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

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