Help with the integral $\int_{0}^{1}u^{-z} x^{u}\left(x^{u}-1 \right )\sin\left(2\pi y\left(x^{u}-1 \right ) \right )du$ I am trying to find the general form of the integral :
$$I(x,y,r)=\int_{0}^{1}u^{-r} x^{u}\left(x^{u}-1 \right )\sin\left(2\pi y\left(x^{u}-1 \right ) \right )du$$
Where $x,y \in \mathbb{R}^{+}$ and $r\in \mathbb{C}$ .
The integral $I(x,y,r)$ cab be written as :
$$I(x,y,r)=\left(\log x \right )^{r-1}\int_{0}^{x-1}\left[\log(1+l) \right ]^{-r}l\sin\left(2\pi y l  \right )dl$$
But that doesn't make it easier to evaluate. Any help is highly appreciated . 
EDIT 
The problem arose in the attempt to give a 'nicer' form for the summation :
$$\sum_{n=1}^{\infty}\mu(n)\frac{d}{dy}\left(\frac{\sin\left(2\pi y\left(x^{1/n}-1 \right ) \right )}{\pi y} \right )$$
Where $\mu(n)$ is the Mobius function. A nicer form is one that doesn't include the function $\mu(n)$. Using the Abel's summation formula, we have : 
$$\sum_{n=1}^{\infty}\mu(n)\frac{d}{dy}\left(\frac{\sin\left(2\pi y\left(x^{1/n}-1 \right ) \right )}{\pi y} \right )=-\int_{1}^{\infty}M(z)\frac{d}{dz}\frac{d}{dy}\left(\frac{\sin\left(2\pi y\left(x^{1/z}-1 \right ) \right )}{\pi y} \right )dz$$
Where $M(z)$ is the  Mertens Function. The rightmost integral may be written as : 
$$-4\pi \log x \int_{0}^{1} M(u^{-1})x^{u}\left(x^{u}-1 \right )\sin\left(2\pi y\left(x^{u}-1 \right ) \right )du$$
Since $M(u^{-1})$ has the integral representation :
$$M(u^{-1})=\frac{1}{2\pi i }\int_{\sigma-i\infty}^{\sigma+i\infty}\frac{u^{-r}}{r\zeta(r)}dr\;\;\;\;\;(\sigma >1\;\; and\;0<u<1)$$
$\zeta(z)$ being the Riemann zeta function, the hope is to express the original summation in terms of a contour integral, and eliminate the need for summing over $\mu(n)$.
 A: Be $\,a:=2\pi y\,$ and $\,z:=-r\,$ with $\,\Re(z)>-1\,$.
$\displaystyle \int\limits_0^1 t^z x^t (x^t  - 1) \sin(a(x^t  - 1)) dt = \sum\limits_{n=1}^2 \int\limits_0^1 t^z (x^t  - 1)^n \sin(a(x^t  - 1)) dt$ 
$\displaystyle \int\limits_0^1 t^z (x^t  - 1)^n \sin(a(x^t  - 1)) dt = \sum\limits_{k=0}^\infty (-1)^k \frac{a^{2k+1}}{(2k+1)!}\int\limits_0^1 t^z (x^t - 1)^{n+2k+1} dt$
$\displaystyle \int\limits_0^1 t^z (x^t  - 1)^m dt = \sum\limits_{j=0}^m (-1)^{m-j} {\binom m j} \int\limits_0^1 t^z x^{tj} dt\enspace$ with $\enspace m:=n+2k+1$
$\displaystyle \int\limits_0^1 t^z x^{tj} dt = \sum\limits_{v=0}^\infty \frac{(j \ln x)^v}{v! (z+1+v)} $
Putting the lines together gives a series for $I(x,y,r)$ with $\,\Re(r) <1\,$ . 
Indeed it’s possible to choose $\,\Re(r)<3\,$ but for $\,1\leq \Re(r)<3\,$ the series solution above is not valid. 
$I(x,y,r)\,$ is divergent for $\,\Re(r)\geq 3\,$ .
A: After a lot of work, it turned out that the integral can be written as :
$$\Gamma(1-r)\sum_{j=0}^{\infty}\sin\left(\frac{\pi}{2}\left[j-4y\right ] \right )\frac{(2\pi y)^{j}}{j!}\left[\gamma^{*}\left(1-r,-(2+j)\log x \right )-\gamma^{*}\left(1-r, -\left(1+j \right )\log x\right ) \right ] $$
where $\gamma^{*}(\cdot)$ is the regularized lower gamma function. 
