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? 

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.
