For the proof of inverse Laplace transform, we change the integral from $\omega$ to $s$. I want to know the reason why we need to change the integral? enter image description here

  • $\begingroup$ Your question is very vague. I have never seen the change of variable you describe. Can you point to an example? Often, the substitution $s = \sigma + j \omega$ is used, but never $s = \omega$. $\endgroup$
    – Peter K.
    Nov 25, 2015 at 12:33
  • $\begingroup$ I have saw this in the book , Signals and Systems by Alan Oppenheim $\endgroup$ Nov 25, 2015 at 12:34
  • $\begingroup$ Nowhere does Oppenheim use the substitution $s = \omega$, because it is nonsensical. Please check the notes here especially page 3 of the PDF. You will see the substitution is $s = \sigma + j \omega$. $\endgroup$
    – Peter K.
    Nov 25, 2015 at 12:38
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    $\begingroup$ The text plainly says it's the inverse Fourier transform, not Laplace that they are performing in equation (9.35). The reason is that the Fourier and Laplace transforms are related: the Fourier transform is the Laplace transform evaluated on the imaginary axis of $s = \sigma + j \omega$ (i.e. $\sigma = 0$. $\endgroup$
    – Peter K.
    Nov 25, 2015 at 13:07
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    $\begingroup$ Got it..and similarly the $d\omega$ part is changed. taking the derivative of complex axis $ds=d\sigma + j d\omega$.. as $\sigma$ is constant so it will be zero so we will left with $ds= j d\omega$ right? $\endgroup$ Nov 25, 2015 at 13:25

1 Answer 1


To summarize the discussion:

  • The usual substitution is $s = \sigma + j \omega$ where $\sigma$ is the real part of the $s$ variable and $\omega$ is the imaginary part.

  • The equation in the image is for the Fourier transform, not the Laplace transform. The Fourier transform can be thought of as the Laplace transform evaluated on the imaginary axis ($\sigma = 0$).

  • The differential $ds$, when looking at real and imaginary parts distinctly, becomes $d\sigma + j d\omega$.

  • Any differential is an infinitesimal (very small) change in that variable. $dx$ is a small change in $x$.


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