How to prove that $\int_{0}^{\infty}{\sin^4(x)\ln(x)}\cdot{\mathrm dx\over x^2}={\pi\over 4}\cdot(1-\gamma)?$ 
How to prove that
  $$\int_{0}^{\infty}{\sin^4(x)\ln(x)}\cdot{\mathrm dx\over x^2}={\pi\over 4}\cdot(1-\gamma).\tag1$$

Here is my attempt:
$$I(a)=\int_{0}^{\infty}{\ln(x)\sin^4(x)\over x^a}\,\mathrm dx\tag2$$
$$I'(a)=\int_{0}^{\infty}{\sin^4(x)\over x^a}\,\mathrm dx\tag3$$
$$I'(2)=\int_{0}^{\infty}{\sin^4(x)\over x^2}\,\mathrm dx\tag4$$
$$I'(2)=\int_{0}^{\infty}{\sin^2(x)\over x^2}\,\mathrm dx-{1\over 4}\int_{0}^{\infty}{\sin^2(2x)\over x^2}\,\mathrm dx=-{\pi\over 2}\tag5$$
Why is this way wrong?
How to prove (1)?

I will outline a self-contained approach, too. By differentiating the integral definition of the $\Gamma$ function, we get the following Lemma:
$$ \mathcal{L}(\log x) = -\frac{\gamma+\log(s)}{s}\tag{1} $$
and it is not difficult to compute from $(1)$ the Laplace transform of $\sin^4(x)\log(x)$.
By Euler/De Moivre's formula we have
$$ \sin^4(x) = \frac{1}{16}\left( e^{4ix}+4 e^{2ix}+6+4 e^{-2ix}+e^{-4ix}\right)\tag{2}$$
and by the shift properties of the Laplace transform and $(1)$ we get
$$ \forall k\in\mathbb{Z},\qquad \mathcal{L}\left(e^{-kix}\log x\right) = -\frac{\gamma+\log(ki+s)}{ki+s}\tag{3} $$
so by $(1),(2),(3)$ and simple algebraic manipulations we get:
$$ \mathcal{L}\left(\sin^4(x)\log x\right) = -\frac{24\gamma}{64s+20s^3+s^5}+\text{LogTerm}$$
$$ {\scriptsize\text{LogTerm} = \frac{1}{16} \left(-\frac{6 \log(s)}{s}+\frac{4 \left(4 \arctan\left(\frac{2}{s}\right)+s \log\left(4+s^2\right)\right)}{4+s^2}-\frac{8\arctan\left(\frac{4}{s}\right)+s \log\left(16+s^2\right)}{16+s^2}\right)}\tag{4} $$
and since $\mathcal{L}^{-1}\left(\frac{1}{x^2}\right)=s$, the computation of the original integral boils down to the computation of elementary integrals by integration by parts. For instance, the term $-\frac{\pi\gamma}{4}$ comes from
$$ \int_0^{+\infty } \frac{24}{(4+s^2)(16+s^2)} \, ds=\frac{\pi}{4}.\tag{5}$$
