# Fourier Transform - Limit not existing

I was wondering what happens when calculating the fourier transform of a fonction, the limit does not exist.

Let me explain this with an example :

I have the function difined by :

f(t) = e^t when -infinity < t < 0


We then calcul the fourier transform :

F(jw) = integral (-infinity -> 0) f(t) e^-jwt dt
F(jw) = integral (-infinity -> 0) e^(1-jw)t dt
F(jw) = [e^(1-jw)t / (1-jw)] between 0 and -infinity.


Here we can't calcul the limit in -infinity. Can we say that :

F(jw) = 1 / (1-jw) ?


If it is wrong, why ? And what can we do ?

(If it can help, the original question is to calcul the fourier transform of the function f(t) defined by e^t between -infinity and 0 AND e^-t between 0 and infinity).

Thank you guys,

PS : I don't know if there is a proper formating of mathematic formulas, sorry about that if it does :(.

## 1 Answer

The integral $$\int_{-\infty}^{0}e^{(1-j\omega) t}dt$$ is defined as the limit of the integral $$\int_{b}^{0}e^{(1-j\omega) t}dt$$ as $b\rightarrow-\infty$. As you said yourself, these can be calculated by a primitive function:

$$\int_{b}^{0}e^{(1-j\omega) t}dt=\Bigg[\frac{e^{(1-j\omega)t}}{1-j\omega} \Bigg]_{t=b}^{0}= \frac{1-e^{(1-j\omega)b}}{1-j\omega}$$

As $b\rightarrow-\infty$, the term $e^{(1-j\omega)b}$ tends to $0$, as: $$|e^{(1-j\omega)b}|=|e^b| |e^{-j\omega t}|=e^b\overset{b\rightarrow-\infty}\longrightarrow0$$

So we get:

$$\int_{-\infty}^{0}e^{(1-j\omega) t}dt=\lim_{b\rightarrow-\infty}\int_{b}^{0}e^{(1-j\omega) t}dt=\lim_{b\rightarrow-\infty} \frac{1-e^{(1-j\omega)b}}{1-j\omega}=\frac{1}{1-j\omega}$$

• I don't get why you can say that |e^(1−jω)b|=|e^b||e^−jωt|=e^b. And I think you wanted to say |e^(1−jω)b|=|e^b||e^−jωb|=e^b. – Xema Nov 18 '14 at 15:02
• I just figured it out myself. The e^-jwb can be written with cos and sin, which vary between -1 and 1. Therefore, the limit is the limit of e^b. – Xema Nov 18 '14 at 18:16