# Confusion in proof of Inverse Laplace Transform

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? • 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 $s = \omega$. – Peter K. Nov 25 '15 at 12:33
• I have saw this in the book , Signals and Systems by Alan Oppenheim – Aadnan Farooq A Nov 25 '15 at 12:34
• 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 $s = \sigma + j \omega$. – Peter K. Nov 25 '15 at 12:38
• 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. $\sigma = 0$. – Peter K. Nov 25 '15 at 13:07
• 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? – Aadnan Farooq A Nov 25 '15 at 13:25

• 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 $x$.