= Solution
The parameter <derivative> of the integrand is $\cos(a\log x)$, whose absolute value is bounded by one on $0<x<1$. This integrable bound justifies <differentiation under the integral sign>, giving
$$
I'(a)=\int_0^1\cos(a\log x)\,dx.
$$
The original <integral> also exists: $|\sin(a\log x)/\log x|\leq|a|$, and its apparent singularity at $x=1$ is removable. Set $x=e^{-t}$ to evaluate the <derivative>:
$$
I'(a)=\int_0^\infty e^{-t}\cos(at)\,dt
=\operatorname{Re}\int_0^\infty e^{-(1-ia)t}\,dt
=\operatorname{Re}\frac1{1-ia}.
$$
Consequently
$$
\boxed{I'(a)=\frac1{1+a^2}}.
$$
Since $I(0)=0$, integration in the parameter fixes the constant and gives the <logarithmic sine integral> identity
$$
\boxed{I(a)=\arctan a,\qquad I(1)=\frac\pi4}.
$$
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