Fourier transform of Wick rotated functions

I am learning the imaginary time formalism of thermal field theory / reviewing the Euclidean formalism of quantum field theory. One thing that appears to be left implicit in many treatments is a statement of the type: the Fourier transform of two functions related by a Wick rotation are also related by a Wick rotation. That is, suppose $$f(t)$$ is a function and $$f_E(\tau) = f(-i\tau)$$. Then the Fourier transform $$\tilde f(\omega)= \int dt e^{i\omega t}f(t)$$ and $$f_E(\omega) = \int d\tau e^{i\omega \tau} f_E(\tau)$$ ought to be related by $$\tilde f_E(\omega) = f(i\omega)$$, or something to that effect.

I can prove such a statement for a real rescaling of the variable: if $$f_a(t) = f(at)$$ then $$\tilde f_a(\omega) = \int dt e^{i\omega t}f(at)= \frac{1}{a}\int_{-\text{sng}(a)\infty}^{\text{sgn}(a)\infty} dt e^{i(\omega/a) t}f(t) = |a|^{-1}\tilde f(\omega/a).$$ But the corresponding steps for a Wick rotation change the domain of integration in a way that I cannot relate to the original Fourier transform: $$\tilde f_E(\omega) = \int d\tau e^{i\omega t}f(-i\tau) = i \int_{t=+i\infty}^{t=-i\infty} dt e^{i(i\omega)t} f(t).$$ The last step to me is not obviously equal to $$\sim\tilde f(i\omega)$$ because the integration is now along the imaginary axis.

Does such a statement actually exist? Am I missing a last step?

• Related: physics.stackexchange.com/q/455387/2451 and links therein. Feb 27, 2022 at 7:14
• @Qmechanic , the statement I am seeking to prove appears to be the third set of equalities in the set of equations (D) in your answer. Should that statement be obvious? I still don't know what to do with the modified contour of integration Feb 27, 2022 at 21:08