Re: Laplace transform of reciprocal?



On 22 May 2006 05:27:29 -0700, 1940LaSalle@xxxxxxxxx wrote:

I need the Laplace transform for 1/t. It would appear at first glance
that this would be Ei(s), where Ei denotes the exponential integral.
Is that right?

No, the transform of 1/t doesn't exist. One source for more
information is the online version of Abramowitz and Stegun at
<http://www.math.sfu.ca/~cbm/aands/>

The Laplace transform of 1/(t+a), for a>0, is given at
<http://www.math.sfu.ca/~cbm/aands/page_1029.htm>

Info on the exponential integral and related functions
is given at
<http://www.math.sfu.ca/~cbm/aands/page_228.htm>

HTH

.



Relevant Pages

  • Re: inverse laplace transform
    ... > The Laplace transform only work with linear ... You simply cannot do a Laplace transform of a nonlinear ... >> follow the impulse invariance IIR filter design method. ...
    (comp.dsp)
  • Re: Why exp(-st) in the Laplace Transform?
    ... If you take the Laplace transform, ... equivalent to multiplying it by a real number ). ... distinguishes the Laplace Transform from the Fourier Transform. ...
    (sci.physics.research)
  • Re: Why exp(-st) in the Laplace Transform?
    ... If you take the Laplace transform, ... equivalent to multiplying it by a real number ). ... distinguishes the Laplace Transform from the Fourier Transform. ...
    (sci.math.research)
  • Re: Why exp(-st) in the Laplace Transform?
    ... The existence of both the Laplace and Fourier transforms ... come from the symmetry of the real line under translations (t -> t + a, ... transform kernels, exp, for w any real number. ... If you take the Laplace transform, ...
    (sci.math.research)
  • Re: Why exp(-st) in the Laplace Transform?
    ... The existence of both the Laplace and Fourier transforms ... come from the symmetry of the real line under translations (t -> t + a, ... transform kernels, exp, for w any real number. ... If you take the Laplace transform, ...
    (sci.physics.research)