How do we prove this conjecture $\int_{0}^{\infty}e^{-\sqrt{x}}\ln{\left(1+{1\over \sqrt{x}}\right)}=2\gamma ?$ I was observing this question by @Brightsun and conjecture $(1)$

$$\int_{0}^{\infty}e^{-\sqrt{x}}\ln{\left(1+{1\over \sqrt{x}}\right)}=2\gamma \tag1$$

An attempt
$x=u^2$ then $(1)$ becomes 
$$2\int_{0}^{\infty}ue^{-u}\ln{\left(1+{1\over u}\right)}\tag2$$
$$2\sum_{n=0}^{\infty}{(-1)^n\over n!}\int_{0}^{\infty}u^{n+1}\ln{\left(1+{1\over u}\right)}\mathrm du\tag3$$
Changing $(2)$ by applying $e^x$, $(3)$ diverges.
How would one prove $(1)?$
 A: By the change of variable 
$$
u=\sqrt{x},\quad x=u^2, \quad dx=2udu,
$$ one has
$$
\begin{align}
\int_{0}^{\infty}e^{-\sqrt{x}}\ln{\left(1+{1\over \sqrt{x}}\right)}dx&=2\int_{0}^{\infty}u\:e^{-u}\ln{\left(1+{1\over u}\right)}du
\\\\&=2\int_{0}^{\infty}u\:e^{-u}\ln{\left(1+u\right)}\:du-2\int_{0}^{\infty}u\:e^{-u}\ln{u}\:du. \tag1
\end{align}
$$ By integrating by parts, the first integral on the right hand side of $(1)$ gives
$$
\begin{align}
&\int_{0}^{\infty}u\:e^{-u}\ln{\left(1+u\right)}\:du
\\\\&=\left[\frac{e^{-u}}{-1}\cdot u\:\ln{\left(1+u\right)}\right]_{0}^{\infty}+\int_{0}^{\infty}e^{-u}\left(\ln{\left(1+u\right)}+\frac{u}{1+u}\right)du
\\\\&=\int_{0}^{\infty}e^{-u}\left(\ln{\left(1+u\right)}+\frac{u+1-1}{1+u}\right)du
\\\\&=\int_{0}^{\infty}e^{-u}\left(\color{red}{\ln{\left(1+u\right)}-\frac{1}{1+u}}\right)du+\int_{0}^{\infty}e^{-u}du
\\\\&=\color{red}{0}+1 \quad (\text{integration by parts})
\end{align}
$$
By integrating by parts, the second integral on the right hand side of $(1)$ gives
$$
\begin{align}
&\int_{0}^{\infty}u\:e^{-u}\ln u\:du
\\\\&=\left[\frac{e^{-u}}{-1}\cdot u\:\ln u\right]_{0}^{\infty}+\int_{0}^{\infty}e^{-u}\left(\ln u+1\right)du
\\\\&=\int_{0}^{\infty}e^{-u}\ln u\:du+1
\end{align}
$$ yielding
$$
\int_{0}^{\infty}e^{-\sqrt{x}}\ln{\left(1+{1\over \sqrt{x}}\right)}dx=-2\int_{0}^{\infty}e^{-u}\ln u\:du,
$$ which is the announced result.
A: Note that
$$\begin{eqnarray*}
  \frac{d}{du}\left[e^{-u}\left(u\ln u-(1+u)\ln(1+u)\right)\right]\hspace{-40mm}&&\\
    &=&e^{-u}\left(\ln u-\ln(1+u)-u\ln u+(1+u)\ln(1+u)\right)\\
    &=&e^{-u}\left(\ln u+u\ln(1+1/u)\right).
\end{eqnarray*}$$
Hence
$$\begin{eqnarray*}
  2\int_0^\infty ue^{-u}\ln(1+1/u)\;du
    &=&2\left[e^{-u}\left(u\ln u-(1+u)\ln(1+u)\right)\right]_0^\infty\\
    &&{}-2\int_0^\infty e^{-u}\ln u\;du\\
    &=&2\gamma
\end{eqnarray*}$$
by the first formula here.
A: Starting from
$$2\int_0^\infty u e^{-u}\ln\left(1+\frac{1}{u}\right)du
=2\int_0^\infty u e^{-u}\left(\ln(1+u)-\ln(u)\right)du$$
Let's find $\int_0^\infty u e^{-u}\ln(u)du$ first. We can introduce the variable $a$ and examine the integral:
$$\int_0^\infty e^{-u} u^adu=\Gamma(a+1)$$
Now we will take the derivative by $a$ to get back to our original integral:
$$\frac{\partial}{\partial a}\int_0^\infty e^{-u} u^adu
=\int_0^\infty e^{-u}u^a\ln(u)du=\Gamma'(a+1)$$
Let $a=1$. We then have 
$$\int_0^\infty e^{-u}u\ln(u)du=\Gamma'(2)=1-\gamma$$
Now
$$\int_0^\infty ue^{-u}\ln(1+u)du=\left[-e^{-u}u\ln(1+u)\right]_0^\infty+
\int_0^\infty e^{-u}\left(\ln(1+u)+\frac{u}{1+u}\right)du$$
by integration by parts. The edge terms are zero, so we are left with (integrating the first integrand by parts again)
$$\int_0^\infty e^{-u}\ln(1+u)du+\int_0^\infty e^{-u}\frac{u}{1+u}du$$
$$=[-e^{-u}\ln(1+u)]_0^\infty+\int_0^\infty \frac{e^{-u}}{1+u}du+\int_0^\infty e^{-u}\frac{u}{1+u}du$$
$$=0+\int_0^\infty e^{-u}du = 1$$
Thus, combining this with our previous result in the first integral:
$$2\int_0^\infty u e^{-u}\ln\left(1+\frac{1}{u}\right)du=2(1-(1-\gamma))=2\gamma$$
A: Let $ I$ your integral (2) (without the factor $2$). As $1-(x+1)\exp(-x)$ is a primitive of $x\exp(-x)$, an integration by parts gives
$$I=\int_0^{+\infty} (\frac {1}{1+x}-\exp(-x))\frac{dx}{x}$$
Then use http://mathworld.wolfram.com/Euler-MascheroniConstant.html 
