Integrating :$\int_{0}^{1} \frac{x^{b}-x^{a}}{\log(x)}\sin(\log(x))dx$ 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?
 A: Your integral consists of two similar part and we only do one of them, namely:
\begin{align}
\int_0^1 \frac{x^a}{\ln(x)}\sin(\ln(x)) dx
\end{align}
Set $x=e^{-u}$ so that $dx = -e^{-u}du$, and we get:
\begin{align}
\int^\infty_0 \frac{e^{-(a+1)u}}{u}\sin(u)du
\end{align}
Now consider: 
\begin{align}
I(z) = \int^\infty_0 \frac{e^{-zu}}{u}\sin(u)du
\end{align}
This is well defined as a real number for $z>0$, otherwise the integral won't converge. We also have:
\begin{align}
I'(z) = -\int^\infty_0 e^{-zu}\sin(u) du
\end{align}
I assume you can evaluate that, two times partial integration and you get:
\begin{align}
I'(z) = -\frac{1}{z^2+1}
\end{align}
Integrating this gives us $I(z)$ namely:
\begin{align}
I(z) = -\arctan(z) + C
\end{align}
Now we would actually determine $C$. But that is not needed, since our original integral is a diference of $I(b+1)$ and $I(a+1)$ so that $C$ will be canceled. So: 
\begin{align}
\int_0^1 \frac{x^b-x^a}{\ln(x)}\sin(\ln(x)) dx =I(b+1) - I(a+1) = \arctan(a+1) - \arctan(b+1)
\end{align}
And we have (indirectly) seen that is only convergent if $a>-1, b>-1$.
Edit:
If you really want to determine the value of $C$:
Note that we know the following:
\begin{align}
\lim_{z\to \infty} I(z) = -\frac{\pi}{2}+C
\end{align}
And we also know:
\begin{align}
\lim_{z\to \infty} I(z) = \lim_{z\to \infty} \int^\infty_0 \frac{e^{-zu}} {u}\sin(u)du \stackrel{\text{DCT}}{=} \int^\infty_0 \lim_{z\to \infty} \frac{e^{-zu}} {u}\sin(u)du = 0
\end{align}
Where DCT means Dominated Convergence Theorem. Now we can solve for $C$ and finally we get:
\begin{align}
I(z) = -\arctan(z)  +\frac{\pi}{2}
\end{align}
A: It suffices to evaluate
$$\int_0^1 \frac{x^a}{\ln x}\sin(\ln x) dx = \int_0^\infty \frac{e^{-(a+1)x}}{x}\sin x dx = \int_0^\infty {e^{-(a+1)x}}\sin x d(\ln x)$$
Integration by parts shows it suffices to evaluate integrals like
$$\int_0^\infty {e^{-(a+1-i)x}}\ln x dx$$
Using the following formula, valid for $\Re(z)> 0$, enables you to finish the problem:
$$\int_0^\infty e^{-zx}\ln x dx = -\frac{\gamma+\ln z}{z}$$
where $\gamma$ is this constant.
A: Observes that: $$\color{red}{\frac{d}{dt}\left(\frac{x^t}{\ln x}\right) =x^t}$$ we have,
\begin{align}\int_{0}^{1} \frac{x^{b}-x^{a}}{\log(x)}\sin(\log(x))dx 
&= \int_{0}^{1} \left[\frac{x^{t}}{\log(x)}\right]_a^b\sin(\log(x))dx   \\ &=\int_a^b\int_{0}^{1} x^t\sin(\log(x))dx dt\\
& \overset{\color{red}{u =-\ln x}}{=}- \int_a^b\int_{0}^{\infty} e^{-ut-u}\sin(u)du dt\\
&= -\int_a^b Im\left(\int_{0}^{\infty} e^{(-t-1+i)u}du \right)dt\\
&=-\int_a^b Im\left[\frac{1}{-t-1+i}e^{(-t-1+i)u} \right]_0^\infty dt \\
&=- \int_a^b \frac{dt}{(t+1)^2+1} \\
&=\color{blue}{\arctan (a+1)-\arctan (b+1)}\end{align}
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\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}
A: We can use a Frullani Integral approach, but in $\mathbb{C}$, it requires Cauchy's Integral Theorem instead of direct cancellation.
$$
\newcommand{\Im}{\operatorname{Im}}
\begin{align}
\int_0^1\frac{x^b-x^a}{\log(x)}\sin(\log(x))\,\mathrm{d}x
&=\int_0^\infty\frac{e^{-(b+1)x}-e^{-(a+1)x}}{x}\sin(x)\,\mathrm{d}x\tag1\\
&=\Im\left(\int_0^\infty\frac{e^{-(b+1-i)x}-e^{-(a+1-i)x}}{x}\,\mathrm{d}x\right)\tag2\\
&=\lim_{\epsilon\to0}\Im\left(\int_\epsilon^{1/\epsilon}\frac{e^{-(b+1-i)x}-e^{-(a+1-i)x}}{x}\,\mathrm{d}x\right)\tag3\\
&=\lim_{\epsilon\to0}\Im\left(\int_{(b+1-i)\epsilon}^{(b+1-i)/\epsilon}\frac{e^{-x}}{x}\,\mathrm{d}x-\int_{(a+1-i)\epsilon}^{(a+1-i)/\epsilon}\frac{e^{-x}}{x}\,\mathrm{d}x\right)\tag4\\
&=\lim_{\epsilon\to0}\Im\left(\int_{(b+1-i)\epsilon}^{(a+1-i)\epsilon}\frac{e^{-x}}{x}\,\mathrm{d}x-\int_{(b+1-i)/\epsilon}^{(a+1-i)/\epsilon}\frac{e^{-x}}{x}\,\mathrm{d}x\right)\tag5\\
&=\Im\left(\log\left(\frac{a+1-i}{b+1-i}\right)\right)\tag6\\[6pt]
&=\tan^{-1}(a+1)-\tan^{-1}(b+1)\tag7
\end{align}
$$
Explanation:
$(1)$: substitute $x\mapsto e^{-x}$
$(2)$: $\sin(x)=\Im\left(e^{ix}\right)$
$(3)$: write improper integral as a limit
$(4)$: separate integrals and substitute $x\mapsto\frac{x}{b+1-i}$ and $x\mapsto\frac{x}{a+1-i}$
$(5)$: apply Cauchy's Integral Theorem
$(6)$: as $\epsilon\to0$, $e^{-x}\to1$ in the first integral and $e^{-x}\to0$ in the second
$(7)$: $\Im(\log(x+iy))=\tan^{-1}(y/x)$
