So here is my proof:
Two important things that i wont proof here are
$\lfloor x\rfloor = \frac {\arctan(\cot(x\pi))}\pi - \frac 12 + x$
and
$\sum_{n=1}^x \frac 1n = \frac {\lfloor x\rfloor}x + \int_1^x \frac {\lfloor t\rfloor}{t^2} \,dt$
But those are well known facts.
From this it follows that
$\sum_{n=1}^x \frac 1n = \frac {\arctan(\cot(x\pi))}{\pi x} - \frac 1{2x} + 1 + \int_1^x \frac {\arctan(\cot(t\pi))}{\pi t^2} - \frac 1{2t^2} + \frac 1t\,dt$
(simply substitute the first into the second)
$\sum_{n=1}^x \frac 1n = \frac {\arctan(\cot(x\pi))}{\pi x} - \frac 1{2x} + 1 + \frac 1{2x} - \frac 12 + \ln(x) + \int_1^x \frac {\arctan(\cot(t\pi))}{\pi t^2} \,dt$
$\sum_{n=1}^x \frac 1n - \ln(x) = \frac 12 + \int_1^x \frac {\arctan(\cot(t\pi))}{\pi t^2} \,dt + \frac {\arctan(cot(x\pi))}{\pi x}$
Now take the limit as x goes to $\infty$
$\gamma = \frac 12 + \int_1^\infty \frac {\arctan(\cot(t\pi))}{\pi t^2} \,dt$
substituting in $u = \frac 1{\pi t}$ gives
$\gamma = \frac 12 + \int_0^{\frac 1{\pi}} \arctan(\cot(\frac 1u)) \,du$
Independent of the Gamma Function, what immediately came to my mind was that we can use the fact that $\frac{d}{dx}a^x=a^x\log a$.
Let us consider $$I(a) = \int_0^{\infty}t^ae^{-t}dt$$
Differentiating under the Integral Sign gives us $I'(a) = \int_0^{\infty}t^ae^{-t}\log tdt$ which is the required integral with $a =1$.
Applying Integration by Parts to $I(a)$,
$$I(a)=-e^{-t}t^a|_0^{\infty}+a\int_0^{\infty}t^{a-1}e^{-t}dt=aI(a-1)$$
$$\therefore I(a) =a!$$
Considering the recurrence relation and differentiating both sides with respect to a,
$$I'(a)=I(a-1)+aI'(a-1)$$
Dividing by $I(a)$ or $aI(a-1)$,
$$\frac{I'(a)}{I(a)}=\frac{I'(a-1)}{I(a-1)}+\frac{1}{a}$$ which gives the connection to the Harmonic Numbers and finally to the Euler-Mascheroni Constant.
Hope you can complete the rest.
Best Answer
A relatively elementary way is to start with known $$\gamma=\int_0^1\frac{1-\cos t}{t}\,dt-\int_1^\infty\frac{\cos t}{t}\,dt.$$
Put $t=ax$ for $a>0$ and do some rearrangements, to get $$\int_0^1\frac{1-\cos ax}{x}\,dx-\int_1^\infty\frac{\cos ax}{x}\,dx=\gamma+\log a.$$
Now the integral representation $J_0(y)=\frac2\pi\int_0^{\pi/2}\cos(y\cos x)\,dx$ yields $$\int_0^1\frac{1-J_0(y)}{y}\,dy-\int_1^\infty\frac{J_0(y)}{y}\,dy=\frac2\pi\int_0^{\pi/2}(\gamma+\log\cos x)\,dx$$ after interchanging integrations (which is not hard to justify).
The result now follows from $\int_0^{\pi/2}\log\cos x\,dx\color{gray}{=\int_0^{\pi/2}\log\sin x\,dx}=-(\pi/2)\log2$.