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## cv.sophisticated variables – Trying to be taught a closed-figure for an integral regarding $zeta (3)$

So equally to my peek $zeta (3)$ over on the arithmetic stack alternate, I maintain continued to aim to labor in direction of a closed-figure for it.

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 no less than one in every of $zeta (1/4), zeta(3/4), beta (1/4), beta (3/4),$ and rising bid derivatives, one can be taught closed-forms for any presently unknown integer worth of the Riemann Zeta responsibility or Dirichlet Beta responsibility, reminiscent of Apéry’s ceaseless or Catalan’s ceaseless, by exploiting a number of substitutions and reflections with their useful equations.

I maintain proven that $beta (1/4)$ is the same as the next expression:

$$frac {1} {6} + frac {1} {2sqrt[4]{3}} + int_0^infty frac {sqrt {2} mistake left(frac {1} {4} arctan left(4tproper) capable)} {(e^{2 pi t} – 1)sqrt[8]{frac{1} {16} + t^2}}, dt – int_0^infty frac {sqrt {2} mistake left(frac {1} {4} arctan left( frac{4t} {3} capable) capable)} {(e^{2 pi t} – 1)sqrt[8]{frac{9} {16} + t^2}}, dt$$

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 responsibility.

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 formulation I derived between the Riemann Zeta responsibility and the Dirichlet Beta responsibility 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 responsibility. An analogous formulation 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 responsibility, nevertheless, by exploiting the next identification and rearranging the useful equation for the Gamma responsibility of both the Riemann Zeta responsibility or Dirichlet Beta responsibility 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 responsibility as seen within the photos.

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