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Question Number 197906 by mnjuly1970 last updated on 03/Oct/23

$$$$$$\mathrm{S}=\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{\Gamma }}^2\left(k\right)}{k\mathrm{\Gamma }\left(2k\right)}=?$$$$----$$

Answered by Dwan last updated on 03/Oct/23

$$\frac{\mathrm{\Gamma }\left(k\right)\mathrm{\Gamma }\left(k\right)}{\mathrm{\Gamma }(k+k)}={\int }_0^1{x}^{k-1}{(1-x)}^{k-1}dx$$$$S={\int }_0^1\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{x^{k-1}{(1-x)}^{k-1}}{k}dx={\int }_0^1\frac{-ln(1-x+x^2)}{x-x^2}dx$$$$=-({\int }_0^1\frac{ln(1-x+x^2)}{1-x}dx+{\int }_0^1\frac{ln(1-x+x^2)}{x}dx)$$$$=-2{\int }_0^1\frac{ln(1-x+x^2)}{x}dx$$$$=2{\int }_0^1\frac{ln(1+x)}{x}dx-2{\int }_0^1\frac{ln(1+x^3)}{x}dx$$$$=2\eta \left(2\right)-\frac{2}{3}{\int }_0^1\frac{ln(1+u)}{u}du$$$$=\frac{2}{3}\eta \left(2\right)=\frac{2}{3}.\frac{{\pi }^2}{12}=\frac{{\pi }^2}{18}$$

Commented by mnjuly1970 last updated on 04/Oct/23

$$thanksalotsir$$$$soexcellentsolution...$$

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