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The integral \begin{align} I_{4} = \int_{0}^{1} \ln(1-x) \ \ln^{2}\left( \ln\left(\frac{1}{x}\right) \right) \ \frac{dx}{x} \end{align} can be expressed as \begin{align} I_{4} = \zeta^{''}(2) - \frac{\gamma^{2} \pi^{2}}{6} - \frac{\gamma \pi^{2}}{3} \ln\left( \frac{2 \pi} {A^{12}} \right) + \frac{\pi^{4}}{36} \end{align} where $A$ is the Glaisher-Kinkelin constant.

The integrals \begin{align} I_{5} = \int_{0}^{1} \frac{\ln(1-x)}{\sqrt{\ln\left(\frac{1}{x}\right)}} \ \ln^{2}\left( \ln \left(\frac{1}{x}\right) \right) \ \frac{dx}{x} \end{align} and \begin{align} I_{6} = \int_{0}^{1} \frac{\ln(1-x)}{\ln^{3/2}\left(\frac{1}{x}\right)} \ \ln^{2}\left( \ln \left(\frac{1}{x}\right) \right) \ \frac{dx}{x} \end{align} can be expressed in terms of $\partial_{s}^{2}\zeta(s)|_{s=3/2}$ and $\partial_{s}^{2}\zeta(s)|_{s=1/2}$, respectively. Can these integrals be evaluated in closed form expression without the use of derivatives of the Zeta function?

If the integrals can be evaluated in such a way what is the resulting value?

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  • $\begingroup$ These are beautiful results. Can you tell me the source, please ? $\endgroup$ – Zaid Alyafeai May 22 '14 at 21:04
  • $\begingroup$ @Zaid There are many known integrals, for example the integrals contained in "Tables of Integrals, series, and products". Many of those integrals can be manipulated to create new integrals. The integrals above are derived from integrals that are known, but do not, as of yet, have completed closed form expressions. $\endgroup$ – Leucippus May 23 '14 at 14:50
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For the first integral, \begin{align} I_{5} &= \int_{0}^{1} \frac{\ln(1-x)}{\sqrt{\ln\left(\frac{1}{x}\right)}} \ \ln^{2}\left( \ln \left(\frac{1}{x}\right) \right) \ \frac{dx}{x}\\&=-\frac{\sqrt{\pi}}{2}(\pi^2+2\gamma+4\log2)^2\zeta\left(\frac32\right)+2\sqrt{\pi}(\gamma+2\log2)\zeta'\left(\frac32\right)-\sqrt{\pi}\zeta''\left(\frac32\right) \end{align} there is only one $\zeta''$ term so it doesn't seem likely to be able to remove it.

The second integral does not converge. But you might be interested in $$\zeta'\left(\frac12\right)=\frac{1}{4}(\pi+2\gamma+2\log8\pi)\zeta\left(\frac12\right)\\\zeta''\left(\frac12\right)=\left(2G+\frac{5\pi^2+4\gamma\pi+4\gamma^2}{16}+\frac{\log8\pi}{4}(\log8\pi+\pi+2\gamma)\right)\zeta\left(\frac12\right)$$

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  • $\begingroup$ The last formula for $\zeta''\left(\tfrac{1}{2}\right)$ does not check numerically. Are you sure it's correct? Do you have a reference or a proof for it? $\endgroup$ – Vladimir Reshetnikov Jul 29 '15 at 2:01

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