The Euler constant arises in the integrals
$$ \int_{0}^{\infty} \ln x ~e^{-x} dx=-\gamma $$
$$ \int_{0}^{\infty} (\ln x)^{2} e^{-x} dx=\gamma^2+\frac{\pi^2}{6}$$
$$ \int_{0}^{\infty} (\ln x)^{3} e^{-x} dx=-\gamma^3-\frac{\gamma \pi^2}{2}-2\zeta(3)$$
$$ \int_{0}^{\infty} (\ln x)^{4} e^{-x} dx=\gamma^4+\gamma^2 \pi^2+\frac{3 \pi^4}{2}+8 \gamma\zeta(3)$$
where $\zeta(\cdot)$ is Riemann zeta function.
Is there general form for the integral$$ \int_{0}^{\infty} (\ln x)^i e^{-x} dx; i=0,1,2,\dots.?$$
Since $\int_0^\infty x^{s-1}e^{-x}dx=\Gamma(s)$, $\int_0^\infty x^{s-1}e^{-x}\ln^i xdx=\Gamma^{(i)}(s)$, so you're evaluating $\Gamma^{(i)}(1)$. Theorem 2.1 here gives the recursion $$\Gamma^{(i)}(1)=-\gamma\Gamma^{(i-1)}(1)+(i-1)!\sum_{k=0}^{i-2}\frac{(-1)^{i-k}\zeta(i-k)\Gamma^{(k)}(1)}{k!}.$$