Re: Bug in LaplaceTransform?
- To: mathgroup at smc.vnet.net
- Subject: [mg97740] Re: [mg97693] Bug in LaplaceTransform?
- From: Daniel Lichtblau <danl at wolfram.com>
- Date: Fri, 20 Mar 2009 02:39:49 -0500 (EST)
- References: <200903190709.CAA23011@smc.vnet.net>
Wieland Brendel wrote:
> Dear reader,
> how can it happen that Mathematica throws out -i for the Laplace
> transform of one,
>
> LaplaceTransform[1, t, I] = -i?
>
> After all the Laplace transformation (with s = I) is defined as
>
> Integrate[Exp[-I t], {t, 0, Infinity}]
>
> and should be undefined. Am I wrong or is this a bug?
>
> Thanks for an answer!
> Wieland
This result is correct and (by now) classical. Quoting from "Generalized
Functions: Theory and Technique" by Ram Kanwal: "The Laplace transform
of the Heaviside function is...1/s"
Some ways to derive this involve regularizing and computing in a
limiting sense. Could do as
In[9]:= Integrate[Exp[-I*s*t], {t,0,Infinity}, Assumptions->Im[s]<0]
Out[9]= -I/s
Now let s approach 1 from below (in the complex plane).
Or as
In[13]:= InputForm[l2 = Integrate[Exp[-I*t]*t^a,
{t,0,Infinity}, Assumptions->-1/1000<a<0]]
Out[13]//InputForm= ((-I)*Gamma[1 + a])/E^((I/2)*a*Pi)
In[14]:= Limit[l2,a->0]
Out[14]= -I
Daniel Lichtblau
Wolfram Research
- References:
- Bug in LaplaceTransform?
- From: Wieland Brendel <wielandbrendel@gmx.net>
- Bug in LaplaceTransform?