Question Number 114541 by mnjuly1970 last updated on 19/Sep/20
Answered by abdomsup last updated on 19/Sep/20
![A=∫_0 ^1 ((x^2 ln(x))/(1−x^2 ))dx ⇒ A =∫_0 ^1 x^2 ln(x)Σ_(n=0) ^∞ x^(2n) dx =Σ_(n=0) ^∞ ∫_0 ^1 x^(2n+2) ln(x)dx u_n =∫_0 ^(1 ) x^(2n+2) ln(x)dx =[(x^(2n+3) /(2n+3))ln(x)]_0 ^1 −∫_0 ^1 (x^(2n+2) /(2n+3))dx =−(1/((2n+3)^2 ))[x^(2n+3) ]_0 ^1 =−(1/((2n+3)^2 )) ⇒A =−Σ_(n=0) ^(∞ ) (1/((2n+3)^2 )) =_(n=p−1) −Σ_(p=1) ^(∞ ) (1/((2p+1)^2 )) we have Σ_(p=1) ^∞ (1/p^2 ) =(1/4)Σ_(p=1) ^∞ (1/p^2 ) +Σ_(p=0) ^∞ (1/((2p+1)^2 )) ⇒ Σ_(p=0) ^∞ (1/((2p+1)^2 )) =(3/4)×(π^2 /6) =(π^2 /8) ⇒Σ_(p=1) ^∞ (1/((2p+1)^2 )) =(π^2 /8)−1 ⇒ A =1−(π^2 /8)](Q114545.png)
$${A}=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{x}^{\mathrm{2}} {ln}\left({x}\right)}{\mathrm{1}−{x}^{\mathrm{2}} }{dx}\:\Rightarrow \\ $$$${A}\:=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}} {ln}\left({x}\right)\sum_{{n}=\mathrm{0}} ^{\infty} \:{x}^{\mathrm{2}{n}} {dx} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{\mathrm{2}{n}+\mathrm{2}} \:{ln}\left({x}\right){dx} \\ $$$${u}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}\:} \:{x}^{\mathrm{2}{n}+\mathrm{2}} {ln}\left({x}\right){dx} \\ $$$$=\left[\frac{{x}^{\mathrm{2}{n}+\mathrm{3}} }{\mathrm{2}{n}+\mathrm{3}}{ln}\left({x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{x}^{\mathrm{2}{n}+\mathrm{2}} }{\mathrm{2}{n}+\mathrm{3}}{dx} \\ $$$$=−\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{3}\right)^{\mathrm{2}} }\left[{x}^{\mathrm{2}{n}+\mathrm{3}} \right]_{\mathrm{0}} ^{\mathrm{1}} \:=−\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow{A}\:=−\sum_{{n}=\mathrm{0}} ^{\infty\:} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$=_{{n}={p}−\mathrm{1}} \:−\sum_{{p}=\mathrm{1}} ^{\infty\:} \:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${we}\:{have}\:\sum_{{p}=\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{{p}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{4}}\sum_{{p}=\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{{p}^{\mathrm{2}} } \\ $$$$+\sum_{{p}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\sum_{{p}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{2}} }\:=\frac{\mathrm{3}}{\mathrm{4}}×\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$$$\Rightarrow\sum_{{p}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{2}} }\:=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}−\mathrm{1}\:\Rightarrow \\ $$$${A}\:=\mathrm{1}−\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 19/Sep/20
$${thank}\:{you}\:.{very}\:{nice}.. \\ $$
Commented by mathmax by abdo last updated on 19/Sep/20
$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome} \\ $$