n-1-1-x-3-n-3-

Question Number 143790 by Dwaipayan Shikari last updated on 18/Jun/21

\underset{n = 1}{\prod^{\infty}} (1 + \frac{x^{3}}{n^{3}})

Commented by Dwaipayan Shikari last updated on 18/Jun/21

I t r i e d

y = \underset{n = 1}{\prod^{\infty}} (1 + \frac{x^{3}}{n^{3}})

\frac{y^{'}}{y} = 3 \underset{n = 1}{\sum^{\infty}} \frac{x^{2}}{(n^{3} + x^{3})}

\frac{y^{'}}{y} = 3 x^{2} Σ \frac{1}{(n - α x) (n - β x) (n - γ x)}

\frac{y^{'}}{y} = \frac{3 x}{β - γ} Σ \frac{1}{(n - α x)} (\frac{1}{(n - β x)} - \frac{1}{(n - γ x)})

\frac{y^{'}}{y} = \frac{3}{(β - γ) (α - β)} {Σ \frac{1}{(n - α x)} - \frac{1}{(n - β x)}} - \frac{3}{(β - γ) (α - γ)} {Σ \frac{1}{(n - α x)} - \frac{1}{(n - γ x)}}

= m (ψ (1 - β x) - ψ (1 - α x)) - n (ψ (1 - γ x) - ψ (1 - α x))

\frac{y^{'}}{y} = m \frac{Γ^{'} (1 - β x)}{Γ (1 - β x)} - m \frac{Γ^{'} (1 - α x)}{Γ (1 - α x)} - n \frac{Γ^{'} (1 - γ x)}{Γ (1 - γ x)} + n \frac{Γ^{'} (1 - α x)}{Γ (1 - α x)}

y = \frac{Γ {(1 - β x)}^{m} Γ {(1 - α x)}^{n}}{Γ {(1 - α x)}^{m} Γ {(1 - γ x)}^{n}}

Commented by Dwaipayan Shikari last updated on 18/Jun/21

α = 1 β = e^{2 i π / 3} γ = e^{- 2 π i / 3}