How to evaluate $\int^\infty_0 \log^2(x)\exp(-x)\,dx$ I attempted to solve this equation using integration by part, but it leads me no where and it gets too complicated to solve. I hope some can give me a hint how to approach this integral
$$\int^\infty_0 \log^2(x)\exp(-x)\,dx$$
Thanks,
 A: $$\Gamma(t+1)=\int_0^\infty x^te^{-x}\,dx.$$
Differentiating twice under the integral sign gives
$$\Gamma''(t+1)=\int_0^\infty (\log x)^2x^te^{-x}\,dx.$$
Take $t=0$.
A: With Gamma function
$$\Gamma(n)=\int_0^\infty e^{-x}x^{n-1}dx$$
then
$$\int_0^\infty e^{-x}\log^2xdx=\Gamma''(1)$$
A: To show the value of $\Gamma''(1)$ does indeed reduce to the result @Robert Israel states, from the definition for the digamma function $\psi(x)$, we have
$$\psi (x) = \frac{\Gamma'(x)}{\Gamma (x)}.$$
Taking the logarithm of both sides followed by differentiating with respect to $x$ gives
$$\frac{\psi^{(1)}(x)}{\psi (x)} = \frac{\Gamma''(x)}{\Gamma' (x)} - \frac{\Gamma' (x)}{\Gamma (x)} = \frac{\Gamma'' (x)}{\Gamma'(x)} - \psi (x). \tag1$$
Here $\psi^{(1)}(x)$ is the polygamma function of order one.
Since $\Gamma'(x) = \psi (x) \Gamma (x)$, after rearranging (1) and solving for $\Gamma''(x)$ we have
$$\Gamma'' (x) = \Gamma (x) \left [\{\psi (x)\}^2 + \psi^{(1)} (x) \right ].$$
Setting $x = 1$ in the above gives
$$\Gamma''(1) = \Gamma (1) \left [\{\psi (1) \}^2 + \psi^{(1)} (1) \right ].$$
Values for each of the functions that appear in the above expression can be found. We have $\Gamma (1) = 1$ which is trivial, $\psi (1) = - \gamma$ where $\gamma$ is the Euler-Mascheroni constant (a proof of this can be found here). Finally, for $\psi^{(1)}(1)$ the polygamma function of order one (or trigamma function) is defined by the series
$$\psi^{(1)}(x) = \sum^\infty_{n = 0} \frac{1}{(x + n)^2}.$$ 
Setting $x = 1$ and shifting the index by $n \mapsto n - 1$ we have
$$\psi^{(1)}(1) = \sum^\infty_{n = 1} \frac{1}{n^2} = \frac{\pi^2}{6},$$
the sum corresponding to the well-known Basel problem.
Thus
$$\Gamma''(1) = \gamma^2 + \frac{\pi^2}{6},$$
as announced. 
