Inverse Fourier transform of $\frac{1-e^{-2\pi ift}}{2\pi if}$ I would like to calculate the inverse Fourier transform of the following
$$H(f) = \frac{1-e^{-2\pi i f t}}{2\pi i f}$$
Can anyone tell me and explain to me how to do that? I don't want just an answer, I would like to understand the method. Thanks a lot in advance.

I multiply by $e^{i 2 \pi ft}$ and integrate
$$\int^\infty _ {-\infty} \frac{e^{i 2 \pi ft}}{i 2 \pi f}-\frac{1}{i 2 \pi f}df$$
Second term is odd, so it will be zero. I am using euler's identity for the first term:
$$\int^\infty _ {-\infty} \frac{\cos (2 \pi ft)}{i 2 \pi f}+i\frac{\sin (2 \pi ft)}{i 2 \pi f}df$$
First term is even function over odd function, so it is odd overall. It will be zero. We have the second term. After cancelling the $i$'s:
$$\int^\infty _ {-\infty} \frac{\sin (2 \pi ft)}{ 2 \pi f}df$$
Rewrite as:
$$t \int^\infty _ {-\infty} \frac{\sin (2 \pi ft)}{ 2 \pi f t}df$$
The inner integral is a famous integral. I will not derive it here. Look here: enter link description here Here is the tricky part:
$$\frac{1}{2\pi} \text{sign}(t) \int^\infty _ {-\infty} \frac{\sin (2 \pi ft)}{ 2 \pi f t}d( 2 \pi f t)$$
You need to have the $sign(t)$ because if $t$ is negative. This is something that comes up very often we you change the variable of integration such that it absorbs a variable which may take negative values. The bounds of the integral must reverse. So we get:
$$\frac{\text{sign} (t)}{2}$$
