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)?
 A: 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
$$ \int_0^{+\infty } \frac{24}{(4+s^2)(16+s^2)} \, ds=\frac{\pi}{4}.\tag{5}$$
A: This is solved essentially in the same way explained in answers to your previous question. As a convenient starting point, I will refer to @Jack D'Aurizio's answer:
$$ \int_{0}^{\infty}\frac{1-\cos(kx)}{x^2}\log(x)\,dx = \frac{k\pi}{2}\left(1-\gamma-\log k\right). \tag{1} $$
Now all you have to do is to write
$$ \sin^4 x = \frac{1}{2}(1 - \cos(2x)) + \frac{1}{8}(1 - \cos(4x)). \tag{2} $$
I hope that the remaining computation is clear to you.

For your attempt, a correct computation would begin with
$$ \frac{d}{da} \int_{0}^{\infty} \frac{\sin^4 x}{x^a} \, dx = - \int_{0}^{\infty} \frac{\sin^4 x}{x^a} \log x \, dx. $$
Notice that you misidentified the derivative of your parametrized integral.
