TITLE: Elementary derivation of certian identites related to the Riemannian Zeta function and the Euler-Mascheroni Constant
QUESTION [6 upvotes]: Is the proof of these identities possible, only using elementary differential and integral calculus? If it is, can anyone direct me to the proofs? ( or give a hint for the solution )
1)$$\int_0^\infty { e^{-x^2} \ln x }\,dx = -\tfrac14(\gamma+2 \ln 2) \sqrt{\pi} $$
2)$$\int_0^\infty { e^{-x} \ln^2 x }\,dx = \gamma^2 + \frac{\pi^2}{6} $$
3) $$\gamma = \int_0^1 \frac{1}{1+x} \sum_{n=1}^\infty x^{2^n-1} \, dx$$
and lastly,
4) $$\zeta(s) = \frac{e^{(\log(2\pi)-1-\gamma/2)s}}{2(s-1)\Gamma(1+s/2)} \prod_\rho \left(1 - \frac{s}{\rho} \right) e^{s/\rho}\!$$
I personally think the last is obtained from a simple use of the Weierstrass factorization theorem. I'm unsure as to what substitution is used.
$\gamma$ is the Euler-Mascheroni constant and $\zeta(s)$ is the Riemannian Zeta function.
Thanks in advance.

REPLY [6 votes]: The general recipe for integrals containing logarithms is to replace $\ln x$ by $\displaystyle\left[\frac{\partial}{\partial s}x^{s}\right]_{s=0}$, then to exchange the order of integration and differentiation, and evaluate the integral in terms of gamma functions. Then the derivatives with respect to $s$ will produce digamma functions and their derivatives (often related to particular values of $\zeta(x)$).
For example, the second integral can be written as
\begin{align}
\int_0^{\infty}e^{-x}\ln^2x\,dx&=\left[\frac{\partial^2}{\partial s^2}\int_0^{\infty}e^{-x}x^{s}dx\right]_{s=0}=\\
&=\left[\frac{\partial^2}{\partial s^2}\Gamma(s+1)\right]_{s=0}=\\&=\gamma^2+\frac{\pi^2}{6}.
\end{align}
A bit more sophisticated example of this technique can be found here.