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Question Number 128175 by bemath last updated on 05/Jan/21

$${\int }_0^1{\int }_0^1\frac{dxdy}{1-{xy}^3}?$$

Commented by liberty last updated on 06/Jan/21

$$\frac{1}{1-{\mathrm{xy}}^3}=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{\left({\mathrm{xy}}^3\right)}^{\mathrm{n}}$$$$\mathrm{I}=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{\int }_0^1{\int }_0^1{\left({\mathrm{xy}}^3\right)}^{\mathrm{n}}\mathrm{dy}\mathrm{dx}=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{\int }_0^1{\mathrm{x}}^{\mathrm{n}}\mathrm{dx}.{\int }_0^1{\mathrm{y}}^{3\mathrm{n}}\mathrm{dy}$$$$=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}{\mid }_0^1.\frac{{\mathrm{y}}^{3\mathrm{n}+1}}{3\mathrm{n}+1}{\mid }_0^1=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{1}{(\mathrm{n}+1)(3\mathrm{n}+1)}$$$$=\frac{3}{2}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}(\frac{1}{3\mathrm{n}+1}-\frac{1}{3\mathrm{n}+3})$$$$=\frac{3}{2}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{\int }_0^1({\mathrm{x}}^{3\mathrm{n}}-{\mathrm{x}}^{3\mathrm{n}+2})\mathrm{dx}$$$$=\frac{3}{2}{\int }_0^1\frac{1-{\mathrm{x}}^2}{1-{\mathrm{x}}^3}\mathrm{dx}=\frac{3}{2}{\int }_0^1\frac{1+\mathrm{x}}{1+\mathrm{x}+{\mathrm{x}}^2}\mathrm{dx}$$$$=\frac{3}{2}{\int }_0^1\frac{\mathrm{x}+\frac{1}{2}}{{(\mathrm{x}+\frac{1}{2})}^2+\frac{3}{4}}+\frac{3}{4}{\int }_0^1\frac{1}{{(\mathrm{x}+\frac{1}{2})}^2+\frac{3}{4}}\mathrm{dx}$$$$={[\frac{3}{2}.\frac{1}{2}\ln ({(\mathrm{x}+\frac{1}{2})}^2+\frac{3}{4})+\frac{3}{4}.\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{\mathrm{x}+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)]}_0^1$$$$=\frac{1}{12}(\pi \sqrt{3}+9\ln 3)$$

Answered by Dwaipayan Shikari last updated on 05/Jan/21

$${\int }_0^1{\int }_0^1\underset{n=0}{\sum ^{\mathrm{\infty }}}{\left({xy}^3\right)}^{n}dxdy$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}{\int }_0^1{y}^{3n}\frac{1}{n+1}dy$$$$=\frac{1}{2}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{(n+1)}-\frac{3}{3n+1}=\frac{1}{2}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{n+1}-\frac{1}{n+\frac{1}{3}}=\frac{1}{2}\psi \left(\frac{1}{3}\right)+\frac{\gamma }{2}$$

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