# Computing $\sum_{n=1}^{\infty} \frac{\psi\left(\frac{n+1}{2}\right)}{ \binom{2n}{n}}$

Here is an interesting series I played with, namely

$$\sum_{n=1}^{\infty} \frac{\displaystyle\psi\left(\frac{n+1}{2}\right)}{\displaystyle \binom{2n}{n}} \approx -0.245969181104090562617616399148$$

where $\psi(x)$ is digamma function

and the closed form I got you may see here, but that was possible because I had some luck.
However, the closed form doesn't look that friendly and maybe we can improve that.
It's also possible that a different approach leads to a shorter closed form, just the intuition.
I'd be interested in an approach that only uses series manipulations, I couldn't do that.

• the closed form doesn't look that friendly - Is there a gold medal for understatements of astronomic proportions awarded on here this site ? – Lucian Nov 7 '14 at 19:36
• May I ask how you obtained the closed form? Simple CAS result or did it require manual manipulations? – user111187 Nov 7 '14 at 21:38
• You make us guess what that $\psi$ function is ... Also did you ask Chuck Norris for the result ? – mick Nov 7 '14 at 22:50
• Also why do you tag with integrals ??? – mick Nov 7 '14 at 22:51
• @mick $\psi$ is standard notation for the digamma function (derivative of $\log \Gamma$). – user111187 Nov 7 '14 at 23:18

A place to begin:

I think the search for a friendlier closed form could benefit from breaking the series up into more manageable components. Start by splitting the sum into the sums of odd and even terms:

\begin{align} S &=\sum_{p=1}^{\infty}\frac{\psi{\left(\frac{p+1}{2}\right)}}{\binom{2p}{p}}\\ &=\sum_{n=1}^{\infty}\frac{\psi{\left(\frac{(2n-1)+1}{2}\right)}}{\binom{2(2n-1)}{(2n-1)}}+\sum_{n=1}^{\infty}\frac{\psi{\left(\frac{(2n)+1}{2}\right)}}{\binom{2(2n)}{(2n)}}\\ &=\sum_{n=1}^{\infty}\frac{\psi{\left(n\right)}}{\binom{4n-2}{2n-1}}+\sum_{n=1}^{\infty}\frac{\psi{\left(n+\frac12\right)}}{\binom{4n}{2n}}\\ &=:S^{(-)}+S^{(+)}.\\ \end{align}

The ratio of the respective denominators of the summands of $S^{(-)}$ and $S^{(+)}$ turns out to be a simple rational function of $n$:

$$\frac{\binom{4n}{2n}}{\binom{4n-2}{2n-1}}=4-\frac{1}{n}.$$

This allows us to rewrite the sum of odd terms as,

\begin{align} S^{(-)} &=\sum_{n=1}^{\infty}\frac{\psi{\left(n\right)}}{\binom{4n-2}{2n-1}}\\ &=\sum_{n=1}^{\infty}\frac{\left(4-\frac{1}{n}\right)\psi{\left(n\right)}}{\binom{4n}{2n}}\\ &=4\sum_{n=1}^{\infty}\frac{\psi{\left(n\right)}}{\binom{4n}{2n}}-\sum_{n=1}^{\infty}\frac{\psi{\left(n\right)}}{n\binom{4n}{2n}}\\ &=:4A-B.\\ \end{align}

The sum of even terms can be further split up using the digamma function identity,

$$\psi{\left(n+\frac12\right)}=2\psi{\left(2n\right)}-\psi{\left(n\right)}-\ln{(4)}.$$

Then,

\begin{align} S^{(+)} &=\sum_{n=1}^{\infty}\frac{\psi{\left(n+\frac12\right)}}{\binom{4n}{2n}}\\ &=\sum_{n=1}^{\infty}\frac{2\psi{\left(2n\right)}-\psi{\left(n\right)}-\ln{(4)}}{\binom{4n}{2n}}\\ &=2\sum_{n=1}^{\infty}\frac{\psi{\left(2n\right)}}{\binom{4n}{2n}}-\sum_{n=1}^{\infty}\frac{\psi{\left(n\right)}}{\binom{4n}{2n}}-\ln{(4)}\sum_{n=1}^{\infty}\frac{1}{\binom{4n}{2n}}\\ &=:2C-A-\ln{(4)}D.\\ \end{align}

We can at least find a nice closed form for the sum $D$ in the last line above:

$$D=\sum_{n=1}^{\infty}\frac{1}{\binom{4n}{2n}}=\frac{1}{15}+\frac{\sqrt{3}\,\pi}{27}-\frac{2\sqrt{5}}{25}\ln{\phi}\approx 0.182118141166556731177.$$

$$-\ln{(4)}\,D\approx -0.25246935215683381486272395984234.$$

It remains to evaluate the three series $A,B,C$. Note that all three series fall under the general form,

$$\sigma{\left(r,s\right)}:=\sum_{n=1}^{\infty}n^{s}\,\frac{\psi{\left(r\,n\right)}}{\binom{4n}{2n}}.$$

Then,

$$\begin{cases} A=\sigma{\left(1,0\right)}\approx -0.08905771667254809;\\ B=\sigma{\left(1,-1\right)}\approx -0.0928236851612787689;\\ C=\sigma{\left(2,0\right)}\approx 0.0904248179545543815129359182;\\ 3A-B+2C\approx 0.0065001710527432522437457174000.\\ \end{cases}$$

Integral representations:

For convenience of notation, we shall define the three parameter auxiliary series $\tau{\left(r,s,t\right)}$ by,

$$\tau{\left(r,s,t\right)}:=\sum_{n=1}^{\infty}\frac{t^{rn-1}n^{s}}{\binom{4n}{2n}}.$$

We'll also go ahead and define the one parameter series $\tau{\left(s\right)}$ in terms of $\tau{\left(r,s,t\right)}$ by taking $r=2$ and $t=1$:

$$\tau{\left(s\right)}:=\tau{\left(2,s,1\right)}=\sum_{n=1}^{\infty}\frac{n^{s}}{\binom{4n}{2n}}.$$

For $\Re{(z)}>0$, the digamma function has the following integral representation:

$$\psi{\left(z\right)}=-\gamma+\int_{0}^{1}\frac{1-t^{z-1}}{1-t}\,\mathrm{d}t.$$

Binomial coefficients can of course be written in terms of either gamma functions or beta functions:

\begin{align} \binom{z}{w} &=\frac{z!}{w!(z-w)!};~z,w\in\mathbb{N}^{+};z\ge w\\ &=\frac{\Gamma{\left(z+1\right)}}{\Gamma{\left(w+1\right)}\Gamma{\left(z-w+1\right)}}\\ &=\frac{1}{w\operatorname{B}{\left(w,z-w+1\right)}}.\\ \end{align}

Then, we can find a single integral representation for reciprocals of binomial coefficients in terms of a that of the beta function:

\begin{align} \frac{1}{\binom{z}{w}} &=w\operatorname{B}{\left(w,z-w+1\right)}\\ &=w\int_{0}^{1}t^{w-1}\left(1-t\right)^{z-w}\,\mathrm{d}t;~\Re{(w)}>0\land\Re{(z)}>\Re{(w)}-1.\\ \end{align}

Starting from the integral representation for the digamma function, we can arrive at a representation for the series $\sigma{\left(r,s\right)}$ (and thus for the series $S$ as well) in terms of the series $\tau{\left(r,s,t\right)}$ via summation under the integral sign:

\begin{align} \sigma{\left(r,s\right)} &=\sum_{n=1}^{\infty}n^{s}\,\frac{\psi{\left(r\,n\right)}}{\binom{4n}{2n}}\\ &=\sum_{n=1}^{\infty}\frac{n^{s}}{\binom{4n}{2n}}\left[-\gamma+\int_{0}^{1}\frac{1-t^{rn-1}}{1-t}\,\mathrm{d}t\right]\\ &=-\gamma\sum_{n=1}^{\infty}\frac{n^{s}}{\binom{4n}{2n}}+\sum_{n=1}^{\infty}\frac{n^{s}}{\binom{4n}{2n}}\int_{0}^{1}\frac{1-t^{rn-1}}{1-t}\,\mathrm{d}t\\ &=-\gamma\sum_{n=1}^{\infty}\frac{n^{s}}{\binom{4n}{2n}}+\int_{0}^{1}\frac{\sum_{n=1}^{\infty}\frac{n^{s}}{\binom{4n}{2n}}-\sum_{n=1}^{\infty}\frac{t^{rn-1}n^{s}}{\binom{4n}{2n}}}{1-t}\,\mathrm{d}t\\ &=-\gamma\,\tau{\left(s\right)}+\int_{0}^{1}\frac{\tau{\left(s\right)}-\tau{\left(r,s,t\right)}}{1-t}\,\mathrm{d}t,\\ \end{align}

\begin{align} \implies S &=3A-B+2C-2\ln{(2)}D\\ &=3\sigma{\left(1,0\right)}-\sigma{\left(1,-1\right)}+2\sigma{\left(2,0\right)}-2\ln{(2)}D\\ &=3\left[-\gamma\,\tau{\left(0\right)}+\int_{0}^{1}\frac{\tau{\left(0\right)}-\tau{\left(1,0,t\right)}}{1-t}\,\mathrm{d}t\right]+2\left[-\gamma\,\tau{\left(0\right)}+\int_{0}^{1}\frac{\tau{\left(0\right)}-\tau{\left(2,0,t\right)}}{1-t}\,\mathrm{d}t\right]\\ &~~~~~ -\left[-\gamma\,\tau{\left(-1\right)}+\int_{0}^{1}\frac{\tau{\left(-1\right)}-\tau{\left(1,-1,t\right)}}{1-t}\,\mathrm{d}t\right]-2\ln{(2)}\tau{\left(0\right)}\\ &=-3\gamma\,\tau{\left(0\right)}+3\int_{0}^{1}\frac{\tau{\left(0\right)}-\tau{\left(1,0,t\right)}}{1-t}\,\mathrm{d}t-2\gamma\,\tau{\left(0\right)}+2\int_{0}^{1}\frac{\tau{\left(0\right)}-\tau{\left(2,0,t\right)}}{1-t}\,\mathrm{d}t\\ &~~~~~ +\gamma\,\tau{\left(-1\right)}-\int_{0}^{1}\frac{\tau{\left(-1\right)}-\tau{\left(1,-1,t\right)}}{1-t}\,\mathrm{d}t-2\ln{(2)}\tau{\left(0\right)}\\ &=-\left[5\gamma+2\ln{(2)}\right]\tau{\left(0\right)}+\gamma\,\tau{\left(-1\right)}+3\int_{0}^{1}\frac{\tau{\left(0\right)}-\tau{\left(1,0,t\right)}}{1-t}\,\mathrm{d}t\\ &~~~~~ +2\int_{0}^{1}\frac{\tau{\left(0\right)}-\tau{\left(2,0,t\right)}}{1-t}\,\mathrm{d}t-\int_{0}^{1}\frac{\tau{\left(-1\right)}-\tau{\left(1,-1,t\right)}}{1-t}\,\mathrm{d}t.\\ \end{align}

An integral representation for $\tau{\left(r,s,t\right)}$ would also be helpful. We can start from the integral representation of the reciprocal binomial coefficient $\frac{1}{\binom{4n}{2n}}$:

\begin{align} \frac{1}{\binom{4n}{2n}} &=2n\operatorname{B}{\left(2n,2n+1\right)}\\ &=\left(4n+1\right)\operatorname{B}{\left(2n+1,2n+1\right)}\\ &=\left(4n+1\right)\int_{0}^{1}u^{2n}\left(1-u\right)^{2n}\,\mathrm{d}u.\\ \end{align}

Then,

\begin{align} \frac{n^{s}}{\binom{4n}{2n}} &=\left(4n+1\right)n^{s}\int_{0}^{1}u^{2n}\left(1-u\right)^{2n}\,\mathrm{d}u\\ &=4n^{s+1}\int_{0}^{1}u^{2n}\left(1-u\right)^{2n}\,\mathrm{d}u+n^{s}\int_{0}^{1}u^{2n}\left(1-u\right)^{2n}\,\mathrm{d}u,\\ \end{align}

and,

$$\frac{t^{rn-1}n^{s}}{\binom{4n}{2n}}=4n^{s+1}t^{rn-1}\int_{0}^{1}u^{2n}\left(1-u\right)^{2n}\,\mathrm{d}u+n^{s}t^{rn-1}\int_{0}^{1}u^{2n}\left(1-u\right)^{2n}\,\mathrm{d}u.$$

Summing over $n$ and switching the order of summation and integration, we arrive at an integral representation for $\tau{\left(r,s,t\right)}$:

\begin{align} \tau{\left(r,s,t\right)} &=\sum_{n=1}^{\infty}\frac{t^{rn-1}n^{s}}{\binom{4n}{2n}}\\ &=\sum_{n=1}^{\infty}4n^{s+1}t^{rn-1}\int_{0}^{1}u^{2n}\left(1-u\right)^{2n}\,\mathrm{d}u+\sum_{n=1}^{\infty}n^{s}t^{rn-1}\int_{0}^{1}u^{2n}\left(1-u\right)^{2n}\,\mathrm{d}u\\ &=4\int_{0}^{1}\sum_{n=1}^{\infty}n^{s+1}t^{rn-1}u^{2n}\left(1-u\right)^{2n}\,\mathrm{d}u+\int_{0}^{1}\sum_{n=1}^{\infty}n^{s}t^{rn-1}u^{2n}\left(1-u\right)^{2n}\,\mathrm{d}u\\ &=\frac{4}{r}\int_{0}^{1}r\sum_{n=1}^{\infty}n^{s}\cdot nt^{rn-1}u^{2n}\left(1-u\right)^{2n}\,\mathrm{d}u+\frac{1}{t}\int_{0}^{1}t\sum_{n=1}^{\infty}n^{s}t^{rn-1}u^{2n}\left(1-u\right)^{2n}\,\mathrm{d}u\\ &=\frac{4}{r}\int_{0}^{1}\sum_{n=1}^{\infty}n^{s}\cdot rnt^{rn-1}u^{2n}\left(1-u\right)^{2n}\,\mathrm{d}u+\frac{1}{t}\int_{0}^{1}\sum_{n=1}^{\infty}n^{s}t^{rn}u^{2n}\left(1-u\right)^{2n}\,\mathrm{d}u\\ &=\frac{4}{r}\int_{0}^{1}\sum_{n=1}^{\infty}n^{s}\frac{\partial}{\partial t}t^{rn}u^{2n}\left(1-u\right)^{2n}\,\mathrm{d}u+\frac{1}{t}\int_{0}^{1}\sum_{n=1}^{\infty}n^{s}t^{rn}u^{2n}\left(1-u\right)^{2n}\,\mathrm{d}u\\ &=\frac{4}{r}\int_{0}^{1}\frac{\partial}{\partial t}\sum_{n=1}^{\infty}n^{s}t^{rn}u^{2n}\left(1-u\right)^{2n}\,\mathrm{d}u+\frac{1}{t}\int_{0}^{1}\sum_{n=1}^{\infty}n^{s}t^{rn}u^{2n}\left(1-u\right)^{2n}\,\mathrm{d}u\\ &=\frac{4}{r}\int_{0}^{1}\frac{\partial}{\partial t}\sum_{n=1}^{\infty}n^{s}\left[t^{r}u^{2}\left(1-u\right)^{2}\right]^n\,\mathrm{d}u+\frac{1}{t}\int_{0}^{1}\sum_{n=1}^{\infty}n^{s}\left[t^{r}u^{2}\left(1-u\right)^{2}\right]^n\,\mathrm{d}u\\ &=\frac{4}{r}\int_{0}^{1}\frac{\partial}{\partial t}\operatorname{Li}_{-s}{\left(t^{r}u^{2}\left(1-u\right)^{2}\right)}\,\mathrm{d}u+\frac{1}{t}\int_{0}^{1}\operatorname{Li}_{-s}{\left(t^{r}u^{2}\left(1-u\right)^{2}\right)}\,\mathrm{d}u\\ &=\frac{4}{r}\int_{0}^{1}\frac{r}{t}\operatorname{Li}_{-s-1}{\left(t^{r}u^{2}\left(1-u\right)^{2}\right)}\,\mathrm{d}u+\frac{1}{t}\int_{0}^{1}\operatorname{Li}_{-s}{\left(t^{r}u^{2}\left(1-u\right)^{2}\right)}\,\mathrm{d}u\\ &=\frac{4}{t}\int_{0}^{1}\operatorname{Li}_{-s-1}{\left(t^{r}u^{2}\left(1-u\right)^{2}\right)}\,\mathrm{d}u+\frac{1}{t}\int_{0}^{1}\operatorname{Li}_{-s}{\left(t^{r}u^{2}\left(1-u\right)^{2}\right)}\,\mathrm{d}u,\\ \end{align}

where $r>0$ and $t>0$.

Calculation of main integral:

For the sake of this problem, we only need to evaluate $\tau{\left(r,s,t\right)}$ in the two particular cases where $s=0$ or $s=-1$:

$$\begin{cases} \tau{\left(r,0,t\right)}=\frac{4}{t}\int_{0}^{1}\operatorname{Li}_{-1}{\left(t^{r}u^{2}\left(1-u\right)^{2}\right)}\,\mathrm{d}u+\frac{1}{t}\int_{0}^{1}\operatorname{Li}_{0}{\left(t^{r}u^{2}\left(1-u\right)^{2}\right)}\,\mathrm{d}u,\\ \tau{\left(r,-1,t\right)}=\frac{4}{t}\int_{0}^{1}\operatorname{Li}_{0}{\left(t^{r}u^{2}\left(1-u\right)^{2}\right)}\,\mathrm{d}u+\frac{1}{t}\int_{0}^{1}\operatorname{Li}_{1}{\left(t^{r}u^{2}\left(1-u\right)^{2}\right)}\,\mathrm{d}u.\\ \end{cases}$$

Define the function $F{\left(z\right)}$ by the integral,

$$F{\left(z\right)}:=\int_{0}^{1}\operatorname{Li}_{1}{\left(zx^{2}\left(1-x\right)^{2}\right)}\,\mathrm{d}x.$$

Then,

$$F^{\prime}{\left(z\right)}=\frac{1}{z}\int_{0}^{1}\operatorname{Li}_{0}{\left(zx^{2}\left(1-x\right)^{2}\right)}\,\mathrm{d}x,$$

and

$$F^{\prime\prime}{\left(z\right)}=\frac{1}{z^2}\int_{0}^{1}\operatorname{Li}_{-1}{\left(zx^{2}\left(1-x\right)^{2}\right)}\,\mathrm{d}x-\frac{1}{z}F^{\prime}{\left(z\right)}.$$

Thus, we can represent the $\tau$ functions we need entirely in terms of the function $F$ and its derivatives:

$$\begin{cases} \tau{\left(r,0,t\right)}=4t^{2r-1}F^{\prime\prime}{\left(t^r\right)}+5t^{r-1}F^{\prime}{\left(t^r\right)},\\ \tau{\left(r,-1,t\right)}=4t^{r-1}F^{\prime}{\left(t^r\right)}+\frac{1}{t}F{\left(t^r\right)}.\\ \end{cases}$$

So we concentrate on finding a closed form for the integral defining $F{(z)}$.

Letting $\sqrt{z}=:a$ and $\sqrt{\frac{a}{4+a}}=:b$, and using the fact that $\operatorname{Li}_{1}{\left(x\right)}=-\ln{\left(1-x\right)}$, we have:

\begin{align} F{\left(z\right)} &=F{\left(a^2\right)}\\ &=\int_{0}^{1}\operatorname{Li}_{1}{\left(a^2x^2\left(1-x\right)^2\right)}\,\mathrm{d}x\\ &=-\int_{0}^{1}\ln{\left(1-a^2x^2\left(1-x\right)^2\right)}\,\mathrm{d}x\\ &=-\int_{0}^{1}\ln{\left(1-ax\left(1-x\right)\right)}\,\mathrm{d}x-\int_{0}^{1}\ln{\left(1+ax\left(1-x\right)\right)}\,\mathrm{d}x\\ &=-\int_{0}^{1}\ln{\left(1-ax\left(1-x\right)\right)}\,\mathrm{d}x-\int_{0}^{1}\ln{\left(1+\frac{4b^2}{1-b^2}x\left(1-x\right)\right)}\,\mathrm{d}x\\ &=-\int_{0}^{1}\ln{\left(1-ax\left(1-x\right)\right)}\,\mathrm{d}x-\int_{0}^{1}\ln{\left(\left(1+\frac{2b}{1-b}x\right)\left(1-\frac{2b}{1+b}x\right)\right)}\,\mathrm{d}x\\ &=-\int_{0}^{1}\ln{\left(1-ax\left(1-x\right)\right)}\,\mathrm{d}x-\int_{0}^{1}\ln{\left(1+\frac{2b}{1-b}x\right)}\,\mathrm{d}x\\ &~~~~~ -\int_{0}^{1}\ln{\left(1-\frac{2b}{1+b}x\right)}\,\mathrm{d}x\\ &=\left[2-2\sqrt{\frac{4-a}{a}}\tan^{-1}{\left(\sqrt{\frac{a}{4-a}}\right)}\right]+\left[1+\frac{1+b}{2b}\ln{\left(\frac{1-b}{1+b}\right)}\right]\\ &~~~~~ +\left[1+\frac{1-b}{2b}\ln{\left(\frac{1-b}{1+b}\right)}\right]\\ &=4-2\sqrt{\frac{4-a}{a}}\tan^{-1}{\left(\sqrt{\frac{a}{4-a}}\right)}+\frac{1}{b}\ln{\left(\frac{1-b}{1+b}\right)}\\ &=4-2\sqrt{\frac{4-a}{a}}\tan^{-1}{\left(\sqrt{\frac{a}{4-a}}\right)}+\sqrt{\frac{4+a}{a}}\ln{\left(1+\frac{a}{2}-\frac12\sqrt{a\left(a+4\right)}\right)}\\ &=4-2\frac{\sqrt{4-\sqrt{z}}}{z^{1/4}}\tan^{-1}{\left(\frac{z^{1/4}}{\sqrt{4-\sqrt{z}}}\right)}+\frac{\sqrt{4+\sqrt{z}}}{z^{1/4}}\ln{\left(1+\frac{\sqrt{z}}{2}-\frac{z^{1/4}}{2}\sqrt{\sqrt{z}+4}\right)}.\\ \end{align}

• $\psi(n)=H_{n-1}~-~\gamma,~$ and $~\displaystyle\sum_1^\infty\frac{H_n}{\displaystyle{2n\choose n}} ~=~ \frac2{27}\bigg[\psi_1\bigg(\frac13\bigg) - \psi_1\bigg(\frac13\bigg) + \pi\sqrt3~\big(2-\ln3\big)\bigg]$. – Lucian Nov 25 '14 at 23:08
• @Lucian As currently written the trigamma terms cancel out, and I'm thinking it can't be that easy. Is there a typo? – David H Nov 26 '14 at 1:49
• The second one is $\psi_1\bigg(\dfrac23\bigg)$. – Lucian Nov 26 '14 at 1:50

I took the expression from your paste-bin and simplified it quite a bit. I did it by first splitting the sum into its main parts and sorting them. A lot of the 201 terms looked similar enough so that I grouped them together. After a lot of simplifying and massaging one gets that

$$sum = A + T + L + P$$

where

$$A = \frac{1}{135} \left(8 \sqrt{3} \pi -5 \gamma \left(9+2 \sqrt{3} \pi \right)\right)$$

and

$$T = -\frac{2}{75} \left(50+13 \sqrt{5}\right) \pi \tan ^{-1}\left(\sqrt{\frac{3}{5}}\right)+\frac{1}{75} \left(50+11 \sqrt{5}\right) \pi \tan ^{-1}\left(\sqrt{15}\right)\\+\frac{1}{240} \left(125+29 \sqrt{5}\right) \sinh ^{-1}(2)-\frac{2 \text{csch}^{-1}(2)}{\sqrt{5}}+\frac{2 (5+\log (2)) \coth ^{-1}\left(\sqrt{5}\right)}{5 \sqrt{5}}$$

and

$$L = \frac{1}{40} \left(5-9 \sqrt{5}\right) \log (2)-\frac{\pi \log (3)}{9 \sqrt{3}}+\log (4)+\frac{1}{80} \left(125+61 \sqrt{5}\right) \log \left(\sqrt{5}-1\right)\\+\frac{\log \left(\frac{1}{2} \left(3+\sqrt{5}\right)\right)}{\sqrt{5}}-\frac{\log (2) \left(800 \sqrt{3} \pi +9 \left(4585+645 \sqrt{5}-48 \sqrt{5} \log \left(\frac{1}{2} \left(3+\sqrt{5}\right)\right)\right)\right)}{10800}$$

and

$$P = \frac{4 \Re\left(\text{Li}_2\left(\frac{1}{8} \left(3+\sqrt{5}-i \sqrt{3} \left(-1+\sqrt{5}\right)\right)\right)-\text{Li}_2\left(\frac{1}{8} \left(3-\sqrt{5}+i \sqrt{6 \left(3+\sqrt{5}\right)}\right)\right)\right)}{5 \sqrt{5}}\\-\frac{4 \Im\text{Li}_2\left(\frac{1}{6} \left(3-i \sqrt{3}\right)\right)}{3 \sqrt{3}}$$

Curiously

$$P - \gamma \approx 0.000220304\ldots$$