Beauty of mathematical constants

Calculus Level 3

For x ( 0 , 1 ) x\in\mathbb(0,1) , if f ( x ) = n k x f(x)=n-k^x and g ( x ) = 1 + k x g(x)=1+k^x for n > 0 , k 0 n>0,k\geq 0 some consecutive integers, n > k n>k n = 1 ( 1 ) n 2 n 2 0 1 ( ln f ( x ) + 2 ln g ( x ) ) d x = η ( 2 ) ζ ( 2 ) ln ( 1 2 ζ ( 4 ) m 1 m 2 m 2 ) = η ( 2 ) ln ( 2 e γ π A 22 ) \sum_{n=1}^{\infty}\frac{(-1)^n}{2n^2}\int_0^1\left(\ln f(x)+2\ln g(x)\right)dx=-\frac{\eta(2)}{\zeta(2)}\ln\left(\frac{1}{2^{\zeta(4)}}\prod_{m\geq 1}\sqrt[m^2]{m^2}\right)=\eta(2)\ln\left(\frac{2e^{\gamma}\pi}{A^{22}}\right) Is the above closed form is correct?


where ζ ( . ) \zeta(.) is Riemann zeta function η ( . ) \eta(.) is Dirichlet eta funtion , A A is Glashier-kinkelin constant , γ \gamma is Euler-Mascheroni constant and e e is Euler-number


Inspiration .

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1 solution

Naren Bhandari
Jul 6, 2020

The correct closed form of it is η ( 2 ) ζ ( 2 ) ln ( 1 2 ζ ( 2 ) m 1 m m 2 ) = η ( 2 ) ln ( 4 e γ π A 12 ) -\frac{\eta(2)}{\zeta(2)}\ln\left(\frac{1}{2^{\zeta(2)}}\prod_{m\geq 1}\sqrt[m^2]{m}\right)=\eta(2)\ln\left(\frac{4e^{\gamma}\pi}{A^{12}}\right)

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