Question Number 61661 by maxmathsup by imad last updated on 05/Jun/19
$$\left.\mathrm{1}\right)\:{calculate}\:\int\int_{{R}^{+^{\mathrm{2}} } } \:\:\:\:\:\frac{{dxdy}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{y}^{\mathrm{2}} \right)} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{ln}\left({x}\right)}{{x}^{\mathrm{2}} −\mathrm{1}}\:{dx}\:. \\ $$
Commented by maxmathsup by imad last updated on 07/Jun/19
![1) ∫∫_R^+^2 ((dxdy)/((1+x^2 )(1+y^2 ))) =∫_0 ^∞ (dx/(1+x^2 )) .∫_0 ^∞ (dy/(1+y^2 )) =(π/2).(π/2) =(π^2 /4) 2) let A =∫_0 ^∞ ((ln(x))/(x^2 −1)) ⇒−A =∫_0 ^∞ ((ln(x))/(1−x^2 ))dx =∫_0 ^1 ((ln(x))/(1−x^2 ))dx +∫_1 ^(+∞) ((ln(x))/(1−x^2 )) dx ∫_0 ^1 ((ln(x))/(1−x^2 ))dx =∫_0 ^1 lnx(Σ_(n=0) ^∞ x^(2n) )dx =Σ_(n=0) ^∞ ∫_0 ^1 x^(2n) ln(x)dx by parts ∫_0 ^1 x^(2n) ln(x)dx =[(1/(2n+1))x^(2n+1) ln(x)]_0 ^1 −∫_0 ^1 (1/((2n+1)))x^(2n) dx =−(1/((2n+1)^2 )) ⇒∫_0 ^1 ((ln(x))/(1−x^2 ))dx =−Σ_(n=0) ^∞ (1/((2n+1)^2 )) Σ_(n=1) ^∞ (1/n^2 ) =(1/4) Σ_(n=1) ^∞ (1/n^2 ) +Σ_(n=0) ^∞ (1/((2n+1)^2 )) ⇒Σ_(n=0) ^∞ (1/((2n+1)^2 )) =(3/4)(π^2 /6) =(π^2 /8) ⇒ ∫_0 ^1 ((ln(x))/(1−x^2 ))dx =−(π^2 /8) ∫_1 ^(+∞) ((ln(x))/(1−x^2 )) dx =_(x =(1/t)) − ∫_0 ^1 ((−ln(t))/(1−(1/t^2 ))) (−(dt/t^2 )) =−∫_0 ^1 ((ln(t))/(t^2 −1))dt =∫_0 ^1 ((ln(t))/(1−t^2 )) dt =−(π^2 /8) ⇒ −A =−(π^2 /8)−(π^2 /8) ⇒A =(π^2 /4) .](https://www.tinkutara.com/question/Q61725.png)
$$\left.\mathrm{1}\right)\:\int\int_{{R}^{+^{\mathrm{2}} } } \:\:\:\:\frac{{dxdy}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{y}^{\mathrm{2}} \right)}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }\:.\int_{\mathrm{0}} ^{\infty} \:\frac{{dy}}{\mathrm{1}+{y}^{\mathrm{2}} }\:=\frac{\pi}{\mathrm{2}}.\frac{\pi}{\mathrm{2}}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\left.\mathrm{2}\right)\:{let}\:{A}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{ln}\left({x}\right)}{{x}^{\mathrm{2}} −\mathrm{1}}\:\Rightarrow−{A}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{ln}\left({x}\right)}{\mathrm{1}−{x}^{\mathrm{2}} }{dx}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left({x}\right)}{\mathrm{1}−{x}^{\mathrm{2}} }{dx}\:+\int_{\mathrm{1}} ^{+\infty} \:\frac{{ln}\left({x}\right)}{\mathrm{1}−{x}^{\mathrm{2}} }\:{dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left({x}\right)}{\mathrm{1}−{x}^{\mathrm{2}} }{dx}\:=\int_{\mathrm{0}} ^{\mathrm{1}} {lnx}\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:{x}^{\mathrm{2}{n}} \right){dx}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{\mathrm{2}{n}} {ln}\left({x}\right){dx}\:\:\:\:{by}\:{parts} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{\mathrm{2}{n}} {ln}\left({x}\right){dx}\:=\left[\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}{x}^{\mathrm{2}{n}+\mathrm{1}} {ln}\left({x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)}{x}^{\mathrm{2}{n}} \:{dx} \\ $$$$=−\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{ln}\left({x}\right)}{\mathrm{1}−{x}^{\mathrm{2}} }{dx}\:=−\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{4}}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:+\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:=\frac{\mathrm{3}}{\mathrm{4}}\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left({x}\right)}{\mathrm{1}−{x}^{\mathrm{2}} }{dx}\:=−\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$$$\int_{\mathrm{1}} ^{+\infty} \:\frac{{ln}\left({x}\right)}{\mathrm{1}−{x}^{\mathrm{2}} }\:{dx}\:=_{{x}\:=\frac{\mathrm{1}}{{t}}} \:\:\:−\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{−{ln}\left({t}\right)}{\mathrm{1}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}\:\left(−\frac{{dt}}{{t}^{\mathrm{2}} }\right)\:=−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left({t}\right)}{{t}^{\mathrm{2}} −\mathrm{1}}{dt}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left({t}\right)}{\mathrm{1}−{t}^{\mathrm{2}} }\:{dt}\:=−\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:\Rightarrow \\ $$$$−{A}\:=−\frac{\pi^{\mathrm{2}} }{\mathrm{8}}−\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:\Rightarrow{A}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\:. \\ $$