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




Question Number 143790 by Dwaipayan Shikari last updated on 18/Jun/21
Π_(n=1) ^∞ (1+(x^3 /n^3 ))
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}+\frac{{x}^{\mathrm{3}} }{{n}^{\mathrm{3}} }\right) \\ $$
Commented by Dwaipayan Shikari last updated on 18/Jun/21
I tried   y=Π_(n=1) ^∞ (1+(x^3 /n^3 ))  ((y′)/y)=3Σ_(n=1) ^∞ (x^2 /((n^3 +x^3 )))  ((y′)/y)=3x^2 Σ(1/((n−αx)(n−βx)(n−γx)))  ((y′)/y)=((3x)/(β−γ))Σ(1/((n−αx)))((1/((n−βx)))−(1/((n−γx))))  ((y′)/y)=(3/((β−γ)(α−β))){Σ(1/((n−αx)))−(1/((n−βx)))}−(3/((β−γ)(α−γ))){Σ(1/((n−αx)))−(1/((n−γx)))}  =m(ψ(1−βx)−ψ(1−αx))−n(ψ(1−γx)−ψ(1−αx))  ((y′)/y)=m((Γ′(1−βx))/(Γ(1−βx)))−m((Γ′(1−αx))/(Γ(1−αx)))−n((Γ′(1−γx))/(Γ(1−γx)))+n((Γ′(1−αx))/(Γ(1−αx)))  y=((Γ(1−βx)^m Γ(1−αx)^n )/(Γ(1−αx)^m Γ(1−γx)^n ))
$${I}\:{tried}\: \\ $$$${y}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}+\frac{{x}^{\mathrm{3}} }{{n}^{\mathrm{3}} }\right) \\ $$$$\frac{{y}'}{{y}}=\mathrm{3}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{\mathrm{2}} }{\left({n}^{\mathrm{3}} +{x}^{\mathrm{3}} \right)} \\ $$$$\frac{{y}'}{{y}}=\mathrm{3}{x}^{\mathrm{2}} \Sigma\frac{\mathrm{1}}{\left({n}−\alpha{x}\right)\left({n}−\beta{x}\right)\left({n}−\gamma{x}\right)} \\ $$$$\frac{{y}'}{{y}}=\frac{\mathrm{3}{x}}{\beta−\gamma}\Sigma\frac{\mathrm{1}}{\left({n}−\alpha{x}\right)}\left(\frac{\mathrm{1}}{\left({n}−\beta{x}\right)}−\frac{\mathrm{1}}{\left({n}−\gamma{x}\right)}\right) \\ $$$$\frac{{y}'}{{y}}=\frac{\mathrm{3}}{\left(\beta−\gamma\right)\left(\alpha−\beta\right)}\left\{\Sigma\frac{\mathrm{1}}{\left({n}−\alpha{x}\right)}−\frac{\mathrm{1}}{\left({n}−\beta{x}\right)}\right\}−\frac{\mathrm{3}}{\left(\beta−\gamma\right)\left(\alpha−\gamma\right)}\left\{\Sigma\frac{\mathrm{1}}{\left({n}−\alpha{x}\right)}−\frac{\mathrm{1}}{\left({n}−\gamma{x}\right)}\right\} \\ $$$$={m}\left(\psi\left(\mathrm{1}−\beta{x}\right)−\psi\left(\mathrm{1}−\alpha{x}\right)\right)−{n}\left(\psi\left(\mathrm{1}−\gamma{x}\right)−\psi\left(\mathrm{1}−\alpha{x}\right)\right) \\ $$$$\frac{{y}'}{{y}}={m}\frac{\Gamma'\left(\mathrm{1}−\beta{x}\right)}{\Gamma\left(\mathrm{1}−\beta{x}\right)}−{m}\frac{\Gamma'\left(\mathrm{1}−\alpha{x}\right)}{\Gamma\left(\mathrm{1}−\alpha{x}\right)}−{n}\frac{\Gamma'\left(\mathrm{1}−\gamma{x}\right)}{\Gamma\left(\mathrm{1}−\gamma{x}\right)}+{n}\frac{\Gamma'\left(\mathrm{1}−\alpha{x}\right)}{\Gamma\left(\mathrm{1}−\alpha{x}\right)} \\ $$$${y}=\frac{\Gamma\left(\mathrm{1}−\beta{x}\right)^{{m}} \Gamma\left(\mathrm{1}−\alpha{x}\right)^{{n}} }{\Gamma\left(\mathrm{1}−\alpha{x}\right)^{{m}} \Gamma\left(\mathrm{1}−\gamma{x}\right)^{{n}} } \\ $$
Commented by Dwaipayan Shikari last updated on 18/Jun/21
α=1  β=e^(2iπ/3)   γ=e^(−2πi/3)
$$\alpha=\mathrm{1}\:\:\beta={e}^{\mathrm{2}{i}\pi/\mathrm{3}} \:\:\gamma={e}^{−\mathrm{2}\pi{i}/\mathrm{3}} \\ $$

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