How to evaluate $\int^\infty_0 \log^2(x)\exp(-x)\,dx$

$$\Gamma(t+1)=\int_0^\infty x^te^{-x}\,dx.$$ Differentiating twice under the integral sign gives $$\Gamma''(t+1)=\int_0^\infty (\log x)^2x^te^{-x}\,dx.$$ Take $t=0$.

With Gamma function $$\Gamma(n)=\int_0^\infty e^{-x}x^{n-1}dx$$ then $$\int_0^\infty e^{-x}\log^2xdx=\Gamma''(1)$$