= Logarithmic sine integral
{title2=$\int_0^1\sin(a\log x)/\log x\,dx=\arctan a$}
For every real $a$, define $I(a)=\int_0^1\sin(a\log x)/\log x\,dx$. The apparent endpoint singularity at $x=1$ is removable, with limit $a$, and the bound $|\sin(a\log x)/\log x|\leq|a|$ ensures integrability. <Differentiation under the integral sign> is justified by the integrable bound $|\cos(a\log x)|\leq1$ on the parameter derivative. Thus
$$
I'(a)=\int_0^1\cos(a\log x)\,dx
=\int_0^\infty e^{-t}\cos(at)\,dt=\frac1{1+a^2},
$$
where $x=e^{-t}$ and the last integral is a <Laplace transform>. Since $I(0)=0$, integrating with respect to $a$ gives $I(a)=\arctan a$. This obtains an integral with a logarithmic denominator by differentiating away that denominator, then restoring the parameter dependence with one elementary antiderivative.
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