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

Solution 1:

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.

Solution 2:

A related problem.

(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 $s=1$.