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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.

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  • $\begingroup$ For the first two identities you can use the Mellin transform and some of its properties. $\endgroup$ – Mhenni Benghorbal Aug 29 '12 at 15:36
  • $\begingroup$ What is your definition of $\gamma$? $\endgroup$ – Sasha Aug 29 '12 at 15:40
  • $\begingroup$ @Sasha, as the "limiting difference between the harmonic series and natural logarithm" $\endgroup$ – Ishihara Sep 4 '12 at 11:11
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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 $s=1$.

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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.

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