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?
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asked Nov 25, 2015 at 11:39
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$\begingroup$ 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
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$\begingroup$ I have saw this in the book , Signals and Systems by Alan Oppenheim $\endgroup$– Aadnan Farooq ANov 25, 2015 at 12:34
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$\begingroup$ 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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1$\begingroup$ 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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1$\begingroup$ 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? $\endgroup$– Aadnan Farooq ANov 25, 2015 at 13:25
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1 Answer
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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 $x$.
