# Fourier transform of a signal with derivative of delta function

I have problems finding the solution of this exercise:

Find the Fourier transfrom of the signal. $$x(t) = e^{-t} \delta'$$

I have tried the following but i'm not sure if it's correct.

$$X(w) = \int_{-\infty}^\infty e^{-t} \delta'e^{-iwt} dt$$

$$X(w) = \int_{-\infty}^\infty \frac d {dt}\delta e^{-(1+iw)t} dt$$

Using integration by parts $$u = e^{-(1+iw)t} \\du = \frac {-1}{1+iw}e^{-(1+iw)t} \\ v = \delta(t) \\ dv = \frac d{dt} \delta(t)$$

$$X(w) = \delta e^{-(1+iw)t} + \int_{-\infty}^\infty \delta(t) \frac {1}{1+iw}e^{-(1+iw)t} dt$$

Evaluating the first parts becomes zero $$X(w) = \int_{-\infty}^\infty \delta(t) \frac {1}{1+iw}e^{-(1+iw)t} dt$$

At this point i don't know how to transform this. I would like to know if my calculations are correct and how to continue.

• The Dirac delta distribution is designed to pull out the value of the function at zero. That is all that is left to do here. – Cameron Williams Jun 28 '17 at 16:44
• The derivative of $e^{-(1+i\omega)t}$ is $-(1+i\omega)e^{-(1+i\omega)t}$. – Mark Viola Jun 28 '17 at 17:27