Integral of $\int_0^1 \frac{\sin(a \cdot \ln(x))\sin (b \cdot \ln(x))}{\ln(x)} dx$ I'm trying to calculate the integral $$\int_0^1 \frac{\sin\Big(a \cdot \ln(x)\Big)\cdot \sin \Big(b \cdot \ln(x)\Big)}{\ln(x)} dx, $$
but am stuck. I tried using Simpsons' rules and got here:
$$\int_0^1 \frac{\cos\Big((a+b) \cdot \ln(x)\Big) - \cos \Big((a-b) \cdot \ln(x)\Big)}{2\ln(x)} dx, $$
but alas, that also got me nowhere. Does anyone have any ideas? 
 A: For $c \in \mathbb{R}$ we have
\begin{align}
\int \limits_0^\infty \frac{1 - \cos(c t)}{t} \, \mathrm{e}^{-t} \, \mathrm{d} t  &= \int \limits_0^\infty \int \limits_0^c \sin(u t) \, \mathrm{d} u \, \mathrm{e}^{-t} \, \mathrm{d} t = \int \limits_0^c \int \limits_0^\infty \sin(u t) \mathrm{e}^{-t} \, \mathrm{d} t \, \mathrm{d} u = \int \limits_0^c \frac{u}{1+u^2} \, \mathrm{d} u \\
&= \frac{1}{2} \ln(1 + c^2) \, ,
\end{align}
so
\begin{align}
\int \limits_0^1 \frac{\sin[a \ln(x)] \sin[b \ln(x)]}{\ln(x)} \, \mathrm{d} x &= \int \limits_0^1 \frac{\cos[(a+b)\ln(x)] - \cos[(a-b) \ln(x)]}{- 2 \ln(x)} \, \mathrm{d} x \\
&\!\!\!\stackrel{x = \mathrm{e}^{-t}}{=} \int \limits_0^\infty \frac{\left(1 - \cos[(a - b) t]\right) - \left(1 - \cos[(a+b) t]\right)}{2t} \, \mathrm{e}^{-t} \, \mathrm{d} t \\
&= \frac{1}{4} \left(\ln[1 + (a-b)^2] - \ln[1 + (a+b)^2]\right) \\
&= \frac{1}{4} \ln \left(\frac{1+(a-b)^2}{1+(a+b)^2}\right) \, .
\end{align}
A: I do not think that it would be very pleasant.
After your simplification, you face two integrals looking like
$$I=\int \frac {\cos(k \log(x))} {\log(x)} \,dx$$ First, let $x=e^t$ to make
$$I=\int \frac{e^t \cos (k t)}{t}\,dt$$ and consider that you need the real part of
$$I=\int \frac{e^t\,e^{ikt}}t\,dt=\int \frac{e^{(1+ik)t}}t\,dt$$ Now, let $(1+ik)t=u$ to make
$$I=\int \frac {e^u} u \,du=\text{Ei}(u)$$ where appears the exponential integral function.
