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Question Number 159143 by mnjuly1970 last updated on 13/Nov/21

Answered by qaz last updated on 14/Nov/21

$$\mathrm{S}=\frac{1}{\mathrm{\Gamma }\left(4\right)}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{\mathrm{\Gamma }(4\mathrm{n}+1)\mathrm{\Gamma }\left(4\right)}{\mathrm{\Gamma }(4\mathrm{n}+5)}$$$$=\frac{1}{6}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{\int }_0^1{\mathrm{x}}^{4\mathrm{n}}{(1-\mathrm{x})}^3\mathrm{dx}$$$$=\frac{1}{6}{\int }_0^1\frac{{(1-\mathrm{x})}^3}{1-{\mathrm{x}}^4}\mathrm{dx}$$$$=\frac{1}{6}{\int }_0^1(\frac{2}{1+\mathrm{x}}-\frac{\mathrm{x}+1}{1+{\mathrm{x}}^2})\mathrm{dx}$$$$=\frac{1}{4}\ln 2-\frac{\pi }{24}$$

Commented by mnjuly1970 last updated on 14/Nov/21

$$bravomrqaz$$

Commented by ArielVyny last updated on 14/Nov/21

$$mrqazjepeuxavoirlaformegeneralde$$$$cegenre{d}^{\prime }expression?$$

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