How to evaluate integral: $ \int_{0}^{1} \operatorname{sgn}(\sin(\ln(x))) \ dx $

The next step could be: $$\int_{0}^{1} \operatorname{sgn}(\sin(\ln(x))) \ dx=\int_{-\infty}^{0} e^{y}\operatorname{sgn}(\sin(y)) \ dy=\sum_{k=0}^{\infty}(-1)^{k+1}\int_{-(k+1)\pi}^{-k\pi} e^{y}\,d y$$ Can you take it from here? BTW the final result is not zero!

If you add $1$ to the integrand and subtract it out again, you get $$ 2 \int_0^1 [ \sin(\ln x) > 0 ]\; dx - 1 $$ where $[\cdots]$ is the Iverson bracket, and the integral here is just the sum of the length of the intervals where $\sin(\ln x)$ is positive. These intervals are $$ [e^{-2k\pi}, e^{-(2k-1)\pi}] \quad \text{for }k=1,2,3\ldots $$ so we want $$ 2 \sum_{k=1}^{\infty} \left( e^{-(2k-1)\pi} - e^{-2k\pi} \right) -1 = 2 \sum_{k=1}^{\infty} (e^{\pi}-1)(e^{-2\pi})^k -1 $$ The sum here is a geometric series.