Elementary derivation of certian identites related to the Riemannian Zeta function and the Euler-Mascheroni Constant

Question

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.

Answer 1

(1) $$ F(s) = \int_{0}^{\infty} x^{s-1} {\rm e}^{-x^2}\,dx \Rightarrow F'(s) = \int_{0}^{\infty }x^{s-1} \ln(x) {\rm e}^{-x^2}\, dx \,, $$

where $F(s)$ is the Mellin transform of $ {\rm e}^{-x^2} $ and it is given by

$$ F(s) = \frac{1}{2} \Gamma\left(\frac{s}{2}\right) \, $$

which implies that

$$ F'(s) = \frac{1}{4}\,\psi \left( \frac{s}{2} \right) \Gamma \left( \frac{s}{2} \right) \,. $$

Substituting $ s=1 $ yields the desired result, $$ -\frac{1}{4}\, \left( \gamma+2\,\ln \left( 2 \right) \right) \sqrt {\pi } $$

(2) You can do the same by taking the Mellin transform of ${\rm e}^{-x}$ and differentiate it twice with respect to $s$ and then substitute $s=1$.