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Question Number 196406 by qaz last updated on 24/Aug/23

$$\underset{n,m=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n+m}nm}{{(n+m)}^2}=?$$

Answered by witcher3 last updated on 26/Aug/23

$$=-\sum _{\mathrm{n},\mathrm{m}}{\int }_0^1{\int }_0^1\mathrm{m}{(-\mathrm{x})}^{\mathrm{m}+\mathrm{n}-1}.{\mathrm{ny}}^{(\mathrm{n}+\mathrm{m}-1)}\mathrm{dxdy}$$$$=\sum _{\mathrm{n},\mathrm{m}}{\int }_0^1{\int }_0^1\mathrm{m}{(-\mathrm{xy})}^{\mathrm{m}}.\mathrm{n}{(-\mathrm{xy})}^{\mathrm{n}-1}\mathrm{dxdy}$$$$\sum _{\mathrm{m}⩾1}\mathrm{m}{(-\mathrm{a})}^{\mathrm{m}}=\frac{\mathrm{a}}{{(1+\mathrm{a})}^2}$$$$={\int }_0^1{\int }_0^1\frac{\left(\mathrm{xy}\right)}{{(1+\mathrm{xy})}^4}.\mathrm{dxdy}$$$$\mathrm{xy}=\mathrm{u}\Rightarrow \mathrm{ydx}=\mathrm{du}$$$$={\int }_0^1\frac{1}{\mathrm{y}}{\int }_0^{\mathrm{y}}\frac{\mathrm{u}}{{(1+\mathrm{u})}^4}\mathrm{du}$$$$={\int }_0^1\frac{1}{\mathrm{y}}.{[-\frac{1}{2{(1+\mathrm{u})}^2}+\frac{1}{3{(1+\mathrm{u})}^3}]}_0^{\mathrm{y}}$$$$={\int }_0^1\frac{1}{\mathrm{y}}[-\frac{1}{2{(1+\mathrm{y})}^2}+\frac{1}{3{(1+\mathrm{y})}^3}+\frac{1}{6}]\mathrm{dy}$$$$={\int }_0^1\frac{{(1+\mathrm{y})}^3+2-3(1+\mathrm{y})}{6{(1+\mathrm{y})}^3\mathrm{y}}$$$$={\int }_0^1\frac{{\mathrm{y}}^2+3\mathrm{y}}{6{(1+\mathrm{y})}^3}={\int }_0^1\frac{{(\mathrm{y}+1)}^2+(\mathrm{y}+1)-2}{6{(1+\mathrm{y})}^3}$$$$=\frac{1}{6}\ln \left(2\right)$$

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