For the Pareto probability distribution, with density function
$$f(x ; x_m, \alpha) = \begin{cases} 0 & \qquad \text{for $x \leq 0$} \\ \alpha \frac{x_m^\alpha}{x^{\alpha+1}} & \qquad \text{for $x > x_m$} \end{cases}$$
we have the characteristic function
$$ \varphi(t) = \alpha (-ix_mt)^\alpha \Gamma(-\alpha,-ix_mt)$$
with $\Gamma(s,x)$ the upper incomplete gamma function. (this is according to the Wikipedia page and Wolfram alpha)
How can I derive this?
I have tried the following steps
$$\begin{array}{} \varphi(t) = E[e^{itX}] &=& \int_{x_m}^\infty \frac{\alpha x_m^\alpha}{x^{\alpha+1}} e^{itx} \,dx \\ &=& \alpha x_m^\alpha \int_{x_m}^\infty x^{-\alpha-1} e^{itx} \,dx \\ &=& \alpha (-i x_m t)^{\alpha} \int_{x_m}^\infty (-itx)^{-\alpha-1} e^{itx} \,(-it)dx \end{array}$$ This implies the following
$$ \int_{x_m}^\infty (-itx)^{-\alpha-1} e^{itx} \,(-it)dx = \Gamma(-\alpha,-ix_mt)$$
That is where I get stuck. How can I prove that step? I know that for positive and real $s$ and $x$ we have $\int_x^\infty x^{s-1} e^{-x}\, dx = \Gamma(s,x)$, but in the case above we have negative and imaginary coefficients.
Edit: I found an article deriving the related Laplace transform for a shifted Pareto distribution (Nadarajah and Kotz 2006 Queueing Systems 54(4):243-244). However, it does not explain the step where I get stuck.
They use the following steps:
$$\begin{array}{rcll} E[e^{-sX}] &=& &\int_{0}^{\infty} \frac{ab^a e^{-sx}}{(b+x)^{a+1}} \,dx \\ &=& a(bs)^a e^{bs}& \int_{bs}^{\infty} u^{-a-1} e^{-u} \,du \qquad \text{substituting $u=b+x$} \\ & = & a(bs)^a e^{bs} &\Gamma(-a,bs) \end{array}$$
Is the last step, changing the integral into a Gamma function, valid for negative $-a$? Can I substitute $s$ with $s = -it$, to get the characteristic function from this Laplace transform?