Consider 2 real functions $f, g: \mathbb{R}\to\mathbb{R}$ and $\alpha \in \mathbb{R}$. First I thought that the following equality was true \begin{align} &\int_{-\infty}^{\infty}dtg(t)\frac{1}{2\pi}\int_{-\infty}^{\infty}d\omega e^{-i\omega t}(i\omega)^\alpha\int_{-\infty}^{\infty} d\tau e^{i\omega \tau}f(\tau) \\ = &(-1)^\alpha \int_{-\infty}^{\infty} d\tau f(\tau)\frac{1}{2\pi}\int_{-\infty}^{\infty} d\omega e^{-i\omega \tau} (i\omega)^\alpha\int_{-\infty}^{\infty} dt e^{i\omega t} g(t) \\ \end{align} where we made the substitution $\omega \to -\omega$.

However on further inspection I noticed that the left side is real (can be checked by taking the complex conjugate) but the right hand side is not (since $(-1)^\alpha = e^{i\pi\alpha}$). So my question is what am I missing?


I think the fault is when I do the substitution $(-1)^\alpha (i\omega)^\alpha \stackrel{?}{=} (-i\omega)^\alpha$. But then how do I isolate the -?

  • $\begingroup$ Shouldn't it be $e^{i\omega \tau}$ after the substitution? Then if you don't factor out the $(-1)^\alpha$ you can use that $-i$ is the conjugate of $i$. $\endgroup$ Oct 29, 2022 at 15:55
  • $\begingroup$ I swapped some integration orders as well (which I am also not sure if I am allowed to do). First the order is $t \to \omega \to \tau$ but after the equal I swapped them such that the order becomes $\tau \to \omega \to t$ and then renamed $t \longleftrightarrow \tau$. $\endgroup$
    – Audrique
    Oct 29, 2022 at 16:02
  • $\begingroup$ If you want to swap the order of integration for a function $f(x)$ you have to ensure $\vert f(x) \vert$ has a finite integral on the same interval. Did you check that? $\endgroup$ Oct 29, 2022 at 18:19
  • $\begingroup$ I came to the first form of this integral with an assumption that it converges for 'nice enough' functions, so I should be able to swap them right. $\endgroup$
    – Audrique
    Oct 30, 2022 at 9:54
  • $\begingroup$ The "nice enough" condition is that $\int \vert f(x) \vert$ has to be bounded on $(-\infty , \infty)$. I don't think that's the case here since $\vert e^{i \omega \tau} \vert = 1$ and the monomial term is unbounded. $\endgroup$ Oct 30, 2022 at 10:02


You must log in to answer this question.

Browse other questions tagged .