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I am struggling to evaluate the following series:

$$\mathcal{S}=\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k^2}\log\left[ \frac{\Gamma\left(1+\frac{k}{2}\right)}{\Gamma\left(\frac12+\frac{k}{2}\right)} \right]=-0.13360...$$

From what I tried, I figured out that, if a closed form does exist, it must be really difficult to find.

Here are my attempts:

Attempt (1):

using the fact that, for $\Re(z)>0$ and $\Re(z-x)>0$ $$\frac{\Gamma(z+1)}{\Gamma(z+1-x)}=e^{-\gamma x}\prod_{n=1}^\infty\left[ \left(1-\frac{x}{z+n}\right)e^{\frac{x}{n}} \right]$$ $$\implies \frac{\Gamma(z+1)}{\Gamma\left(z+\frac12\right)}=e^{-\frac{\gamma}{2}}\prod_{n=1}^\infty\left[ \left(1-\frac{1}{2(z+n)}\right)e^{\frac{1}{2n}} \right]$$ $$\implies \frac{\Gamma\left(1+\frac{k}{2}\right)}{\Gamma\left(\frac12+\frac{k}{2}\right)}=e^{-\frac{\gamma}{2}}\prod_{n=1}^\infty\left[ \left(1-\frac{1}{k+2n}\right)e^{\frac{1}{2n}} \right]$$ $$\implies \log\left[ \frac{\Gamma\left(1+\frac{k}{2}\right)}{\Gamma\left(\frac12+\frac{k}{2}\right)} \right]=-\frac{\gamma}{2}+\sum_{n=1}^\infty \left[\frac{1}{2n}+\log\left(1-\frac{1}{k+2n}\right)\right]$$ and here the tough part is the double sum that we end up with after plugging this value into the original $\mathcal{S}$. Notice that all these steps where valid only for $k>1$, so we would get $$\mathcal{S}=\log\left(\frac{\sqrt{\pi}}{2}\right)+\frac{\gamma}{2}\left(1-\frac{\pi^2}{12} \right) +\sum_{k=2}^\infty \sum_{n=1}^\infty \frac{(-1)^{k+1}}{k^2}\left[\frac{1}{2n}+\log\left(1-\frac{1}{k+2n}\right)\right]$$ which had me stuck.

Attempt (2):

I tried splitting the sum into even and odd terms, in order to get rid of the gamma functions. For instance, the odd terms form the following series: $$\mathcal{S_1}=\sum_{k=0}^\infty\frac{1}{(2k+1)^2}\log\left[ \frac{\Gamma(k+1+\frac12)}{\Gamma(k+1)} \right]$$ Now, using the fact that $$\frac{\Gamma(n+\frac12)}{\Gamma (n)}={{2n} \choose n}\frac{n}{4^n}\sqrt{\pi}$$ we have $$\mathcal{S_1}=\sum_{k=0}^\infty\frac{1}{(2k+1)^2}\log\left[{{2k} \choose k}\frac{2k+1}{4^k}\frac{\sqrt{\pi}}{2} \right]$$ $$=\frac{\pi^2}{8}\log\left(\frac{\sqrt{\pi}}{2}\right)+\sum_{k=0}^\infty\frac{1}{(2k+1)^2}\log\left[{{2k} \choose k}\frac{2k+1}{4^k} \right]$$ but now this series has this $\log$ term that is difficult to handle. We can't even split it using $\log$ properties, since the resulting series wouldn't converge at all.

Attempt (3):

It can be shown that $$\int_0^1\int_0^1\frac{\log(1+x^a)}{1+x}\text{d}x \text{d}a=2\log^22+\frac{\pi^2}{24}\log\pi+\mathcal{S}$$ and notice that here, changing the order of integration is allowed. I couldn't evaluate this double integral either in the end.

EDIT:

Attempt (4):

Using this we can rewrite $\mathcal{S}$ as: $$\mathcal{S}=-\frac{\pi^2}{24}\log\pi-\int_0^1\frac{\text{Li}_2(-x)+\frac{\pi^2}{12}}{(x+1)\log x}\text{d}x$$ and this integral is similar to the ones that appear here and here, so maybe it can be expressed as a series involving harmonic numbers as well?

Zima
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  • The use of square brackets for grouping expressions is becoming obsolete and now is mostly used only in very elementary Mathematics. You should try to use only $($ and $)$, and leave $[$ to $]$ for other uses. – jjagmath Jul 06 '23 at 22:58
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    Not really a closed form, but I found $$ \mathcal{S} = - \frac{{\zeta '(2)}}{4} - \frac{{\pi ^2 }}{{12}}\log 2 - \frac{1}{2}\int_0^{ + \infty } {\tanh \left( {\frac{t}{2}} \right)\frac{{{\mathop{\rm Li}\nolimits} _2 ( - {\rm e}^{ - t} )}}{t}{\rm d}t} . $$ – Gary Jul 07 '23 at 00:39
  • The last formulation is really interesting . – Claude Leibovici Jul 07 '23 at 11:36
  • @ClaudeLeibovici using the last formulation, together with Gary's result, I arrived at the following integral: $$\int_0^1\frac {\left(1+\frac{1}{x}\right)\text{Li}_2(-x)+\frac{\pi^2}{6}} {(x+1)\log x}\text{d}x=\pi^2\left( \frac{\log2}{4}+\frac{\gamma}{12}-\log A \right)$$ which I think it's beautiful, but useless in this context unfortunately. – Zima Jul 07 '23 at 15:25
  • (3) would be extremely hard since even some fixed values of $a$, the closed form expression is really horrid to look at https://math.stackexchange.com/questions/426325 – oO_ƲRF_Oo Sep 14 '23 at 17:15
  • @oO_ƲRF_Oo yea I lost all my hope on this one too long ago – Zima Sep 14 '23 at 21:27

1 Answers1

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If you would accept to stay with an infinite summation, you could use $$S=\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k^2}\log\left[ \frac{\Gamma\left(1+\frac{k}{2}\right)}{\Gamma\left(\frac12+\frac{k}{2}\right)} \right]$$ $$S=\log \left(\frac{\sqrt{\pi }}{2}\right)+\sum_{k=2}^\infty\frac{(-1)^{k+1}}{k^2}\log\left[ \frac{\Gamma\left(1+\frac{k}{2}\right)}{\Gamma\left(\frac12+\frac{k}{2}\right)} \right]$$ Now, using Stirling approximation $$\log\left[ \frac{\Gamma\left(1+\frac{k}{2}\right)}{\Gamma\left(\frac12+\frac{k}{2}\right)} \right]=\frac{1}{2} (\log (k)-\log (2))+\frac 14 \sum_{n=0}^\infty \frac {a_n} {b_n} \,k^{-(2n+1)}$$ where the $a_n$ and $b_n$ form respectively sequences $A275994$ and $A219931$ in $OEIS$.

This would make $$S=\log \left(\frac{\sqrt{\pi }}{2}\right)-\frac{\pi ^2-6}{12} \log (2)-\frac{1}{4} \zeta '(2)+$$ $$\frac 14 \sum_{n=0}^\infty\sum_{k=2}^\infty (-1)^{k+1}\frac {a_n} {b_n} \,k^{-(2n+3)}$$ and $$\sum_{k=2}^\infty (-1)^{k+1}\,k^{-(2n+3)}=\left(1-2^{-2 (n+1)}\right) \zeta (2 n+3)-1$$

Claude Leibovici
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