[Largesum_{k=1}^{infty}frac{(2^{2k-1}-2)(4^{2k+1}-3^{2k+1})}{144^k\,k\,(2k+1)}zeta(2k)
]
(Largemathbf{Solution:})
Within the interval (displaystyle 0 < x < pi/2\,), the logtangent function has the series representation
[ln( an x)=ln x -sum_{k=1}^{infty}(-1)^kfrac{2^{2k}(2^{2k-1}-1)B_{2k}}{k\,(2k)!}x^{2k}
]
So, let's integrate this series over a suitable interval, such as:
[egin{align*}
&int_{pi/4}^{pi/3}ln( an x)\,mathrm{d}x=int_{pi/4}^{pi/3}ln x\,mathrm{d}x- sum_{k=1}^{infty}(-1)^kfrac{2^{2k}(2^{2k-1}-1)B_{2k}}{k\,(2k)!}int_{pi/4}^{pi/3}x^{2k} \,mathrm{d}x\
&=left(xln x-x
ight)Biggr|_{pi/4}^{pi/3}- sum_{k=1}^{infty}(-1)^kfrac{2^{2k}(2^{2k-1}-1)B_{2k}}{k\,(2k)!}left(frac{x^{2k+1}}{2k+1 }
ight)Biggr|_{pi/4}^{pi/3}\
&=frac{pi}{3}lnfrac{pi}{3}-frac{pi}{4}lnfrac{pi}{4}-frac{pi}{12} -sum_{k=1}^{infty}(-1)^kfrac{2^{2k}(2^{2k-1}-1)B_{2k}pi^{2k+1}}{k\,(2k+1)\,(2k)!}left(frac{1}{3^{2k+1}}-frac{1}{4^{2k+1}}
ight)\
&=pilnleft(frac{sqrt{2}\,pi^{1/12}}{3^{1/3}}
ight)-frac{pi}{12} -sum_{k=1}^{infty}(-1)^kfrac{2^{2k}(2^{2k-1}-1)B_{2k}pi^{2k+1}}{k\,(2k+1)\,(2k)!}left(frac{4^{2k+1}-3^{2k+1}}{12^{2k+1}}
ight)\
&=pilnleft(frac{sqrt{2}\,pi^{1/12}}{3^{1/3}\,e^{1/12}}
ight) -sum_{k=1}^{infty}(-1)^kfrac{2^{2k}(2^{2k-1}-1)B_{2k}pi^{2k+1}}{k\,(2k+1)\,(2k)!}left(frac{4^{2k+1}-3^{2k+1}}{12^{2k+1}}
ight)
end{align*}]
Next, we use the classic Zeta function result:
[zeta(2n)=(-1)^{n+1}\,frac{B_{2n}\,pi^{2n}2^{2n-1}}{(2n)!}
]
In the re-arranged form:
[B_{2n}= (-1)^{n+1}frac{(2n)!}{pi^{2n}\,2^{2n-1}}zeta(2n)
]
to obtain
[pilnleft(frac{sqrt{2}\,pi^{1/12}}{3^{1/3}\,e^{1/12}}
ight) +frac{pi}{6}\,sum_{k=1}^{infty}frac{(2^{2k-1}-2)(4^{2k+1}-3^{2k+1})}{144^k\,k\,(2k+1)}zeta(2k)
]
The logtangent integral is also solvable in terms of the extbf{Clausen Function}:
[int_0^{phi}ln( an x)\,mathrm{d}x=-frac{1}{2} ext{Cl}_2(2phi)-frac{1}{2} ext{Cl}_2(pi-2phi)
]
So the previous integral is equivalent to
[int_{pi/4}^{pi/3}ln( an x)mathrm{d}x=-frac{1}{2} ext{Cl}_2left(frac{2pi}{3}
ight)-frac{1}{2} ext{Cl}_2left(frac{pi}{3}
ight)+ ext{Cl}_2left(frac{pi}{2}
ight)= mathbf{G}-frac{5}{6} ext{Cl}_2left(frac{pi}{3}
ight)
]
Since
[ ext{Cl}_2left(frac{2pi}{3}
ight)=frac{2}{3} ext{Cl}_2left(frac{pi}{3}
ight)~,~ ext{Cl}_2left(frac{pi}{2}
ight)=mathbf{G}
]
Equating the two different evaluations of the logtangent integral, we get the final result:
[Largeoxed{displaystyle sum_{k=1}^{infty}frac{(2^{2k-1}-2)(4^{2k+1}-3^{2k+1})}{144^k\,k\,(2k+1)}zeta(2k)=color{Blue} {frac{6mathbf{G}}{pi}-frac{5}{pi} ext{Cl}_2left(frac{pi}{3}
ight)+ 6lnleft(frac{3^{1/3}\,e^{1/12}}{sqrt{2}\,pi^{1/12}}
ight)}}
]
If expressed in terms of the Eta function (alternating Zeta function), since the logtangent series then reduces to:
[ln( an x) = ln x + sum_{k=1}^{infty}frac{2^{2k}\,eta(2k)}{pi^{2k}\,k}x^{2k}
]
So, for (displaystyle 0 < p < q <1/2\,) and (displaystyle p\, ,qin mathbb{Q}\,), we have
[egin{align*}
int_{ppi}^{qpi}ln( an x)\,mathrm{d}x&=frac{1}{2}left[ ext{Cl}_2(2ppi)+ ext{Cl}_2(pi-2ppi)- ext{Cl}_2(2qpi)- ext{Cl}_2(pi-2qpi)
ight]\
&=left(xln x-x
ight)Biggr|_{ppi}^{qpi}+pi sum_{k=1}^{infty}frac{2^{2k}(q^{2k+1}-p^{2k+1})}{k\,(2k+1)}eta(2k)
end{align*}]
Or
[egin{align*}
&sum_{k=1}^{infty}frac{2^{2k}(q^{2k+1}-p^{2k+1})}{k\,(2k+1)}eta(2k)\
&=frac{1}{2pi}left[ ext{Cl}_2(2ppi)+ ext{Cl}_2(pi-2ppi)- ext{Cl}_2(2qpi)- ext{Cl}_2(pi-2qpi)
ight]-frac{1}{pi}left(xln x-x
ight)Biggr|_{ppi}^{qpi}
end{align*}]
setting (p=1/4) and (q=3/10) gives the series:
[largecolor{DarkGreen} {sum_{k=1}^{infty}frac{(6^{2k+1}-5^{2k+1})}{100^k\,k\,(2k+1)}eta(2k)=frac{20 mathbf{G}}{pi}-frac{10}{pi}left[ ext{Cl}_2left(frac{2pi}{5}
ight)+ ext{Cl}_2left(frac{3pi}{5}
ight)
ight]+20lnleft(frac{5^{3/10}\,e^{1/20}}{2^{1/5}\,3^{3/10}\,pi^{1/20}}
ight)}
]