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نشان دهید: $$ \int _0^ \infty \int _0^ \infty \frac{Arccos( \frac{1- x^{2} }{1+ x^{3} } )+Arccos( \frac{1- y^{2} }{1+y^{3} } )}{(1+ x^{2} )(1+ y^{2} )( x^{2} + y^{2} )} dxdy= \pi C- \frac{ \pi ^{3} }{8} + \frac{3 \pi ^{2} }{4}ln(2) $$ C ثابت" کاتالان"

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$$I= \int _0^ \infty  \int _0^ \infty (Arccos ( \frac{1- x^{2} }{1+ x^{3} } )+Arccos ( \frac{1- y^{2} }{1+ y^{3} } ) \frac{1}{(1+ x^{2} )(1+ y^{2} )( x^{2} + y^{2} )} dxdy=2 \int _0^ \infty  \int _0^ \infty  \frac{Arccos( \frac{1- x^{2} }{1+ y^{2} } )}{(1+ x^{2} )(1+ y^{2} )( x^{2} + y^{2} )} dxdy= \frac{1}{2}  \int _0^ \infty  \int _0^ \infty  \frac{Arccos( \frac{1-x}{1-y} )}{ \sqrt{xy} (1+x)(1+y)(x+y)}dxdy= \frac{1}{2}  \int _0^ \infty  \frac{Arccos( \frac{1-x}{1+x} )}{ \sqrt{x} (1+x)} \underbrace{ \int _0^ \infty  \frac{dy}{ \sqrt{y} (1+y)(x+y)}dx= I_{2}  }  \Longrightarrow  I_{2}=  \int _0^ \infty  \frac{ y^{ -\frac{1}{2} } }{(1+y)(x+y)}= \frac{1}{1-x}  \int _0^ \infty  y^{ -\frac{1}{2} } ( \frac{1}{x+y} - \frac{1}{y+1} )dy= \frac{1}{1-x} ( \frac{1}{x} \int _0^ \infty  \frac{ y^{ -\frac{1}{2} } }{1+ \frac{y}{x} } dy- \beta ( \frac{1}{2} , \frac{1}{2} ) )= \frac{1}{1-x} ( \frac{1}{ \sqrt{x} }  \int _0^ \infty  \frac{ y^{ -\frac{1}{2} } }{1+y} dy- \pi )= \frac{ \pi }{1-x} ( \frac{1}{ \sqrt{x} } -1)= \frac{ \pi (1- \sqrt{x} )}{b} = \frac{ \pi }{(1+x) \sqrt{x} }$$$$I= \frac{ \pi }{2} \int _0^ \infty  \frac{Arccos( \frac{1-x}{1+x} )}{x(1+x)(1+ \sqrt{x} )} dx \Longrightarrow u=Arccos ( \frac{1-x}{1+x} ) \wedge du= \frac{dx}{ \sqrt{x} (1+x)}  \wedge dv= \frac{dx}{x(1+x)(1+ \sqrt{x} )}  \wedge v=-Arctan ( \sqrt{x} )-ln(1+ \sqrt{x} )+lnx- \frac{1}{2}ln(x+1) \Longrightarrow  \frac{ \pi }{2} [(uv | _0^ \infty - \int _0^ \infty  \frac{-Arctan( \sqrt{x})-ln(1+ \sqrt{x} ) +lnx- \frac{1}{2} ln(x+1)}{ \sqrt{x}(1+x) } dx]=- \frac{  \pi ^{3}}{4} + \frac{ \pi }{2}  \int _0^ \infty  \frac{Arctan( \sqrt{x} )+ln(1+ \sqrt{x} )-lnx+ \frac{1}{2} ln(x+1)}{ \sqrt{x}(1+x) } =- \frac{  \pi ^{3} }{4} + \frac{ \pi }{2} ( I_{2} + I_{3} - I_{4} + \frac{1}{2}  I_{5} ) $$$$I= \frac{ \pi }{2} \int _0^ \infty  \frac{Arccos( \frac{1-x}{1+x} )}{x(1+x)(1+ \sqrt{x} )} dx \Longrightarrow u=Arccos ( \frac{1-x}{1+x} ) \wedge du= \frac{dx}{ \sqrt{x} (1+x)}  \wedge dv= \frac{dx}{x(1+x)(1+ \sqrt{x} )}  \wedge v=-Arctan ( \sqrt{x} )-ln(1+ \sqrt{x} )+lnx- \frac{1}{2}ln(x+1) \Longrightarrow  \frac{ \pi }{2} [(uv | _0^ \infty - \int _0^ \infty  \frac{-Arctan( \sqrt{x})-ln(1+ \sqrt{x} ) +lnx- \frac{1}{2} ln(x+1)}{ \sqrt{x}(1+x) } dx]=- \frac{  \pi ^{3}}{4} + \frac{ \pi }{2}  \int _0^ \infty  \frac{Arctan( \sqrt{x} )+ln(1+ \sqrt{x} )-lnx+ \frac{1}{2} ln(x+1)}{ \sqrt{x}(1+x) } =- \frac{  \pi ^{3} }{4} + \frac{ \pi }{2} ( I_{2} + I_{3} - I_{4} + \frac{1}{2}  I_{5} )$$$$ \ast  I_{2}= \int _0^ \infty  \frac{Arctan( \sqrt{x} )}{ \sqrt{x} (1+x)} dx=2 \int _0^ \infty Arctan ( \sqrt{x} ).(Arctan( \sqrt{x} ))'dx =2 (Arctan1)^{2} = \frac{  \pi ^{2} }{4} $$ $$I_{3}= \int _0^ \infty  \frac{ln(1+ \sqrt{x} )}{ \sqrt{x} (1+x)}dx= \int _0^ \infty  \frac{ln(1+k \sqrt{x} ) | _0^1}{ \sqrt{x} (1+x)} dx= \int _0^1 \int _0^ \infty  \frac{1}{(1+k \sqrt{x} )(1+x)} dxdk= \int _0^1 \frac{1}{ k^{2} +1}  \int _0^ \infty ( \frac{1}{ x^{2} +1} - \frac{ax}{ x^{2} +1} + \frac{1}{ax+1} )dxdk= \int _0^1 \frac{ \pi k}{ k^{2} +1} dx-2 \int _0^1  \frac{lnk}{ k^{2} +1} dk=2C+ \frac{ \pi }{2}ln2 $$ $$$$$$ I_{4} = \int _0^ \infty  \frac{lnx}{ \sqrt{x}(1+x) } dx= \lim_{a\to 0}  \frac{d}{da}  \int _0^ \infty  \frac{ x^{a- \frac{1}{2} } }{1+x} dx= \lim_{a\to 0}  \beta (a+ \frac{1}{2} ,-a+ \frac{1}{2} )= \lim_{a\to 0}   \pi ^{2} tan (\pi a).sec( \pi a)=0$$ $$I_{5}= \int _0^ \infty  \frac{ln(x+1)}{ \sqrt{x} (1+x)} dx= \lim_{a\to 0} \frac{d}{da} \int _0^ \infty  \frac{ x^{ -\frac{1}{2} } }{(1+ x)^{1-a} } dx= \lim_{a\to 0}   \frac{d}{da}  \beta ( \frac{1}{2} , \frac{1}{2} -a) =- \lim_{a\to 0} \frac{d}{da}  [  ( \psi ^{(0)}( \frac{a}{b} -a) -  \psi ^{(0)} (1-a)) \beta ( \frac{1}{2} , \frac{1}{2} -a)]=- \pi (  \psi ^{(0)}( \frac{1}{2} ) -  \psi ^{(0)}(1) )=2 \pi ln2$$