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.
Best Answer
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.