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 :(.


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:

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

  • $\begingroup$ 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. $\endgroup$ – Xema Nov 18 '14 at 15:02
  • $\begingroup$ 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. $\endgroup$ – Xema Nov 18 '14 at 18:16

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