Closed form for $\sum_{n=1}^{\infty}\frac{(2\log\phi)^{2n+3}B_{2n}}{2n(2n+3)!}$

Question

I need a closed form for the sum $$\sum_{n=1}^{\infty}\frac{(2\log\phi)^{2n+3}B_{2n}}{2n(2n+3)!} $$ where $\phi=\frac{1+\sqrt{5}}{2}$ is the golden ratio and $B_n$ are the Bernoulli numbers.

I tried denoting $$S:=\sum_{n=1}^{\infty}\frac{(2\log\phi)^{2n+3}B_{2n}}{2n(2n+3)!} $$ Using the integral representation of $B_{2n}$ we have $$B_{2n}=\frac{4(-1)^n n (2n-1)}{(2\pi)^{2n}}\int_{0}^{1}\frac{\log(1-t)\log^{2n-2}(t)}{t}\ dt $$ So we get by inserting this above representation of $B_{2n}$ in $S$ and interchanging the order of sum and integral (which needs to be justified) we get $$S=4\int_{0}^{1}\frac{\log(1-t)}{t\log^2 t}\left(\sum_{n=1}^{\infty}\frac{(-1)^n (2n-1)}{(2n+3)!} \left(\frac{\log\phi\log t}{\pi}\right)^{2n}\right)\ dt $$ Now we have (see here) $$\sum_{n=1}^{\infty}\frac{(-1)^n(2n-1)a^{2n}}{(2n+3)!}=\frac{a^3-6a\cos a-18 a+24\sin a}{6a^3} $$ So putting $a=\frac{\log\phi\log t}{\pi}$ we obtain

$$S=4\int_{0}^{1}\frac{\log(1-t)}{t\log^2 t}\left(\frac{1}{6}-\frac{\pi^2\cos (\frac{\log\phi\log t}{\pi})}{\log^2\phi\log^2 t}-\frac{3\pi^2}{\log^2\phi\log^2 t}+\frac{\pi^3\sin (\frac{\log\phi\log t}{\pi})}{\log^3\phi\log^3 t}\right)\ dt $$ The above integral is convergent as I tried on Mathematica cloud with the code

NIntegrate[Log[1-t]/(tLog[t]^2)*(1-6Cos[Log[GoldenRatio]*Log[t]/Pi]/(Log[GoldenRatio]*Log[t]/Pi)^2-18/(Log[GoldenRatio]Log[t]/Pi)^2+24Sin[Log[GoldenRatio]*Log[t]/Pi]/(Log[GoldenRatio]Log[t]/Pi)^3), {t, 0, 1}]

The answer I got is 0.00184137

I also know that $$\coth x=2\sum_{n=1}^{\infty}\frac{B_{2n}(2x)^{2n-1}}{(2n)!} \ \ \ \ \, 0<|x|<\pi$$ Now $\log\phi <1<\pi $, but I am unable to use this above formula either by integrating or differentiating it. Any help would be highly appreciated.

Answer 1

Good news

Mathematica provides a closed form for

$$S=\sum_{n=1}^{\infty}\frac{x^{2n+3}B_{2n}}{2n(2n+3)!}$$

Bad news

It involves the derivative of the generalized Riemann $\zeta$ function with respect to it first argument

$$S=x \zeta (3)-\frac{6 \log (2 \pi )-11}{36} x^3 +\frac{4i \pi ^3}{3} \left(\zeta ^{(1,0)}\left(-3,1-\frac{i x}{2 \pi }\right)-\zeta ^{(1,0)}\left(-3,1+\frac{i x}{2 \pi }\right)\right)$$

Numerically, for $x=2\log(\phi)$

$$S=0.000572157938618908\cdots$$

Edit

If we rewrite your second expression of $S$

$$S=4\int_{0}^{1}\frac{\log(1-t)}{t\log^2 (t)}\left(\sum_{n=1}^{\infty}\frac{(-1)^n (2n-1)}{(2n+3)!} \left(\frac{\log(\phi)\log (t)}{\pi}\right)^{2n}\right)\, dt$$ as, with $x=\frac{\log(\phi)} \pi$, $$S=4\sum_{n=1}^{\infty}(-1)^n \frac{ (2n-1)}{(2n+3)!} x^{2n} \int_{0}^{1} \frac{\log(1-t)}{t}\, \log^{2n-2} (t)\,dt$$ Using Feynman' trick $$I(a)=\int_{0}^{1} \frac{\log(1-at)}{t}\, \log^{2n-2} (t)\,dt$$ $$I'(a)=-\int_{0}^{1}\frac{\log ^{2 n-2}(t)}{1-a t}\,dt=-\frac{\Gamma (2 n-1) \,\text{Li}_{2 n-1}(a)}{a}$$ $$I(a)=-\frac{\Gamma (2 n-1)\, \text{Li}_{2 n-1}(a)}{a} \quad\implies \quad I(1)=-\Gamma (2 n-1)\, \text{Li}_{2 n}(1)$$

$$S=4\sum_{n=1}^{\infty}(-1)^{n+1} \frac{ (2n-1)}{(2n+3)!} \Gamma (2 n-1)\, \text{Li}_{2 n}(1)\,x^{2n}$$ $$S=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{\text{Li}_{2 n}(1)}{n (n+1) (2 n+1) (2 n+3)}\, x^{2n}$$

Assuming $x>0$

$$S=\frac{1}{18} \left(\frac{9 \zeta (3)}{\pi ^2 x^2}+11-6 \log (2 \pi )\right)+\frac{i \left(\zeta ^{(1,0)}(-3,1-i x)-\zeta ^{(1,0)}(-3,1+i x)\right)}{3 x^3}$$ which, with $x=\frac{\log(\phi)} \pi$, gives

$$S=0.001283651297033247\cdots$$