Suppose $X(jw)$ is the Fourier Transform of $x(t)$. We know that the Fourier Transform of $\frac{dx(t)}{dt}$ is $jwX(jw)$. Now, why the following integration by parts does not give the same result?

$\int_{-\infty}^{\infty}\frac{dx(t)}{dt} e^{-jwt}dt = x(t)e^{-jwt}]_{-\infty}^{\infty}+ \int_{-\infty}^{\infty}jw x(t)e^{-jwt}dt =x(t)e^{-jwt}]_{-\infty}^{\infty}+ jwX(jw)$.

What to do with the first term $x(t)e^{-jwt}]_{-\infty}^{\infty}$?

Presumably $\int_{-\infty}^{\infty}x(t)dt$ exists. Therefore $\lim{x\to \pm\infty}=0$, so the term you are worried about $=0$.

  • Do all integrable functions vanish at infinity, though? I know Dirichlet conditions hold here. – Elnaz Nov 8 at 20:17
  • Integrability usually means $\int_{-\infty}^{\infty}||x(t)|dt \lt \infty$, in which case the integrand $\to 0$. This is presumably require if you want to take the derivative of the Fourier transform. – herb steinberg Nov 8 at 21:52
  • It is possible that the limits at infinity are not $0$. However in that case the limits do not exist. For example the function could have a non-zero value on an unbounded set of measure zero . – herb steinberg Nov 9 at 4:41

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.