integration - The partial expectation $mathbb{E}(X;_{X>K})$ for an alpha-stable distributed random variable

The partial expectation $\mathbb{E}(X;_{X>K})$ for an alpha-stable distributed random variable:

By playing with convolutions of Characteristic Functions of alpha-Stable distributions $S(\alpha, \beta, \mu, \sigma)$ and a payoff $K$, assuming $\mu=0$ and deriving under the integral sign, found the partial expectation $\mathbb{E}(X;_{X>K})$ , i.e., where F(x) is the distribution function of X, $\int_K^\infty x\, \mathrm{d}F(x)$ (which is not to be confused with the conditional expectation).

I am ending up with a difficult integral (easy to evaluate numerically but hard to get explicitly). With $1<\alpha\leq 2$:

$$\psi (\alpha, \beta, \sigma, K) = \frac{1}{2 \pi }\int_{-\infty }^{\infty } \alpha \sigma ^{\alpha } \left| u\right| ^{\alpha -2} \left(1+i \beta \tan \left(\frac{\pi \alpha }{2}\right) \text{sgn}(u)\right) \exp \left(\left| u \sigma \right| ^{\alpha } \left(-1-i \beta \tan \left(\frac{\pi \alpha }{2}\right) \text{sgn}(u)\right)+i K u\right) du$$

The solutions is easy for $K=0$, so
$$\psi(\alpha,\beta,\sigma,0)=-\sigma\frac{\Gamma \left(-\frac{1}{\alpha }\right) \left(\left(1+i \beta \tan \left(\frac{\pi \alpha }{2}\right)\right)^{1/\alpha }+\left(1-i \beta \tan \left(\frac{\pi \alpha }{2}\right)\right)^{1/\alpha }\right)}{\pi \alpha } .$$
Also, there is a well known solution for symmetric cases in Zolotarev's book. But it is the $K \ne 0$ that is critical.