Consider these two similar integrals
$$\int_{0}^{\infty}\sin\left({1\over 4x^2}\right)\ln x\cdot{\mathrm dx\over x^2}=-\sqrt{\pi\over 2}\cdot\left({\pi-2\gamma \over 4}\right)\tag1$$
$$\int_{0}^{\infty}\sin\left({1\over 4x^2}\right)\ln^2x\cdot{\mathrm dx\over x^2}=\sqrt{\pi\over 2}\cdot\left({\pi-2\gamma \over 4}\right)^2\tag2$$
How does one prove $(1)$ and $(2)$?
An attempt:
Using the $\sin x$ series, and let $u=\ln x$, then $(1)$ becomes
$$\sum_{n=0}^{\infty}{(-1)^n\over 2^{2n+1}(2n+1)!}\int_{-\infty}^{\infty}ue^{-2(n+1)u}\mathrm du\tag3$$
Integral $(3)$ diverges.
Another attempt:
$u={1\over 4x^2}$ then $(1)$ becomes
$$-{1\over 8}\int_{0}^{\infty}\sin u\ln{(4u)}\cdot{\mathrm du\over u^{3/2}}\tag4$$
Using $\ln u$ series, then $(4)$ becomes
$$-{1\over 4}\sum_{n=0}^{\infty}{1\over 2n+1}\int_{0}^{\infty}\left({4u-1\over 4u+1}\right)^{2n+1}\cdot{\sin u\over u^{3/2}}\mathrm du\tag5$$
Using $\coth^{-1} x={1\over 2}\ln{x-1\over x+1}$
$$-{1\over 4}\sum_{n=0}^{\infty}{1\over 2n+1}\int_{0}^{\infty}e^{2(2n+1)\coth^{-1}4x}\cdot{\sin u\over u^{3/2}}\mathrm du\tag6$$
How else can we proceed?