Differentiation of Fourier Transform

Question

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 $x(t)e^{-jwt}]_{-\infty}^{\infty}$?

Answer 1

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

Answer 2

I like this derivation avoiding integration by parts step.

If $x(t) = \frac{1}{2\pi} \left( \int_{-\infty}^{+\infty} \hat{x}(w) e^{i\omega t} \,d\omega\right)$ than we can write $dx(t)/dt = \frac{1}{2\pi} \left( \int_{-\infty}^{+\infty} \hat{x}(w) (i\omega) e^{i\omega t} \,d\omega\right)$. From last equation we can deduce $\hat{dx/dt} = \hat{x}(w) (i\omega)$.