I am trying to prove that
$$\int_{\mathbb{R}} x e^{-x} \cdot e^{-e^{-x}-x} = 1 - \gamma$$ and
$$\int_{\mathbb{R}} x^2 e^{-x} \cdot e^{-e^{-x}-x} = \pi^2/6 - 2\gamma + \gamma^2$$
I have tried to attack the integral using all possible calculus laws, but it brings me nowhere. We have derived that $\int_{\mathbb{R}} x \cdot e^{-e^{-x}-x} = \gamma$
On
What I have done so far is:
First integral Denote $u(x) = xe^{-x}$ and $f(x) := e^{-e^{-x}-x}$, let $F(x)$ be the antiderivative of $f(x)$. Then \begin{align*} \int_{\mathbb{R}} x e^{-x} \cdot e^{-e^{-x}-x} &= \int_{\mathbb{R}} u(x)f(x) \end{align*} Intergration by parts yields \begin{align*} \hspace{1cm} &= - \int_{\mathbb{R}} u'(x)F(x) \end{align*} We have $u'(x) = e^{-x} - u(x)$; thus, \begin{align*} \hspace{3.5cm} &= - \int_{\mathbb{R}} [e^{-x} - u(x)]F(x)\\ &= - \int_{\mathbb{R}} e^{-x} F(x) + \int_{\mathbb{R}} u(x)F(x) \end{align*} At last note that $f(x) = e^{-x}F(x)$; hence \begin{align*} \hspace{2cm} &= - \int_{\mathbb{R}} f(x) + \int_{\mathbb{R}} xf(x)\\ &= - 1 + \gamma \end{align*}
$$I(n)=\int_{-\infty}^\infty x^{n} e^{-x} e^{-e^{-x}-x}dx$$ Let $u=e^{-x}$, $x=-\ln(u)$, $du=-e^{-x}dx$ $$I(n)=-\int_\infty^0 (-\ln(u))^{n} ue^{-u}du=(-1)^n \int_0^\infty \ln^n(u)u e^{-u}du$$
Now the Gamma function is defined as $$\Gamma(s)=\int_0^\infty x^{s-1}e^{-x}dx$$ So $$\left(\frac{d}{ds}\right)^n\Gamma(s)=\int_0^\infty \ln^n(x) x^{s-1} e^{-x}dx$$ Thus $$I(n)=(-1)^n \left(\frac{d}{ds}\right)^n\Gamma(s)|_{s=2}$$ Now let $n=1,2$ to get the values of the two original integrals. It could help to know that $$\Gamma'(2)=-\gamma+H_1=1-\gamma$$ and that $$\psi_1(2)=\frac{\pi^2}{6}-1$$ where $\psi_1(s)$ is the Trigamma Function.
EDIT: $$\psi_1(s)=\left(\frac{d}{ds}\right)^2 \ln(\Gamma(s))=\frac{\Gamma(s)\Gamma''(s)-\left(\Gamma'(s)\right)^2}{\Gamma^2(s)}$$ Evaluating at $s=2$ we get $$\Gamma''(2)-\left(\Gamma'(2)\right)^2=\frac{\pi^2}{6}-1$$ So $$\Gamma''(2)=\frac{\pi^2}{6}-1 +\left(\Gamma'(2)\right)^2=\frac{\pi^2}{6}-1 +(1-\gamma)^2=\frac{\pi^2}{6}-2\gamma+\gamma^2$$