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I have to find the Fourier coefficients of $$ \frac{1}{1+ t^{2}} $$ I tried with $$ \frac{1}{T}\int_{0}^{T} \frac{1}{1+it}e^{-i 2 \pi f_0 T } $$ but I should do at least two integrals by parts , so I tried to calculate first the even part and the odd part of x(t). I obtained that the even part is $$ x_p = \frac{1}{1 + t^2} $$ and the odd part is $$ x_d = \frac{-2it}{1 + t^2} $$ now I should find the Fourier coefficients of $x_p$ and of $x_d$ ( $ X_p(k)$ and $ X_d(k)$) and the Fourier coefficients should be the sum of these two. The only problem is that I that I still have integrals by parts.

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    $\begingroup$ You need the Fourier coefficients? So, that is a periodic signal described as $1/(1+t^2)$ in one period? It looks to me as a non-periodic signal and you need its Fourier Transform to check its frequency content. $\endgroup$ – GKH Jan 16 at 14:21
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Assuming that you want the Fourier Transform of this signal, you can easily obtain it by using a very useful property and a well known FT pair in continuous time. The pair is

$$x(t) = e^{-at}u(t), \: \: a>0 \longleftrightarrow X(f) = \frac{1}{a+j2\pi f}$$

and the so-called duality property says that if $$x(t) \longleftrightarrow X(f)$$ is an FT pair, then:

$$X(t) \longleftrightarrow x(-f)$$

For $a=1$, the first equation gives $$x(t)=e^{-t}u(t) \longleftrightarrow X(f)= \frac{1}{1+j2\pi f}$$

Now you have to apply the duality property to get your result quickly and nicely. That is, since $$x(t) = e^{-at}u(t) \longleftrightarrow X(f) = \frac{1}{a+j2\pi f}$$ the duality property suggests that $$X(t) = \frac{1}{a+j2\pi t} \longleftrightarrow x(-f) = e^{af}u(-f)$$ From that, we can extract the signal we're interested in: $$X(t)= \frac{\frac{1}{2\pi}}{\frac{a}{2\pi} + jt}= \frac{1}{2\pi}\frac{1}{\frac{a}{2\pi} + jt}\longleftrightarrow x(-f)=e^{af}u(-f)$$ and then $$\frac{1}{\frac{a}{2\pi} + jt} \longleftrightarrow 2\pi e^{af}u(-f)$$ by multiplying both terms by $2\pi$. Finally, setting $a=2\pi$ yields $$\frac{1}{1+jt} \longleftrightarrow 2\pi e^{2\pi f}u(-f)$$

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  • $\begingroup$ I’m trying to understand and (Maybe , I hope ) I understand how you found X(f) but I don’t know why k=2pi f. I mean , we don’t know k value. 2pi I shouldn’t be w? And after finding X(f) I don’t know how to find the coefficients. I thought that the definition of X(f) it’s an integral from -oo to +oo and that x_k is a very similar integral but with*1/T . but again we don’t know the extremes of integration ( we don’t have period ) so I can’t use the value of X(f) to find coefficients with the classic definition $\endgroup$ – Elena Martini Jan 16 at 17:09
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    $\begingroup$ Hmm, OK, we have to clear things up... :) I know you know about the Fourier Series. Do you know anything about the Fourier Transform? In brief, the Fourier Series is an expansion of a periodic signal in complex harmonic exponentials, that is, $$x(t) = \sum_{k} X_k e^{j2\pi kf_0 t}$$ with $$X_k = \frac{1}{T_0}\int_{<T_0>} x(t)e^{-j2\pi kf_0t}dt$$ As such, the Fourier Series are useful for periodic signals only. On the other hand, the Fourier Transform is more or less the same expansion, but for non-periodic signals. $\endgroup$ – GKH Jan 16 at 18:29
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    $\begingroup$ The Fourier Transform is defined as $$X(f) = \int_{-\infty}^{+\infty}x(t)e^{-j2\pi ft}dt$$ and as you can see it is defined for all $f$ in $(-\infty, +\infty)$ and the integration is performed not over a single period but for all $t$. The Fourier Series is defined only for certain frequencies that are integer multiples of a fundamental $f_0$. In your case, it seems you have a non-periodc signal, so the Fourier Transform is suitable for representing your signal in the frequency domain. Is it true that your signal is non-periodic indeed? If yes, are all these clear? $\endgroup$ – GKH Jan 16 at 18:32
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    $\begingroup$ Also, what is your signal? Is it $$\frac{1}{1+t^2}$$ or $$\frac{1}{1 + jt}?$$ Because it is not clear from your question... $\endgroup$ – GKH Jan 16 at 18:38
  • $\begingroup$ thank you so much for having clarified the links between series, coefficients and fourier transform!!! I spend my days doing exercises to understand because sometimes I find theory on my book difficult. But your explanations always help me ! The original signal was $$ \frac{1}{1+ it} $$ and I have to calculate Fourier transform that should be $$ 2 \pi e^{2 \pi f } u(-f) $$ I wrote that I had to find Fourier coefficients because my brain is totally out 🤯 $\endgroup$ – Elena Martini Jan 17 at 1:18

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