Helo everybody.
I would like to know how to get the Discrete Fourier Transform of the following expression:
$$a[n] = \frac{1}{n}$$
If we try to apply the definition of the DTFT, we get:
$$A(\omega)=\sum_{k = -\infty}^{\infty} \frac{1}{k}e^{i\omega k}$$
Which does not classically converge.
Nevertheless, it may be possible to obtain a result involving Dirac's Delta Function which is in fact my main purpose with this post. I realized that:
$$\frac{dA(\omega)}{dw} = i \sum_{k = -\infty}^{\infty} e^{i\omega k} = i\delta(\omega + 2n\pi)$$
However, I am having trouble expressing $A$ directly in terms of the integral of $\delta$. This is because:
$$\int_a^bd\omega \ \delta(\omega + 2n\pi) = \int_a^bd\omega \sum_{k = -\infty}^{\infty} e^{i\omega k} = \sum_{k = -\infty}^{\infty} \frac{-i}{k}e^{i\omega k} \bigg\rvert^{b}_a$$
And I am not able to find any suitable value for $a$. Furthermore, the integral of $\delta$ corresponds in fact to the Heaviside Step Function $H(\omega)$, and it seems so counterintuitive to me relating $A(\omega)$, which is symmetrical with respect to the origin, with this last function.
Any help will be welcomed!
Thank you in advance :)
Edit: The Continuous Time Fourier trasnform of $a(x) = \frac{1}{x}$ is:
$$A_c(\omega) = \text{sgn}(\omega)$$
Up to a factor of $2\pi$ due to normalization. That is, it corresponds to a step function
The immediate problem with the calculation in question is that the series and derivative is wrong or improperly written. One would have to guess, but the most natural setting is with $a[0]=0$ so that the series have to exclude the corresponding term, $$ A(w) =\sum_{k\in\Bbb Z,\,k\ne 0}\frac{e^{ikw}}{k},\\ A'(w)=i\sum_{k\in\Bbb Z,\,k\ne 0}e^{ikw}=-i+\frac{i}{2\pi}\sum_{n\in\Bbb z}\delta(w-2n\pi). $$ The last holds in the distributional sense, test functions have to be at least continuous, for easier results demand compact support.
Then you get for the integral of $iA'(w)$ the diagonal from the first term, while the second term provides periodic jumps down towards a base line.
One can obtain a more direct derivation by applying convergence-enhancing methods. One such method is to consider $$ \sum_{k\in\Bbb Z}c_k=\lim_{|q|<1,\, q\to 1}\sum_{k\in\Bbb Z}q^{|k|}c_k $$ For convergent series this reproduces its value, and for some non-convergent series the limit gives a sensible value.
Applying this method to the series in question, and using the series expansion of the main branch of the complex logarithm for $|z|<1$, $$ -{\rm Ln}(1-z) = \sum_{k=1}^\infty\frac{z^k}{k}. $$ Applying this expansion with $z=qe^{\pm iw}$, the modified series evaluates to \begin{align} \sum_{k\ne 0}\frac{q^{|k|}}{k}e^{ikw} &=\sum_{k=1}^\infty\frac{q^ke^{ikw}}k-\frac{q^ke^{-ikw}}k \\ &=-{\rm Ln}(1-qe^{iw})+{\rm Ln}(1-qe^{-iw}) \\ &=-2i\arg(1-qe^{iw})\tag{a} \\ &=-2i\arg(1-q+q(1-e^{iw})) \\ &=-2i\arg\left(1-q+2iqe^{i(w-\pi)/2}\sin(w/2)\right)\tag{b} \end{align} By (a) the result will be $2\pi$ periodic. By (b) on the period $(0,2\pi)$ with $\sin(w/2)>0$ and for $q\approx 1$, this gives a value close to $i(\pi-w)$. The periodic continuation is a saw-tooth pattern of period $2\pi$ oscillating between $i\pi$ and $-i\pi$.