Let $X$ be a lognormal random variable with pdf $f(X)$ and cdf $F(X)$. The mean and variance of $X$ are assumed to be $\mu$ and $\sigma^2$ respectively. If we assume any constant $k \ge 0$ and $0 < \alpha < 1$, how can we calculate following partial expectation?
$$\int_k^\infty x^{1-\alpha} f(x) dx$$
We assume that the the mean and variance of $\ln X$ are $\mu$ and $\sigma^2$ respectively. That is, $X=e^{\mu+\sigma\xi}$, where $\xi \sim N(0, 1)$ is a standard normal random variable. Then \begin{align*} \int_k^\infty x^{1-\alpha} f(x) dx &= \mathbb{E}\left(\mathbb{I}_{X\ge k} X^{1-\alpha} \right)\\ &=e^{(1-\alpha)\mu}\mathbb{E}\left(\mathbb{I}_{\xi\ge \frac{\ln k-\mu}{\sigma}} e^{(1-\alpha)\sigma \xi} \right)\\ &=e^{(1-\alpha)\mu}\int_{\frac{\ln k-\mu}{\sigma}}^{\infty}\frac{1}{\sqrt{2\pi}}e^{(1-\alpha)\sigma x} e^{-\frac{x^2}{2}}dx\\ &=e^{(1-\alpha)\mu}\int_{\frac{\ln k-\mu}{\sigma}}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac{(x-(1-\alpha)\sigma)^2}{2}+ \frac{(1-\alpha)^2\sigma^2}{2}}dx\\ &=e^{(1-\alpha)\mu + \frac{(1-\alpha)^2\sigma^2}{2}}\Phi\left(\frac{-\ln k +\mu + (1-\alpha)\sigma^2}{\sigma} \right), \end{align*} where $\Phi$ is the cumulative distribution function of a standard normal random variable.