Question Number 78707 by abdomathmax last updated on 20/Jan/20
![let I =∫_0 ^1 ((ln(1+x))/(1+x^2 ))dx and J =∫∫_([0,1]^2 ) (x/((1+x^2 )(1+xy)))dxdy find J by two method and deduce the valueof I](https://www.tinkutara.com/question/Q78707.png)
$${let}\:{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:{and}\: \\ $$$${J}\:=\int\int_{\left[\mathrm{0},\mathrm{1}\right]^{\mathrm{2}} } \:\:\:\:\frac{{x}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{xy}\right)}{dxdy} \\ $$$${find}\:{J}\:{by}\:{two}\:{method}\:{and}\:{deduce}\:\:{the}\:{valueof}\:{I} \\ $$
Answered by mind is power last updated on 20/Jan/20
![J=∫_0 ^1 (1/(1+x^2 ))∫_0 ^1 (x/(1+xy))dydx =∫_0 ^1 (1/(1+x^2 ))[ln(1+xy)]_0 ^1 dx =∫_0 ^1 ((ln(1+x))/(1+x^2 ))=I By Fubini J=∫_0 ^1 ∫_0 ^1 (x/((1+x^2 )(1+xy)))dxdy (x/((1+x^2 )(1+xy)))=((−y)/(1+y^2 )).(1/(1+xy^ ))+((cx+d)/(1+x^2 )) d=(y/(1+y^2 )) (1/(1+y^2 ))=c⇒(x/((1+x^2 )(1+xy)))=(1/(1+y^2 ))(((−y)/(1+xy))+((x+y)/(1+x^2 ))) J=∫_0 ^1 ((1/(1+y^2 ))(∫_0 ^1 {((−y)/(1+xy))+((x+y)/(1+x^2 ))}dx)dy) =∫_0 ^1 ((1/(1+y^2 )){[_0 ^1 −ln(1+xy)+((ln(1+x^2 ))/2)+yarctan(x)]dy =∫_0 ^1 {((−ln(1+y))/(1+y^2 ))+((ln(2))/(2(1+y^2 )))+(y/(1+y^2 )).(π/4)}dy J=−∫_0 ^1 ((ln(1+y)dy)/(1+y^2 ))+((ln(2))/2)∫_0 ^1 (dy/(1+y^2 ))+(π/4)∫_0 ^1 (y/(1+y^2 ))dy=I by first ⇒−I+((ln(2))/2).(π/4)+(π/4)[((ln(1+y^2 ))/2)]_0 ^1 =I ⇒2I=((πln(2))/4)⇒I=((πln(2))/8)=πln(2^(1/8) )](https://www.tinkutara.com/question/Q78769.png)
$$\mathrm{J}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{x}}{\mathrm{1}+\mathrm{xy}}\mathrm{dydx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\left[\mathrm{ln}\left(\mathrm{1}+\mathrm{xy}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }=\mathrm{I} \\ $$$$\mathrm{By}\:\mathrm{Fubini} \\ $$$$\mathrm{J}=\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{x}}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{xy}\right)}\mathrm{dxdy} \\ $$$$ \\ $$$$\frac{\mathrm{x}}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{xy}\right)}=\frac{−\mathrm{y}}{\mathrm{1}+\mathrm{y}^{\mathrm{2}} }.\frac{\mathrm{1}}{\mathrm{1}+\mathrm{xy}^{} }+\frac{\mathrm{cx}+\mathrm{d}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} } \\ $$$$\mathrm{d}=\frac{\mathrm{y}}{\mathrm{1}+\mathrm{y}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+\mathrm{y}^{\mathrm{2}} }=\mathrm{c}\Rightarrow\frac{\mathrm{x}}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{xy}\right)}=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{y}^{\mathrm{2}} }\left(\frac{−\mathrm{y}}{\mathrm{1}+\mathrm{xy}}+\frac{\mathrm{x}+\mathrm{y}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\right) \\ $$$$\mathrm{J}=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{1}+\mathrm{y}^{\mathrm{2}} }\left(\int_{\mathrm{0}} ^{\mathrm{1}} \left\{\frac{−\mathrm{y}}{\mathrm{1}+\mathrm{xy}}+\frac{\mathrm{x}+\mathrm{y}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\right\}\mathrm{dx}\right)\mathrm{dy}\right) \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{1}+\mathrm{y}^{\mathrm{2}} }\left\{\left[_{\mathrm{0}} ^{\mathrm{1}} −\mathrm{ln}\left(\mathrm{1}+\mathrm{xy}\right)+\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{2}}+\mathrm{yarctan}\left(\mathrm{x}\right)\right]\mathrm{dy}\right.\right. \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left\{\frac{−\mathrm{ln}\left(\mathrm{1}+\mathrm{y}\right)}{\mathrm{1}+\mathrm{y}^{\mathrm{2}} }+\frac{\mathrm{ln}\left(\mathrm{2}\right)}{\mathrm{2}\left(\mathrm{1}+\mathrm{y}^{\mathrm{2}} \right)}+\frac{\mathrm{y}}{\mathrm{1}+\mathrm{y}^{\mathrm{2}} }.\frac{\pi}{\mathrm{4}}\right\}\mathrm{dy} \\ $$$$\mathrm{J}=−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{y}\right)\mathrm{dy}}{\mathrm{1}+\mathrm{y}^{\mathrm{2}} }+\frac{\mathrm{ln}\left(\mathrm{2}\right)}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{dy}}{\mathrm{1}+\mathrm{y}^{\mathrm{2}} }+\frac{\pi}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{y}}{\mathrm{1}+\mathrm{y}^{\mathrm{2}} }\mathrm{dy}=\mathrm{I}\:\mathrm{by}\:\mathrm{first} \\ $$$$\Rightarrow−\mathrm{I}+\frac{\mathrm{ln}\left(\mathrm{2}\right)}{\mathrm{2}}.\frac{\pi}{\mathrm{4}}+\frac{\pi}{\mathrm{4}}\left[\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{y}^{\mathrm{2}} \right)}{\mathrm{2}}\right]_{\mathrm{0}} ^{\mathrm{1}} =\mathrm{I} \\ $$$$\Rightarrow\mathrm{2I}=\frac{\pi\mathrm{ln}\left(\mathrm{2}\right)}{\mathrm{4}}\Rightarrow\mathrm{I}=\frac{\pi\mathrm{ln}\left(\mathrm{2}\right)}{\mathrm{8}}=\pi\mathrm{ln}\left(\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{8}}} \right) \\ $$$$ \\ $$