Proving $\int_{0}^{\infty}\sin\left({1\over 4x^2}\right)\ln x\cdot{\mathrm dx\over x^2}=-\sqrt{\pi\over 2}\cdot\left({\pi-2\gamma \over 4}\right)$ 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?
 A: Recall
$$ \int_0^\infty u^te^{-su}du=\frac{\Gamma(t+1)}{s^{t+1}}. $$
So
\begin{eqnarray}
&&\int_{0}^{\infty}u^{t}\sin u\mathrm du\\
&=&\lim_{\epsilon\to0^+}\Im \int_{\epsilon}^{\infty}u^{t} e^{iu}\mathrm du=\Im \frac{\Gamma(t+1)}{(-i)^{t+1}}\\
&=&\Im \Gamma(t+1)(i)^{t+1}=\Gamma(t+1)\sin(\frac{(t+1)\pi}{2}).
\end{eqnarray}
Let
$$ I(a)=\int_0^{\infty}x^{-a}\sin(\frac1{4x^2})dx $$
and then
$$I(a)=-2^{-2+a}\int_0^{\infty}u^{\frac{-3+a}2}\sin udu=-2^{a-2}\sin(\frac{(a-1)\pi}{4})\Gamma(\frac{a-1}{2}). $$
So
\begin{eqnarray}
I'(2)&=&-2^{a-2}\ln2\sin(\frac{(a-1)\pi}{4})\Gamma(\frac{a-1}{2})-2^{a-4}\pi\cos(\frac{(a-1)\pi}{4})\Gamma(\frac{a-1}{2})\\
&&-2^{a-3}\sin(\frac{(a-1)\pi}{4})\Gamma'(\frac{a-1}{2})\bigg|_{a=2}\\
&=&-\ln2\cdot\frac1{\sqrt2}\sqrt{\pi}-\frac\pi 4\frac{1}{\sqrt2}\sqrt\pi-\frac{1}{2\sqrt2}\Gamma'(\frac12)
\end{eqnarray}
Noting
$$ \frac{\Gamma'(z+1)}{\Gamma(z+1)}=-\gamma+\sum_{n=1}^\infty\left(\frac{1}{n}-\frac{1}{z+n}\right)$$
one has
$$ \Gamma'(\frac12)=\Gamma(\frac12)\left[-\gamma+\sum_{n=1}^\infty\left(\frac{1}{n}-\frac{1}{n-\frac12}\right)\right]=\sqrt{\pi}(-\gamma-2\ln2).$$
Thus
$$ I'(2)=-\sqrt{\pi\over 2}\left({\pi-2\gamma \over 4}\right). $$
You can treat $I''(2)$ in the same way and the calculation will be longer.
