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Coming across the calculation of a special integral I get stuck on the following series, which I have given its integral representation :

$$\text{J}=\sum_{n\geq1}\frac{1}{n}\ln\bigg(1+\frac{1}{n}\bigg)=\int_{0}^{1}\frac{1}{x}\bigg(\psi(1+x)+\gamma \bigg)dx\,$$
Where $\psi$ denotes the digamma function and $\gamma$ represents the Euler-Mascheroni constant. I am wondering if such a series has a closed-form.

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    $\begingroup$ Related Integral of digamma function and Infinite product experimental mathematics question. $\endgroup$ – Sil Aug 17 '18 at 20:37
  • $\begingroup$ @Sil Manythanks, from the links above it seems that this integral has no closed-form. $\endgroup$ – Kays Tomy Aug 17 '18 at 20:57
  • $\begingroup$ Possibly the sign infront of $\gamma$ should be $+$ ? $\endgroup$ – Diger Aug 17 '18 at 21:01
  • $\begingroup$ @Diger, just a typo, it is edited, thanks $\endgroup$ – Kays Tomy Aug 17 '18 at 21:06
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    $\begingroup$ The integral representation $$ J = \int_{0}^{1}\frac{(1-u)\log(1-u)}{u\log(u)}\,du$$ follows from Frullani's theorem and allows a simple numerical evaluation. $\endgroup$ – Jack D'Aurizio Aug 17 '18 at 22:57
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I deleted my comment, since I found a better approximation. By the use of Binet's formula (see Binet), one can derive $${\rm J}=\sum_{k=2}^\infty \frac{(-1)^k}{k-1} \, \zeta(k)$$ which converges somewhat slowly. By partial summation this can be brought to $$\zeta(2) \log 2 + \sum_{k=2}^\infty \frac{(-1)^k}{2} \left[\Psi\left(\frac{k+1}{2}\right) - \Psi\left(\frac{k}{2}\right)\right] \left[\zeta(k)-\zeta(k+1)\right] \, .$$ Another partial summation yields $$\zeta(2)\log 2 + \frac{1}{2} \sum_{n=1}^\infty \Big\{ \Psi(n+1) \left[\zeta(2n+2)-\zeta(2n+1)\right] \\ - \Psi(n+1/2) \left[\zeta(2n+2)-\zeta(2n)\right] + \Psi(n) \left[\zeta(2n+1)-\zeta(2n)\right] \Big\}$$ which converges a lot faster with exponential order ${\cal O}\left(\frac{1}{n4^{n+2}}\right)$. Cutting the series after only 5 terms yields \begin{align}{\rm J}&\approx 1.257743 \tag{approximation} \\ {\rm J} &= 1.257746... \tag{exact}\end{align} correct up to 6 digits.

Interestingly the number is somewhat similar to the alike looking sum $$\gamma=\sum_{k=2}^\infty \frac{(-1)^k}{k} \, \zeta(k)$$ from which an alternative expression arises $${\rm J} = \gamma + \sum_{k=2}^\infty \frac{(-1)^k \, \zeta(k)}{k(k-1)} \, .$$

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