I'm interested in the quantity: $\int_0^\infty \omega \hat f (\omega) d\omega$ where $\hat f$ is the (one-sided) cosine transform of $f$, that is $f(x) = \int_0^\infty \hat f(\omega)cos(x \omega) d \omega$.
A more or less equivalent way of looking at it is as the integral over the Hilbert-transform of the derivative of $f$, or the integral over the $|\omega| \hat f$, where here $\hat f$ denotes the Fourier transform.
In particular, I am trying to find a closed form for the case $f(x) = \exp(-x^{2k})$, where for the case $k=1$ this integral comes out to be $\frac{2}{\sqrt{\pi}}$.
It seems like this has to have been looked at before, I would be very grateful for any suggestions.
After some further investigation, it seems like the answer is fairly straightforward, and given by $\frac{-2}{\pi}\int_0^\infty \frac{f'(x)}{x} dx$, this can be obtained by considering the Fourier-transform of the absolute value function.
If we look at the situation when $\hat f$ is the Fourier-transform, then $ \int_{-\infty}^\infty |\omega| \hat f(\omega) d \omega$ is (on the level of distributions) equal to $\frac{-i}{2\pi} ( g \ast f')(0)$, where $g$ is the inverse Fourier-Transform of the step function and the scaling factor comes from the way we have normalized the Fourier-transform. By e.g. Theorem 7.19 of Rudin's Functional analysis, this is well-defined. Now it remains to find the inverse Fourier transform of a step function, which is given by the principal value of $\frac{2 i}{x}$, giving the claim (the additional factor 2 comes if we consider the cosine transform instead of the Fourier-transform).