A closed-form for $\int_1^\infty x^{-\frac{a}{2}}e^{-x}\cos(a b\log x)dx, $ when $a,b>1$

Question

Yesterday I've found this problem from some calculations that I did

Question. For $a>1$ and $b>1$, calculate in a closed-form (if it is possible find such closed-form) in terms of $a$ and $b$ the integral $$\int_1^\infty x^{-\frac{a}{2}}e^{-x}\cos(a\cdot b\log x)dx. $$

Thanks in advance.

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\int_{1}^{\infty}x^{-a/2}\expo{-x}\cos\pars{ab\ln\pars{x}}\,\dd x = \Re\int_{1}^{\infty}x^{-a/2}\expo{-x}\expo{\ic ab\ln\pars{x}}\,\dd x = \Re\int_{1}^{\infty}{\expo{-x} \over x^{a\pars{1/2 - b\ic}}}\,\dd x \\[5mm] & = \Re\pars{\mrm{E}_{a\pars{1/2 - b\ic}}\pars{1}} \end{align}

$\ds{\mrm{E}_{p}\pars{z}}$ is the Generalized Exponential Integral.