Question Number 196406 by qaz last updated on 24/Aug/23
$$\underset{{n},{m}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}+{m}} {nm}}{\left({n}+{m}\right)^{\mathrm{2}} }=? \\ $$
Answered by witcher3 last updated on 26/Aug/23
![=−Σ_(n,m) ∫_0 ^1 ∫_0 ^1 m(−x)^(m+n−1) .ny^((n+m−1)) dxdy =Σ_(n,m) ∫_0 ^1 ∫_0 ^1 m(−xy)^m .n(−xy)^(n−1) dxdy Σ_(m≥1) m(−a)^m =(a/((1+a)^2 )) =∫_0 ^1 ∫_0 ^1 (((xy))/((1+xy)^4 )).dxdy xy=u⇒ydx=du =∫_0 ^1 (1/y)∫_0 ^y (u/((1+u)^4 ))du =∫_0 ^1 (1/y).[−(1/(2(1+u)^2 ))+(1/(3(1+u)^3 ))]_0 ^y =∫_0 ^1 (1/y)[−(1/(2(1+y)^2 ))+(1/(3(1+y)^3 ))+(1/6)]dy =∫_0 ^1 (((1+y)^3 +2−3(1+y))/(6(1+y)^3 y)) =∫_0 ^1 ((y^2 +3y)/(6(1+y)^3 ))=∫_0 ^1 (((y+1)^2 +(y+1)−2)/(6(1+y)^3 )) =(1/6)ln(2)](https://www.tinkutara.com/question/Q196494.png)
$$=−\underset{\mathrm{n},\mathrm{m}} {\sum}\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{m}\left(−\mathrm{x}\right)^{\mathrm{m}+\mathrm{n}−\mathrm{1}} .\mathrm{ny}^{\left(\mathrm{n}+\mathrm{m}−\mathrm{1}\right)} \mathrm{dxdy} \\ $$$$=\underset{\mathrm{n},\mathrm{m}} {\sum}\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{m}\left(−\mathrm{xy}\right)^{\mathrm{m}} .\mathrm{n}\left(−\mathrm{xy}\right)^{\mathrm{n}−\mathrm{1}} \mathrm{dxdy} \\ $$$$\underset{\mathrm{m}\geqslant\mathrm{1}} {\sum}\mathrm{m}\left(−\mathrm{a}\right)^{\mathrm{m}} =\frac{\mathrm{a}}{\left(\mathrm{1}+\mathrm{a}\right)^{\mathrm{2}} } \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{xy}\right)}{\left(\mathrm{1}+\mathrm{xy}\right)^{\mathrm{4}} }.\mathrm{dxdy} \\ $$$$\mathrm{xy}=\mathrm{u}\Rightarrow\mathrm{ydx}=\mathrm{du} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{y}}\int_{\mathrm{0}} ^{\mathrm{y}} \frac{\mathrm{u}}{\left(\mathrm{1}+\mathrm{u}\right)^{\mathrm{4}} }\mathrm{du} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{y}}.\left[−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}+\mathrm{u}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}\left(\mathrm{1}+\mathrm{u}\right)^{\mathrm{3}} }\right]_{\mathrm{0}} ^{\mathrm{y}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{y}}\left[−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}+\mathrm{y}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}\left(\mathrm{1}+\mathrm{y}\right)^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{6}}\right]\mathrm{dy} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{1}+\mathrm{y}\right)^{\mathrm{3}} +\mathrm{2}−\mathrm{3}\left(\mathrm{1}+\mathrm{y}\right)}{\mathrm{6}\left(\mathrm{1}+\mathrm{y}\right)^{\mathrm{3}} \mathrm{y}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{y}^{\mathrm{2}} +\mathrm{3y}}{\mathrm{6}\left(\mathrm{1}+\mathrm{y}\right)^{\mathrm{3}} }=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{y}+\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{y}+\mathrm{1}\right)−\mathrm{2}}{\mathrm{6}\left(\mathrm{1}+\mathrm{y}\right)^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\mathrm{ln}\left(\mathrm{2}\right) \\ $$