# Differentiation of Fourier Transform

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 $$=0$$.
• 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 '18 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 '18 at 4:41
• In fact $\liminf$ of the integrand must go to 0 but not necessarily the $\limsup$ to be more precise with Steinberg statement. – Malik Dec 19 '18 at 0:13