TITLE: Integrating :$\int_{0}^{1} \frac{x^{b}-x^{a}}{\log(x)}\sin(\log(x))dx$
QUESTION [6 upvotes]: Consider $$\int_{0}^{1} \frac{x^{b}-x^{a}}{\log(x)}\sin(\log(x))dx$$
I thought about making some substitution like : $x^{y}$(it's easy to calculate $$\int_{0}^{1}\frac{x^{b}-x^{a}}{\log(x)}dx = \int_{a}^{b}dy\int_{0}^{1}x^{y}dx=\log(b-1)-\log(a-1)$$).
But I couldn't find substitution like that. Any hints?

REPLY [3 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[10px,#ffd]{\int_{0}^{1}{x^{b} - x^{a} \over \ln\pars{x}}\,\sin\pars{\ln\pars{x}}\,\dd x} =
\int_{0}^{1}\pars{x^{b} - x^{a}}{1 \over 2}\int_{-1}^{1}\expo{\ic k\ln\pars{x}}\,\dd k\,\dd x
\\[5mm] = &\
{1 \over 2}\int_{-1}^{1}\int_{0}^{1}\pars{x^{b + \ic k} -
x^{a + \ic k}}\dd x\,\dd k =
{1 \over 2}\int_{-1}^{1}\pars{{1 \over b + 1 + \ic k} -
{1 \over a + 1 + \ic k}}\dd k
\\[5mm] = &\
\int_{0}^{1}\bracks{{b + 1 \over \pars{b + 1}^{2} + k^{2}} -
{a + 1 \over \pars{a + 1}^{2} + k^{2}}}\,\dd k
\\[5mm] = &\
\int_{0}^{1/\pars{b + 1}}{\dd k \over k^{2} + 1} -
\int_{0}^{1/\pars{a + 1}}{\dd k \over k^{2} + 1}
\\[5mm] = &\
\bbx{\arctan\pars{1 \over b + 1} - \arctan\pars{1 \over a + 1}}
\\ &\
\end{align}