Evaluate $\int_{1}^{\infty}e^{-x}\ln^{2}\left(x\right)dx$ Evaluate :$$\int_{1}^{\infty}e^{-x}\ln^{2}\left(x\right)\mathrm{d}x$$
I've tried to solve this with some elegant substitutions like $t=e^x$ or $t=\ln\left(x\right)$ . I've also tried to integrate by parts without any success. any help would be good.

Let's define (for $\Re(a)>-1$) the function $$ f(a):=\int_{1}^{\infty}x^a\,e^{-x}\left(x\right)\,\mathrm{d}x$$ then this is the incomplete gamma function $\,f(a)=\Gamma(a+1,1)$.
But since $\,\displaystyle x^a=e^{a\ln(x)}\,$ we have :
$$f''(a)=\int_{1}^{\infty}x^a\;\ln^{2}(x)\;e^{-x}\,\mathrm{d}x$$
So that your function is the second derivative of $\,\Gamma(a+1,1)\,$ at $\,a=0$ (so that $x^a$ disappears) or the second derivative relatively to $a$ of $\,\Gamma(a,1)\,$ at $\,a=1$ :  $$f''(0)=\lim_{a\to 1}\Gamma(a,1)''$$
(this answer was also given by Mr.G in the comments)
Let's add that the derivatives are not much simpler so that it is probably better to keep it this way...
Of course as noted by Eric Naslund (+1) the answer is simpler when the lower bound is $0$ because in this case we get simply $\;\lim_{a\to 0}\Gamma(a+1,0)''=\Gamma''(1)$.
