I am not sure if you can obtain the other three from the first integral through some substitution.
The $(3)$ one can be obtained from the definition of constant..
Notice that
$$\sum_{i=1}^N\frac{1}{i}=\int_0^1 \frac{1-t^N}{1-t}\,dt$$
and
$$\ln N=\int_0^1 \frac{t^{N-1}-1}{\ln t}\,dt$$
The above can be proved using Frullani's integral.
Therefore
$$\begin{aligned}
\lim_{N\rightarrow \infty} \left(\sum_{i=1}^N\frac{1}{i}-\ln N\right) &=\lim_{N\rightarrow \infty}\int_0^1\left(\frac{1-t^N}{1-t}-\frac{t^{N-1}-1}{\ln t}\,dt\right)\,dt \\
&=\int_0^1 \left(\frac{1}{1-t}+\frac{1}{\ln t}\right)\,dt
\end{aligned}$$
Make the substitution $\ln x=-t$ to obtain the $(2)$.
I will suggest some (maybe) interesting constants for which irrationality or transcendence can be studied. First let me recall some well known constants (this list could also be useful to formulate some other conjectures here):
$$G:=\frac{\Gamma^2 (1/4)}{2 \sqrt{2 \pi^3} }$$
known as Gauss's constant, which is transcendent. Then, we can define the following two lemniscate constants, which are transcendental:
$$ L_1:=\pi G $$
$$L_2 := \frac{1}{2 G}$$
(they both arise when looking for a formula for the arc lenght of a lemniscate).
Another interesting number, which arises in a similar way to the golden ratio, is the plastic constant $\rho$, which is the unique real solution of the equation $x^3=x+1$. This constant is irrational.
A well known number, which has been proven to be irrational by Apéry, is $\zeta(3)$, which is also called Apéry's constant.
Now we can study the irrationality/transcendence of the following numbers:
$$ \Omega + \pi, \, \Omega + G, \, \Omega^{\rho}, \, \Omega^{\zeta(3)}, \, \Omega e^{L_1}, \Omega^{\ln \pi}, \sqrt{2 \Omega}, ... $$
This list can become really long: you just have to combine in many other ways all the above constants with $\Omega$, $e$, $\pi$, $i$, ...
Best Answer
$x^{1/x}$ has a maximum at $x=e$ with e as euler‘s number. (can be shown by taking the derivative of the function)
Because the function is increasing before and decreasing after $x=e$, if x and y are two real numbers, which are bigger than 0 and $x<y$, than if x and y are smaller than e:
$$x^{1/x}<y^{1/y}$$
Raise it to the power of $xy$.
$$x^y<y^x$$
This holds true because x and y are positive numbers.
In your case the two numbers are smaller than e, so this proof applies.