Finding a solutions for an equation

analytic quantity idea – Trying to be taught a closed-figure for this integral referring to $zeta (3)$ retort

Hello pricey customer to our community We will proffer you an answer to this query analytic quantity idea – Trying to be taught a closed-figure for this integral referring to $zeta (3)$ ,and the retort will breathe typical by means of documented info sources, We welcome you and proffer you recent questions and solutions, Many customer are questioning concerning the retort to this query.

analytic quantity idea – Trying to be taught a closed-figure for this integral referring to $zeta (3)$

The following integral is expounded to a search of mine for a closed-figure to Apéry’s ceaseless $zeta (3)$.
$$int_0^infty frac {mistake left(frac {1} {4} arctan left(frac {4t} {3} capable) capable)}{left( e^{2 pi t} – 1 capable) sqrt[4]{t}} , dt$$

I maintain managed to array that with at the least one in all $zeta (1/4), zeta(3/4), beta (1/4), beta (3/4),$ and growing bid derivatives, one can be taught closed-forms for any at present unknown integer worth of the Riemann Zeta obligation or Dirichlet Beta obligation, akin to Apéry’s ceaseless or Catalan’s ceaseless, by exploiting a number of substitutions and reflections with their purposeful equations.

I maintain proven that $beta (1/4)$ is the same as the next integral:
$$frac {1} {6} + frac {1} {2sqrt[4]{3}} + int_0^infty frac {sqrt {2} mistake left(frac {1} {4} arctan left(4tproper) capable)} {(e^{2 pi t} – 1)sqrt[8]{frac{1} {16} + t^2}}, dt – int_0^infty frac {sqrt {2} mistake left(frac {1} {4} arctan left( frac{4t} {3} capable) capable)} {(e^{2 pi t} – 1)sqrt[8]{frac{9} {16} + t^2}}, dt$$

I maintain proven that $zeta (3)$ is the same as the next expression:
$$frac {3 sqrt{2}pi^3} {112} + alpha$$
Where $alpha$ is the next immense expression, continuing $frac {3 sqrt{2}pi^3} {112}$ within the following picture, stirring the regularised Hurwitz Zeta obligation.

I surmise that any potential future closed-figure for Apéry’s ceaseless will breathe of this determine.

Similarly, Catalan’s ceaseless is the next smaller expression.

I did this by exploiting the next relationship method I derived between the Riemann Zeta obligation and the Dirichlet Beta obligation that applies to constructive integers:
$$zeta (s) = frac {2^s} {2^s – 1} beta (s) + (-1)^s frac {2} {2^s (2^s – 1) Gamma (s))} psi^{(s-1)} left( frac {3} {4} capable)$$
Where $psi$ is the Polygamma obligation. An identical method can breathe limpid by utilizing $psi^{(s-1)} left( frac {1} {4} capable)$ as an alternative:
$$zeta (s) = -frac {2^s} {2^s – 1} beta (s) + (-1)^s frac {2} {2^s (2^s – 1) Gamma (s))} psi^{(s-1)} left( frac {1} {4} capable)$$

These formulae endure closed-forms stirring the Polygamma obligation, nonetheless, by exploiting the next identification and rearranging the purposeful equation for the Gamma obligation of both the Riemann Zeta obligation or Dirichlet Beta obligation for airplane or weird $s$ respectively,
$$psi^{(s-1)} (z) = frac {d^s} {dz^s} ln (Gamma (z))$$
one can endure expressions that assassinate not contain the Polygamma obligation as seen within the pictures.

we are going to proffer you the answer to analytic quantity idea – Trying to be taught a closed-figure for this integral referring to $zeta (3)$ query through our community which brings all of the solutions from a number of reliable sources.

Add comment