How to show $\int_{0}^{\infty}e^{-x}\ln^{2}x\:\mathrm{d}x=\gamma ^{2}+\frac{\pi ^{2}}{6}$? How to show this equation below is true?
$$\int_{0}^{\infty}e^{-x}\ln^{2}x\:\mathrm{d}x=\gamma ^{2}+\frac{\pi ^{2}}{6}$$
Where $\gamma$ is the Euler-Mascheroni constant....

As already mentioned, the integral evaluates to $\Gamma''(1)$.
Recall the digamma and trigamma functions:
$$\psi_{0} (x) = \frac{d}{dx} \ln \Gamma(x) = \frac{\Gamma'(x)}{\Gamma(x)}$$
$$\psi_{1}(x) = \frac{d}{dx} \psi_{0}(x) = \frac{\Gamma''(x)\Gamma(x)- (\Gamma'(x))^{2}}{\Gamma^{2}(x)}$$
Then
$$\psi_{1}(1) = \frac{\Gamma''(1)\Gamma(1) + (\Gamma'(1))^{2}}{\Gamma^{2}(1)} = \Gamma''(1) - (\Gamma'(1))^{2} $$
The digamma and trigamma functions also have series definitions (which can be derived from the infinite product representation of the gamma function):
$$ \psi_{0}(x) = - \gamma + \sum_{n=0}^{\infty} \Big( \frac{1}{n+1} - \frac{1}{n+x}\Big)$$
$$\psi_{1}(x) = \sum_{n=0}^{\infty} \frac{1}{(n+x)^{2}}$$
So
$$ \psi_{1}(1) = \sum_{k=0}^{\infty} \frac{1}{(k+1)^2} = \zeta(2)  = \frac{\pi^{2}}{6}$$
And 
$$\Gamma'(1) = \psi_{0}(1) \Gamma(1) = \psi_{0}(1) = - \gamma +  \sum_{n=0}^{\infty} \left(\frac{1}{n+1} - \frac{1}{n+1} \right) = - \gamma$$
Therefore,
$$\Gamma''(1) = \psi_{1}(1) + (\Gamma'(1))^{2} = \frac{\pi^{2}}{6} + \gamma^{2}$$
