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Question Number 139101 by mnjuly1970 last updated on 22/Apr/21

$$.......nicecalculus.....$$$$\mathit{\varphi }\stackrel{???}{=}{\int }_0^1{\int }_0^1\frac{1-x}{1-xy}{(-ln\left(xy\right))}^{2019}dxdy$$$$.........$$

Answered by Dwaipayan Shikari last updated on 22/Apr/21

$$\vartheta \left(\alpha \right)={\int }_0^1{\int }_0^1\frac{1-x}{1-xy}{\left(xy\right)}^{\alpha }dxdy$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}{\int }_0^1{\int }_0^1{\left(xy\right)}^{n+\alpha }-x^{n+\alpha +1}{y}^{n+\alpha }dxdy$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}{\int }_0^1\frac{y^{n+\alpha }}{n+\alpha +1}-\frac{y^{n+\alpha }}{(n+\alpha +2)}dy$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{(n+\alpha +1)}^2}-\frac{1}{(n+\alpha +1)(n+\alpha +2)}$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{(n+\alpha +1)}^2}-\psi (\alpha +1)+\psi (\alpha +2)={\psi }^{\prime }(\alpha +1)-\psi (\alpha +1)+\psi (\alpha +2)$$$${\vartheta }^{2019}\left(\alpha \right)=2020!{(-1)}^{2019}{\psi }^{2021}(\alpha +1)-2019!{\psi }^{2020}(\alpha +1)+2019!{\psi }^{2020}(\alpha +2)$$$$-{\vartheta }^{2019}\left(0\right)=2020!\zeta \left(2020\right)+2019!\zeta \left(2020\right)-2019!(\zeta \left(2021\right)-1)$$

Commented by mnjuly1970 last updated on 22/Apr/21

$$thanksalot$$

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