Question Number 28200 by abdo imad last updated on 21/Jan/18
![let give I= ∫_0 ^1 ((ln(1+x))/(1+x^2 ))dx and J=∫∫_([0,1]^2 ) (x/((1+x^2 )(1+xy)))dxdy calculate J by two methods then find the value of I.](https://www.tinkutara.com/question/Q28200.png)
$${let}\:{give}\:{I}=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{ln}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:\:{and}\:{J}=\int\int_{\left[\mathrm{0},\mathrm{1}\right]^{\mathrm{2}} } \:\:\:\frac{{x}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{xy}\right)}{dxdy} \\ $$$${calculate}\:{J}\:{by}\:{two}\:{methods}\:{then}\:{find}\:{the}\:{value}\:{of}\:{I}. \\ $$
Commented by abdo imad last updated on 22/Jan/18
![we have J= ∫_0 ^1 (∫_0 ^1 (x/(1+xy))dy)(dx/(1+x^2 )) but ∫_0 ^1 (x/(1+xy))dy= [ln(1+xy)]_(y=0) ^(y=1) =ln (1+x) so J= ∫_0 ^1 ((ln(1+x))/(1+x^2 ))dx from another side J= ∫_0 ^1 ( ∫_0 ^1 (x/((1+x^2 )(1+xy)))dx)dy let decompose F(x)= (x/((1+x^2 )(1+xy)))= (α/(1+xy))+ ((ax+b)/(1+x^2 )) α= lim_(x→((−1)/y)) (1+xy)F(x)= ((−1)/(y( 1+(1/y^2 ))))= ((−1)/(y+(1/y)))= ((−y)/(1+y^2 )) lim_(x→+∝) xF(x)=0= (α/y) +a ⇒a= −(α/y) = (1/(1+y^2 )) F(x)= ((−y)/((1+y^2 )(1+xy))) + (((1/(1+y^2 ))x +b)/(1+x^2 )) F(0)=0 = ((−y)/(1+y^2 )) +b⇒ b=(y/(1+y^2 )) and F(x)= ((−y)/((1+y^2 )(1+xy))) + (1/(1+y^2 ()) ((x+y)/(1+x^2 ))) ∫_0 ^1 F(x)dx= ((−1)/(1+y^2 )) ∫_0 ^1 ((ydx)/(1+xy)) +(1/(2(1+y^2 ))) ∫_0 ^1 ((2x)/(1+x^2 ))dx +(y/(1+y^2 ))∫_0 ^1 (dx/(1+x^2 )) = ((−1)/(1+y^2 )) [ln(1+xy)]_(x=0) ^(x=1) +(1/(2(1+y^2 ))) [ln(1+x^2 )]_0 ^1 +(π/4) (y/(1+y^2 )) −((ln(1+y))/(1+y^2 )) +((ln2)/(2(1+y^2 ))) + ((πy)/(4(1+y^2 ))) J=−∫_0 ^1 ((ln(1+y))/(1+y^2 ))dy +((ln2)/2) ∫_0 ^1 (dy/(1+y^2 )) +(π/8) ∫_0 ^1 ((2ydy)/(1+y^2 )) 2J= (π/8)ln2 +(π/8) ln2= (π/4)ln2 ⇒J= (π/8)ln2 .](https://www.tinkutara.com/question/Q28246.png)
$${we}\:{have}\:\:{J}=\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{x}}{\mathrm{1}+{xy}}{dy}\right)\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }\:{but} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{x}}{\mathrm{1}+{xy}}{dy}=\:\left[{ln}\left(\mathrm{1}+{xy}\right)\right]_{{y}=\mathrm{0}} ^{{y}=\mathrm{1}} ={ln}\:\left(\mathrm{1}+{x}\right)\:{so} \\ $$$${J}=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{ln}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:\:\:{from}\:{another}\:{side} \\ $$$${J}=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\frac{{x}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{xy}\right)}{dx}\right){dy}\:{let}\:{decompose} \\ $$$${F}\left({x}\right)=\:\:\frac{{x}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{xy}\right)}=\:\:\:\frac{\alpha}{\mathrm{1}+{xy}}+\:\:\frac{{ax}+{b}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$\alpha=\:{lim}_{{x}\rightarrow\frac{−\mathrm{1}}{{y}}} \:\left(\mathrm{1}+{xy}\right){F}\left({x}\right)=\:\:\:\frac{−\mathrm{1}}{{y}\left(\:\mathrm{1}+\frac{\mathrm{1}}{{y}^{\mathrm{2}} }\right)}=\:\frac{−\mathrm{1}}{{y}+\frac{\mathrm{1}}{{y}}}=\:\frac{−{y}}{\mathrm{1}+{y}^{\mathrm{2}} } \\ $$$${lim}_{{x}\rightarrow+\propto} {xF}\left({x}\right)=\mathrm{0}=\:\frac{\alpha}{{y}}\:+{a}\:\Rightarrow{a}=\:−\frac{\alpha}{{y}}\:=\:\:\frac{\mathrm{1}}{\mathrm{1}+{y}^{\mathrm{2}} } \\ $$$${F}\left({x}\right)=\:\:\frac{−{y}}{\left(\mathrm{1}+{y}^{\mathrm{2}} \right)\left(\mathrm{1}+{xy}\right)}\:+\:\:\frac{\frac{\mathrm{1}}{\mathrm{1}+{y}^{\mathrm{2}} }{x}\:+{b}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$${F}\left(\mathrm{0}\right)=\mathrm{0}\:=\:\frac{−{y}}{\mathrm{1}+{y}^{\mathrm{2}} }\:+{b}\Rightarrow\:{b}=\frac{{y}}{\mathrm{1}+{y}^{\mathrm{2}} }\:\:{and} \\ $$$${F}\left({x}\right)=\:\frac{−{y}}{\left(\mathrm{1}+{y}^{\mathrm{2}} \right)\left(\mathrm{1}+{xy}\right)}\:+\:\frac{\mathrm{1}}{\mathrm{1}+{y}^{\mathrm{2}} \left(\right.}\:\frac{{x}+{y}}{\left.\mathrm{1}+{x}^{\mathrm{2}} \right)} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {F}\left({x}\right){dx}=\:\frac{−\mathrm{1}}{\mathrm{1}+{y}^{\mathrm{2}} }\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{ydx}}{\mathrm{1}+{xy}}\:\:+\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}+{y}^{\mathrm{2}} \right)}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:+\frac{{y}}{\mathrm{1}+{y}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$=\:\frac{−\mathrm{1}}{\mathrm{1}+{y}^{\mathrm{2}} }\:\left[{ln}\left(\mathrm{1}+{xy}\right)\right]_{{x}=\mathrm{0}} ^{{x}=\mathrm{1}} \:\:\:+\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}+{y}^{\mathrm{2}} \right)}\:\left[{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$+\frac{\pi}{\mathrm{4}}\:\frac{{y}}{\mathrm{1}+{y}^{\mathrm{2}} } \\ $$$$−\frac{{ln}\left(\mathrm{1}+{y}\right)}{\mathrm{1}+{y}^{\mathrm{2}} }\:\:\:+\frac{{ln}\mathrm{2}}{\mathrm{2}\left(\mathrm{1}+{y}^{\mathrm{2}} \right)}\:\:+\:\frac{\pi{y}}{\mathrm{4}\left(\mathrm{1}+{y}^{\mathrm{2}} \right)} \\ $$$${J}=−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{y}\right)}{\mathrm{1}+{y}^{\mathrm{2}} }{dy}\:\:+\frac{{ln}\mathrm{2}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dy}}{\mathrm{1}+{y}^{\mathrm{2}} }\:\:+\frac{\pi}{\mathrm{8}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{2}{ydy}}{\mathrm{1}+{y}^{\mathrm{2}} } \\ $$$$\mathrm{2}{J}=\:\frac{\pi}{\mathrm{8}}{ln}\mathrm{2}\:\:+\frac{\pi}{\mathrm{8}}\:{ln}\mathrm{2}=\:\frac{\pi}{\mathrm{4}}{ln}\mathrm{2}\:\Rightarrow{J}=\:\frac{\pi}{\mathrm{8}}{ln}\mathrm{2}\:\:. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$