Let $\sigma>0$ , $1<\alpha\leq 2$, and $-1\leq \beta \leq 1$. I am looking for a closed-form solution (or something near) for the following integral.
$$\frac{1}{2 \pi } \text{PV}\int_{-\infty }^{\infty } \left(\pi \delta (u)-\frac{i e^{-i K u}}{u}\right) \left(\mu +\alpha \beta \sigma \tan \left(\frac{\pi \alpha }{2}\right) \left(\sigma \sqrt{u^2}\right)^{\alpha -1}-\frac{i \alpha \left(\sigma \sqrt{u^2}\right)^{\alpha }}{u}\right) \exp \left(i \left(-\mu u+\left(\sigma \sqrt{u^2}\right)^{\alpha } \left(-\frac{\beta \sqrt{u^2} \tan (\frac{\pi \alpha }{2})}{u}+i\right)\right)\right) \, du$$ (where PV indicates the Cauchy principal value and $\delta(.)$ is the Dirac delta function).
This is an improper integral near $u=0$. The numerator goes to 1 near $u=0$ but the denominator makes the integral divergent like $\int_0^1 u^{-2}du$ . I suggest revisiting the two original functions whose Fourier transforms yielded your integral to see that they are in fact integrable and that their convolutions therefore is defined. (For example, are the two functions $L^1$ integrable?) Another suggestion: calculate or estimate your integral over the real line but with the interval $[-\delta,\delta]$ removed. Then see if the limit as $\delta\to0$ exists.
Good luck! Sam
PS, to make matters simpler, take $\beta=K=\mu=0$ and $\alpha=2, \sigma=1$. Might be easier to see what's going on in order to determine whether it can exist.
PSS, my comment has become obsolete since the integral keeps changing.