Laplace transform problem: > > I have 3 parameters h,m,n. Parameters m and n are integers. I tried > to find inverse Laplace transform of function: > InverseLaplaceTransform[-1/h - ((((-h+p)/(h+p))^(m-2n)) * (h+p)^2) / > (h*(h-p)*p),p,x] > > *-*-*-*-*- Mathematica code *-*-*-*-* > > \!\(InverseLaplaceTransform[\(-\(1\/h\)\) - \(\((\(\(-h\) + p\)\/\(h + > p\))\)\ > \^\(m - 2\ n\)\ \((h + p)\)\^2\)\/\(h\ \((h - p)\)\ p\), p, x]\) > > -*-*-*-*-*--*-*-*-*-*--*-*-*-*-*--*-*-*-*-*--* > > but mathematica can not give me an answer. Can it be solved by > Mathematica? > > Thank you very much! > > Alexey. You can express the result as a finite sum. For 2 n - m + 2 >= 0 we have In[1]:= HAp[p_] := -1/h - ((-h + p)/(h + p))^(m - 2 n)* (h + p)^2/(h (h - p) p) In[2]:= c[i_] := 1/(h p (p - h)^(2 n - m + 1))* Binomial[2 n - m + 2, i] p^i h^(2 n - m + 2 - i) In[3]:= FullSimplify[HAp[p] == -1/h + Sum[c[i], {i, 0, 2 n - m + 2}], Element[{m, n}, Integers]] Out[3]= True In[4]:= Sum[InverseLaplaceTransform[c[2 n - m + 2 - i], p, x], {i, 0, 2 n - m + 2}] // Refine[#, Element[{m, n}, Integers]] /. Binomial[n_, k_] :> Binomial[n, n - k]& Out[4]= Sum[h^(-1 + i) x^(-1 + i) Binomial[2 - m + 2 n, i] Hypergeometric1F1Regularized[1 - m + 2 n, i, h x], {i, 0, 2 - m + 2 n}] In[5]:= Block[{n = 5, m = 2}, % == InverseLaplaceTransform[HAp[p], p, x] // FullSimplify] Out[5]= True In fact the transform of c[2 n - m + 2] is wrong because the image is asymptotic to 1/h at infinity and the original contains a DiracDelta term but it conveniently cancels out. Maxim Rytin m.r at inbox.ru