# Why can't I use the differentiation property of the Fourier transform?

I have some question about the function in frequency domain and I'd like to know its inverse fourier transform (IFT)

$$G(jw) = \dfrac{jw\cdot (jw+1)}{(2+jw)(3+jw)}$$

I know that:

$$\dfrac{d}{dt}x(t)\Leftrightarrow jw\cdot X(jw)$$

And easy to figure out:

$$G(jw) = jw\cdot \left(\dfrac{2}{jw+3}-\dfrac{2}{jw+2}\right)\\g(t) = \dfrac{d}{dt}2\cdot(e^{-3t}-e^{-2t}) = \boxed{-6\cdot e^{-3t}+4\cdot e^{-2t}}$$

ok. But i tried to compute this way:

lets call $$jw = s$$:

$$G(s) = \dfrac{s\cdot (s+1)}{(s+2)(s+3)} = \dfrac{s^2+s}{s^2+5s+6} = 1-\dfrac{4s+6}{(s+2)(s+3)}$$

Using partial fraction:

$$\dfrac{4s+6}{(s+2)(s+3)} = \dfrac{-2}{s+2}+\dfrac{6}{s+3}$$

So:

$$G(s) = 1-\left(\dfrac{-2}{s+2}+\dfrac{6}{s+3}\right)\\G(jw) = 1+\dfrac{2}{2+jw}-\dfrac{6}{3+jw}\Rightarrow g(t) = 2\pi \cdot\delta(t)+2\cdot e^{-2t}-6\cdot e^{-3t}$$

What I did wrong?

1. the "easy to figure out" part is wrong: $$G(j\omega)=j\omega\cdot(\ldots)$$. Correct the part in parentheses.
2. you forget the step function $$u(t)$$. The function $$g(t)$$ is not equal to the derivative of $$c_1e^{-2t}+c_2e^{-3t}$$ but it is the derivative of$$\big(c_1e^{-2t}+c_2e^{-3t}\big)u(t)$$. This makes a difference, because when you take the derivative, you must also consider the step function, which results in a discontinuity at $$t=0$$, hence the Dirac impulse.
3. your result obtained by the second method has the correct constants, but it also misses the step function. Furthermore, with the usual EE definition of the Fourier transform, there shouldn't be a factor of $$2\pi$$ before the Dirac impulse.