evaluation of parametric integral I do not know how to compute integral $\int_0^1\frac{x^a-x^b}{\ln\;x}\cos(\ln\;x)\;\mbox{d}x$ where $a,b>0$. I only know that I should use $\frac{x^a-x^b}{\ln\;x}=\int_b^a x^y \;\mbox{d}y$ and Fubini's formula.
Thanks for any help.
 A: Note that 
$$\cos(\log(x))+i\sin(\log(x)) =  \,e^{i\log(x)} = x^i$$
Think about the extension such that $a,b \in \mathbb{C}$
$$\int^1_0 \frac{x^a-x^b}{\log(x)}\,dx = \log \left( \frac{b+1}{a+1}\right)$$
Let $a=c+i$ and $b = d+i$
$$\int^1_0 \frac{(x^c-x^d)(\cos(\log(x))+i\sin(\log(x)))}{\log(x)}\,dx = \log \left( \frac{c+i+1}{d+i+1}\right)$$
By computing the real part 
\begin{align}\int^1_0 \frac{(x^c-x^d)\cos(\log(x))}{\log(x)}\,dx &= \log \left|\frac{c+i+1}{d+i+1}\right|\\
&=\frac{1}{2}\log\left[ \frac{(cd+1)^2}{(d^2+1)^2}+\frac{(d-c)^2}{(d^2+1)^2}\right]\end{align}
Note we used that $\Re \log|z| = \log\sqrt{x^2+y^2}$, hence we reach to 

\begin{align}\int^1_0
 \frac{(x^{c-1}-x^{d-1})\cos(\log(x))}{\log(x)}\,dx 
 &=\frac{1}{2}\log\left[\frac{c^2+1}{d^2+1}\right]\end{align}


Another approach
Consider 
$$f(a) = \int^1_0 \frac{x^{c-1}-x^{d-1}}{\log (x)} \cos(a\log(x))\,dx$$
$$f'(a) = -\int^1_0 (x^{c-1}-x^{d-1}) \sin(a\log(x))\,dx$$
Let $t = -\log(x)$
$$f'(a) = \int^\infty_0 (e^{-ct}-e^{-dt}) \sin(at)\,dt = \frac{a}{a^2+c^2}-\frac{a}{d^2+a^2}$$
By integration 
$$f(a)= \frac{1}{2}\log \left(\frac{a^2+c^2}{a^2+d^2} \right)$$

Third approach
$$\int^1_0 \frac{(x^a-x^b)\cos(\log(x)}{\log(x)}dx = \int^1_0\int_b^a x^y \cos(\log(x))\,dy dx = \int^a_b \int^1_0x^y \cos(\log(x))\, dx dy $$
Note that 
$$ \int_b^a\int^1_0x^y \cos(\log(x))\, dx \,dy= \int_b^a\frac{(y+1)}{(y+1)^2+1}\,dy = \frac{1}{2}\log \left[ \frac{(a+1)^2+1}{(b+1)^2+1}\right]$$
