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

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

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

$$Itried$$$$y=\underset{n=1}{\prod ^{\mathrm{\infty }}}(1+\frac{x^3}{n^3})$$$$\frac{y^{\prime }}{y}=3\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{x^2}{(n^3+x^3)}$$$$\frac{y^{\prime }}{y}=3x^2\mathrm{\Sigma }\frac{1}{(n-\alpha x)(n-\beta x)(n-\gamma x)}$$$$\frac{y^{\prime }}{y}=\frac{3x}{\beta -\gamma }\mathrm{\Sigma }\frac{1}{(n-\alpha x)}(\frac{1}{(n-\beta x)}-\frac{1}{(n-\gamma x)})$$$$\frac{y^{\prime }}{y}=\frac{3}{(\beta -\gamma )(\alpha -\beta )}\{\mathrm{\Sigma }\frac{1}{(n-\alpha x)}-\frac{1}{(n-\beta x)}\}-\frac{3}{(\beta -\gamma )(\alpha -\gamma )}\{\mathrm{\Sigma }\frac{1}{(n-\alpha x)}-\frac{1}{(n-\gamma x)}\}$$$$=m(\psi (1-\beta x)-\psi (1-\alpha x))-n(\psi (1-\gamma x)-\psi (1-\alpha x))$$$$\frac{y^{\prime }}{y}=m\frac{{\mathrm{\Gamma }}^{\prime }(1-\beta x)}{\mathrm{\Gamma }(1-\beta x)}-m\frac{{\mathrm{\Gamma }}^{\prime }(1-\alpha x)}{\mathrm{\Gamma }(1-\alpha x)}-n\frac{{\mathrm{\Gamma }}^{\prime }(1-\gamma x)}{\mathrm{\Gamma }(1-\gamma x)}+n\frac{{\mathrm{\Gamma }}^{\prime }(1-\alpha x)}{\mathrm{\Gamma }(1-\alpha x)}$$$$y=\frac{\mathrm{\Gamma }{(1-\beta x)}^m\mathrm{\Gamma }{(1-\alpha x)}^n}{\mathrm{\Gamma }{(1-\alpha x)}^m\mathrm{\Gamma }{(1-\gamma x)}^n}$$

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

$$\alpha =1\beta =e^{2i\pi /3}\gamma =e^{-2\pi i/3}$$

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