If $F(k)$ is the Fourier Transform of $f(x)$, then the derivative $F'(k)$ is $-i$ times the Fourier Transform of $xf(x)$. Let $f(x)=\frac{1}{x^2+1}$. Then, $$F(k)=\int_{-\infty}^{\infty} \frac{1}{x^2+1}e^{-ikx}dx$$and $$F'(k)=-i\int_{-\infty}^{\infty} \frac{x}{x^2+1}e^{-ikx}dx$$Caution is sugge...
I use the usual convention $\hat f(k) = \int e^{-ikx} f(x) dx$, so your argument is not correct, I think. Note also that $\frac{x}{1+x^2}\notin L^1$ so your second integral is not really well defined.
The derivative of this example does exist and in the theory of Generalized Functions, the relationship maintains. And there is no "usual" convention for Fourier Transforms. If you change $k$ to $-k$, the approach is still sound.
Peter, the integrals converge. The integrals are most certainly well-defined as both improper Riemann integrals or Lebesgue integrals.
I know what you mean. This integral doesn't exists in Lebesguesense but converges as an improper integral. However, I would like to consider Lebesgue integration only. The rest goes with distributions. Anyway, I still don't see an answer to my question. But thanks so far.
I gave you a way to find the FT. And my approach yields the Wolfram Alpha solution that you were trying to match. What is yet to be answered?
The integral DOES exist as a Lebesgue integral. What is the source of confusion? The oscillatory nature of the complex exponential term provides convergence when $f$ tend to zero as $x$ tends to infinity. Well known.
Apart from the fact that, as I already said $\frac{x}{1+x^2}e^{-ikx}$ is not Lebesgue integrable, could you please make your answer a bit more detailed, I used your techniques (just in a rigorous way) but got the same result.
Peter. The integrand is Lebesgue integrable. I don't know why you're insisting it isn't. A function does not have to be absolutely integrable to be integrable.
What theorem are you using to believe that the integrand isn't LI?
It's just a basic definition of Lebesgue integrable.
f is Lebesgue integrale if and only if |f| is Lebesgue integrable.
it is not a theorem, it comes from the construction of the lebesgue integral
I downvoted because you didn't read my attempts and just put on a wrong answer. I did not use your ideas, these are already in my attempts (just in a rigourous way).
What was your question? What did you do wrong? It is not easy to tell since your posting does not define the notation that you used. I'll edit and show you step by step that I recover WA.
Discussion between Dr. MV and Peter
Discussion between Dr. MV and Peter