Solving an integral for a characteristic function For $L>0, H>0,\alpha>0,\sigma>0,$
$$f(t)=\int_L^H \frac{ e^{i t x} \alpha  H \left(\frac{\sigma -H \log \left(\frac{H-x}{H-L}\right)}{\sigma }\right)^{-\alpha -1}}{\sigma  (H-x)} \, \mathrm{d}x$$
$\textbf{Background}$ This is the characteristic function of a nonstandard probability distribution.
 A: HINT
Substitution
$$y=\frac{\sigma -H \log \left(\dfrac{H-x}{H-L}\right)}{\sigma },\tag1$$
$$dy = \dfrac H{\sigma(H-x)},$$
$$x = H-(H-L)e^{\sigma(1-y)/H}$$
allows to write the issue integral
$$f(t)=\int\limits_L^H \frac{ e^{itx} \alpha  H \left(\frac{\sigma -H \log \left(\frac{H-x}{H-L}\right)}{\sigma}\right)^{-\alpha -1}}{\sigma  (H-x)} \, \mathrm{d}x\tag2$$
in the form of
$$f(t)=e^{it\left(H-(H-L)e^{\sigma/H}\right)}\int\limits_1^\infty e^{-it(H-L)e^{-\sigma y/H}} \alpha  y^{-\alpha-1}\, \mathrm{d}y.\tag3$$
Integration by parts leads to the forms of
\begin{align}
&f(t)=e^{it(H-(H-L)e^{\sigma/H})}\times\\
&\left(-e^{-it(H-L)e^{-\sigma y/H}}y^{-\alpha}\Big|_{y=1}^\infty + it\dfrac{H-L}H\sigma\int\limits_1^\infty y^{-\alpha} e^{-it(H-L)e^{-\sigma y/H}}e^{-\sigma y/H}\, \mathrm{d}y\right)\\
&=e^{it\left(H-2(H-L)\cosh\left(\frac\sigma H\right)\right)}+it\frac{H-L}H\sigma e^{it(H-(H-L)e^{\sigma/H})}\int\limits_1^\infty y^{-\alpha} e^{-it(H-L)e^{-\sigma y/H}}e^{-\sigma y/H}\, \mathrm{d}y.\tag4\\
\end{align}
Note that
\begin{align}
&\int\limits_1^\infty y^{-\alpha} e^{-it(H-L)e^{-\sigma y/H}}e^{-\sigma y/H}\, \mathrm{d}y=\\
&\sum\limits_{k=0}^\infty (-it(H-L))^k\int\limits_0^\infty y^{-\alpha}e^{-\frac{\sigma(k+1)y}H}\, \mathrm{d}y\\
\end{align}
can be calculated using incomplete gamma function.
