Show us how to prove that $\int_{0}^{\infty}{\sin(e^{-\gamma}x)}\cdot{\ln{x}\over x}\mathrm dx=0$ I saw these two intgerals and would like to know if they are correct.

$$\int_{0}^{\infty}{\sin(e^{-\gamma}x)}\cdot{\ln{x}\over x}\mathrm dx=0\tag1$$
$$\int_{0}^{\infty}{\sin\left(\sqrt{x}^{\sqrt{2}}\right)}\cdot{\ln{x}\over x}\mathrm dx=-\pi\gamma\tag2$$

This is the well know $\gamma$ Euler-Mascheroni's constant
Here I ignored the limits
I apply sub: to $(1)$
$u=\ln{x}$
$xdu=dx$
$$I = \int{ue^{-u}\sin(e^{u-\gamma})}du\tag3$$
Apply integration by parts to $(3)$
$$\int{ue^{-u}}du=-e^{-u}(1+u)$$
$$I={-e^{-u}(1+u)\sin(e^{u-\gamma})}+\int{e^{-u}(1+u)\cos(e^{u-\gamma})}du$$
Encounter more harder than before.
Please show us how to prove $(1)$ and $(2)$
 A: First note that 

$$\int_{0}^{\infty}{\sin(x)}\cdot{\ln{x}\over x}\mathrm
 dx=-\frac{\pi \gamma}{2}$$

Note the Mellin transform  of the sine function 
$$\int^\infty_0 x^{s-1}\sin(x)\,dx =  \Gamma (s) \sin\left( \frac{\pi s}{2} \right)$$
Which can be written as 
$$\int^\infty_0 x^{s-1}\sin(x)\,dx = \frac{\pi \Gamma(s+1)}{2\Gamma(s/2+1)\Gamma(1-s/2)}$$
Now differentiate with respect to $s$ and let $s \to 0$ to obtain the result.
First integral
\begin{align}
\int_{0}^{\infty}{\sin(ax)}\cdot{\ln{x}\over x}\mathrm dx&=\int_{0}^{\infty}{\sin(t)}\cdot{\ln{t}-\ln a\over t}\mathrm dt\\&= \int_{0}^{\infty}{\sin(t)}\cdot{\ln{t}\over t}\mathrm dt-(\ln a) \frac{\pi}{2} \\
&=\frac{-\pi \gamma}{2}-\ln(a) \frac{\pi}{2}
\end{align}
Now letting $a =e^{-\gamma}$ the result follows.
Second Integral
$$\int_{0}^{\infty}{\sin\left(x^{a}\right)}\cdot{\ln{x}\over x}\mathrm dx =\frac{1}{a^2}\int_{0}^{\infty}{\sin\left(t\right)}\cdot{\ln{t}\over t}\mathrm dt  = -\frac{\gamma \pi}{2a^2}$$
For $a = \sqrt{2}/2$ the result follows.
