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در دانشگاه توسط mansour (392 امتیاز)

نشان دهید که: $$ \int _0^ \infty \frac{lnx+ ln^{2} x+ ln^{3} x}{(1+x)(1+ x^{2} )} dx= -\frac{ \pi ^{2} }{128} .(7 \pi ^{2} -8 \pi +8)$$

توسط mansour (392 امتیاز)
ویرایش شده توسط mansour
$$ \int _0^ \infty  \frac{ x^{k} }{(1+x)(1+ x^{2} )} dx= \frac{1}{2}  \int _0^ \infty  \frac{ x^{k} [( x^{2} +1)-(1- x^{2} )}{(1+x)(1+ x^{2} )} dx= \frac{1}{2}  \int _0^ \infty  \frac{ x^{k} }{1+x} dx+ \frac{1}{2}  \int _0^ \infty  \frac{ x^{k} (1+x)(1-x)}{(1+x)(1+ x^{2} )} dx= \frac{1}{2}  \int _0^ \infty  \frac{ x^{k} }{1+x} dx+ \frac{1}{2}  \int _0^ \infty  \frac{ x^{k} (1-x)}{1+ x^{2} } dx= \frac{1}{2}  \int_0^ \infty   \frac{ x^{k} }{1+x} dx+ \frac{1}{2}  \int _0^ \infty  \frac{ x^{k} }{1+ x^{2} } dx- \frac{1}{2}  \int _0^ \infty  \frac{ x^{k+1} }{1+ x^{2} } dx= \frac{1}{2}  \int _0^ \infty  \frac{ x^{k} }{1+x} dx+ \frac{1}{2}  \int _0^ \infty  \frac{ ( x^{2} )^{ \frac{k-1}{2} } }{1+ x^{2} } dx- \frac{1}{2}  \int _0^ \infty  \frac{ ( x^{2} )^{ \frac{k}{2} } -x }{1+ x^{2} } dx \wedge y= x^{2}  \Longrightarrow dy=2xdx \Longrightarrow  \frac{1}{2} dy=xdx \Longrightarrow = \frac{1}{2}  \int _0^ \infty  \frac{ x^{k} }{1+x} dx+ \frac{1}{4}  \int _0^ \infty  \frac{ y^{ \frac{k-1}{2} } }{1+y} dy- \frac{1}{4}  \int _0^ \infty  \frac{ y^{ \frac{k}{2} } }{1+y} dy or= \frac{1}{2}  \int _0^ \infty  \frac{ x^{k} }{1+x} dx+ \frac{1}{4}  \int _0^ \infty  \frac{ x^{ \frac{k-1}{2} } }{1+x} dx- \frac{1}{4}  \int _0^ \infty  \frac{ x^{ \frac{k}{2} } }{1+x} dx$$$$from: \beta (m,n)= \int  _0^ \infty \frac{ x^{m-1} }{ (1+x)^{m+n} } dx= \frac{ \Gamma (m) \Gamma (n)}{ \Gamma (m+n)}  \Longrightarrow = \frac{1}{2}  \Gamma (1+k) \Gamma (-k)+ \frac{1}{4}  \Gamma ( \frac{k+1}{2} ) \Gamma ( \frac{1-k}{2} )- \frac{1}{4}  \Gamma (1+ \frac{k}{2} ) \Gamma (-  \frac{k}{2} ) from  \Gamma (a) \Gamma (1-a)= \frac{ \pi }{sin( \pi a)}  \Longrightarrow  \Gamma (-a) \Gamma (1+a)=- \frac{ \pi }{sin( \pi a)}  \wedge  \Gamma ( \frac{a+1}{2} ) \Gamma ( \frac{1-a}{2} )= \frac{ \pi }{sin[ \frac{ \pi }{2} (a+1)]} = \frac{ \pi }{cos( \frac{ \pi a}{2} )}  \wedge  \Gamma ( \frac{-a}{2} ) \Gamma (1+ \frac{a}{2} )= \frac{ -\pi }{sin( \frac{ \pi a}{2} )}  \Longrightarrow = \frac{- \pi }{2} . \frac{1}{sin( \pi k)} + \frac{ \pi }{4} . \frac{1}{cos( \frac{ \pi }{2} k)} + \frac{ \pi }{4} . \frac{1}{sin( \frac{ \pi k}{2} )} = \frac { \pi }{4} . \frac{1}{sin( \frac{ \pi k}{2} )cos( \frac{ \pi k}{2} )} + \frac{ \pi }{4} . \frac{sin( \frac{ \pi k}{2} )}{sin( \frac{ \pi k}{2} )cos( \frac{ \pi k}{2} )} + \frac{ \pi }{4} . \frac{cos( \frac{ \pi k}{2} )}{sin( \frac{ \pi k}{2} )cos( \frac{ \pi k}{2} )} $$$$ \frac{ -\pi }{2} . \frac{1}{sin( \pi k)} + \frac{ \pi }{4} . \frac{1}{cos( \frac{ \pi k}{2} )} + \frac{ \pi }{4} . \frac{cos( \frac{ \pi k}{2} )}{sin( \frac{ \pi k}{2} )cos( \frac{ \pi k }{2} )} = \frac{- \pi }{2} . \frac{1}{sin( \pi k)} + \frac{ \pi }{2} . \frac{sin( \frac{ \pi k }{2} )}{sin( \pi k)} + \frac{ \pi }{2} . \frac{cos( \frac{ \pi k}{2} )}{sin( \pi k)} = \frac{- \pi }{2} .  \frac{1}{sin( \pi k)} + \frac{ \pi }{2} . \frac{sin( \frac{ \pi k}{2} )}{sin( \pi k)} + \frac{ \pi }{2} . \frac{cos( \frac{ \pi k}{2} )}{sin( \pi k)} = \frac{ \pi }{2} . \frac{(sin( \frac{ \pi k}{2} )+cos( \frac{ \pi k}{2} )-1)}{sin( \pi k)} = \frac{ \pi }{2} . \frac{sin \pi k}{(sin \pi k)[1+sin( \frac{ \pi k}{2} )+cos( \frac{ \pi k}{2} )]} = \frac{ \pi }{2} . \frac{1}{1+sin( \frac{ \pi k}{2} )+cos( \frac{ \pi k}{2} )}  \wedge -1 <  k <  2 \Longrightarrow  \int _0^ \infty  \frac{ x^{k} }{(1+x)(1+ x^{2} )} dx= \frac{ \pi }{2} . \frac{1}{1+sin( \frac{ \pi k}{2} )+cos( \frac{ \pi k}{2} )} = \frac{\partial}{\partial k}  \int _0^ \infty  \frac{ x^{k} }{(1+x)(1+ x^{2} )} dx= \frac{ \pi }{2}  \frac{\partial}{\partial k}  [(1+sin( \frac{ \pi k}{2} )+cos ( \frac{ \pi k}{2} ))]^{-1}  \Longrightarrow  \int _0^ \infty  \frac{ x^{k} lnx}{(1+x)(1+ x^{2} )} dx= \frac{- \pi }{2} . \frac{ \pi }{2} . \frac{cos( \frac{ \pi k}{2} )-sin( \frac{ \pi k}{2} )}{ [1+sin ( \frac{ \pi k}{2} )+cos ( \frac{ \pi k}{2} )]^{2} } = \frac{  \pi ^{2} }{4} . \frac{cos( \frac{ \pi k}{2} )-sin( \frac{ \pi k}{2} )}{ (1+sin ( \frac{ \pi k}{2} )+cos ( \frac{ \pi k}{2} ))^{2} } $$$$ \int _0^ \infty  \frac{ x^{k}  ln^{2}x }{(1+x)(1+ x^{2} )} dx= -\frac{  \pi ^{2} }{4}  \frac{\partial}{\partial k} [ \frac{cos( \frac{ \pi k}{2} )-sin( \frac{ \pi k}{2} )}{ (1+sin ( \frac{ \pi k}{2} )+cos ( \frac{ \pi k}{2} ))^{2} } ]=- \frac{  \pi ^{2} }{4} [ \frac{ \frac{- \pi }{2} (sin( \frac{ \pi k}{2} )+cos( \frac{ \pi k}{2} ))( 1+sin( \frac{ \pi k}{2} )+cos ( \frac{ \pi k}{2} ))^{2} - \pi  (cos ( \frac{ \pi k}{2} )-sin ( \frac{ \pi k}{2} ))^{2} (1+sin( \frac{ \pi k}{2} )+cos( \frac{ \pi k}{2} ))}{ (1+sin ( \frac{ \pi k}{2} )+cos ( \frac{ \pi k}{2} ))^{4} } ]= \frac{  \pi ^{3} }{4} [ \frac{ \frac{1}{2}(sin ( \frac{ \pi k}{2} )+cos( \frac{ \pi k}{2} ))}{ (1+sin ( \frac{ \pi k}{2} )+cos ( \frac{ \pi k}{2} ))^{2} }+ \frac{ (cos( \frac{ \pi k}{2} )-sin( \frac{ \pi k}{2} ))^{2} }{ (1+sin( \frac{ \pi k}{2} )+cos( \frac{ \pi k}{2} ))^{3} }  ]= \frac{  \pi ^{3} }{4} [ \frac{1}{2} . \frac{(sin( \frac{ \pi k}{2} )+cos( \frac{  \pi k}{2} ))}{ (1+sin( \frac{ \pi k}{2} )+cos ( \frac{ \pi k}{2} ))^{2} } + \frac{1-sin \pi k}{ (1+sin ( \frac{ \pi k}{2} )+cos ( \frac{ \pi k}{2} ))^{3} } ] \wedge  \int _0^ \infty  \frac{ x^{k}  ln^{3} x}{(1+x)(1+ x^{2} )} dx= \frac{  \pi ^{3} }{8}  \frac{\partial}{\partial k} [ \frac{sin( \frac{ \pi k}{2} )+cos( \frac{ \pi k}{2} )}{ (1+sin( \frac{ \pi k}{2} )+cos( \frac{ \pi k}{2} ))^{2} } ]+ \frac{  \pi ^{3} }{4}  \frac{\partial}{\partial k} [ \frac{1-sin \pi k}{ (1+sin ( \frac{ \pi k}{2} )+cos ( \frac{ \pi k}{2} ))^{3} } ]= \frac{  \pi ^{3} }{8} [ \frac{ \pi }{2} ( \frac{ (cos( \frac{ \pi k}{2} )-sin( \frac{ \pi k}{2} )) (1+sin( \frac{ \pi k}{2} )+cos ( \frac{ \pi k}{2} ))^{2} )- \pi (sin( \frac{ \pi k}{2} )+cos( \frac{ \pi k}{2} )(cos( \frac{ \pi k}{2} )-sin( \frac{ \pi k}{2} ))(1+sin( \frac{ \pi k}{2} )+cos( \frac{ \pi k}{2} ))] }{ (1+sin ( \frac{ \pi k}{2} )+cos ( \frac{ \pi k}{2} ))^{6} } )+ \frac{  \pi ^{3} }{4} [ \frac{- \pi cos \pi k (1+sin ( \frac{ \pi k}{2} )+cos ( \frac{ \pi k}{2} ))^{3} - \frac{3 \pi }{2}  (1-sin  \pi k)^{2} (1+sin( \frac{ \pi k}{2} )+cos ( \frac{ \pi k}{2} ))^{2}  }{ (1+sin ( \frac{ \pi k}{2} )+cos ( \frac{ \pi k}{2} ))^{6} } ]$$$$ \frac{  \pi ^{4} }{8} [ \frac{1}{2}  \frac{(cos( \frac{ \pi k}{2} )-sin( \frac{ \pi k}{2} ))}{ (1+sin ( \frac{ \pi k}{2} )+cos ( \frac{ \pi k}{2} ))^{2} } - \frac{cos( \frac{ \pi k}{2} )}{ (1+sin ( \frac{ \pi k}{2} )+cos ( \frac{ \pi k}{2} ))^{3} } ]- \frac{  \pi ^{4} }{4} [ \frac{cos \pi k}{ (1+sin ( \frac{ \pi k}{2} )+cos ( \frac{ \pi k}{2} ))^{3} } + \frac{3}{2}  \frac{ (1-sin \pi k)^{2} }{ (1+sin ( \frac{ \pi k}{2} )+cos ( \frac{ \pi k}{2} ))^{4} } ] \Longrightarrow  \int _0^ \infty  \frac{ x^{k}  ln^{2} x}{(1+x)(1+ x^{2} )} dx= \frac{  \pi ^{4} }{16}  \frac{[cos( \frac{ \pi k}{2} )-sin( \frac{ \pi k}{2} )]}{ (1+sin ( \frac{ \pi k}{2} )+cos ( \frac{ \pi k}{2} ))^{2} } - \frac{3  \pi ^{4} }{8}  \frac{cos \pi k}{ (1+sin ( \frac{ \pi k}{2} )+cos ( \frac{ \pi k}{2} ))^{3} } - \frac{3  \pi ^{4} }{8}  \frac{ (1-sin \pi k)^{2} }{ (1+sin ( \frac{ \pi k}{2} )+cos ( \frac{ \pi k}{2} ))^{4} }  \wedge  \int _0^ \infty  \frac{ x^{k}  ln^{2}x }{(1+x)(1+ x^{2} )} dx= \frac{  \pi ^{3} }{8} . \frac{sin( \frac{ \pi k}{2} )+cos( \frac{ \pi k}{2} )}{ (1+sin ( \frac{ \pi k}{2} )+cos ( \frac{ \pi k}{2} ))^{2} } + \frac{  \pi ^{3} }{6}  \frac{(1-sin \pi k)}{ (1+sin ( \frac{ \pi k}{2} )+cos ( \frac{ \pi k}{2} ))^{3} } $$$$ \Longrightarrow k=0: \int _0^ \infty  \frac{ ln^{3}x }{(1+x)(1+ x^{2} )} dx= \frac{  \pi ^{4} }{16} . \frac{1}{4} - \frac{3  \pi ^{4} }{8} . \frac{1}{8} - \frac{3  \pi ^{4} }{8} . \frac{1}{16} = \frac{  \pi ^{4} }{64} - \frac{ 3 \pi ^{4} }{64} - \frac{3  \pi ^{4} }{128} = -\frac{7  \pi ^{4} }{128}  \wedge k=0: \int _0^ \infty  \frac{ ln^{2} x}{(1+x)(1+ x^{2} )} dx= \frac{  \pi ^{3}}{8} . \frac{1}{4} + \frac{  \pi ^{3} }{4} . \frac{1}{8} = \frac{  \pi ^{3} }{16}  \Longrightarrow  \int _0^ \infty  \frac{a}{b} dx= \frac{a}{b} . \frac{a}{b}  \Longrightarrow k=0, \int _0^ \infty  \frac{lnx}{(1+x)(1+ x^{2} )} dx=- \frac{ x^{2} }{4} . \frac{1}{4} =- \frac{ x^{2} }{16}  \Longrightarrow ?= -\frac{  \pi ^{2} }{16} + \frac{  \pi ^{3} }{16} - \frac{7  \pi ^{4} }{128} =- \frac{8  \pi ^{2} }{128} + \frac{8  \pi ^{3} }{128} - \frac{7  \pi ^{4} }{128} = -\frac{  \pi ^{2} }{128} (7  \pi ^{2} -8 \pi +8)$$

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توسط mansour (392 امتیاز)
 
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$$ \int _0^ \infty \frac{ x^{k} }{(1+x)(1+ x^{2} )} dx= \frac{1}{2} \int _0^ \infty \frac{ x^{k} [( x^{2} +1)-(1- x^{2} )}{(1+x)(1+ x^{2} )} dx= \frac{1}{2} \int _0^ \infty \frac{ x^{k} }{1+x} dx+ \frac{1}{2} \int _0^ \infty \frac{ x^{k} (1+x)(1-x)}{(1+x)(1+ x^{2} )} dx= \frac{1}{2} \int _0^ \infty \frac{ x^{k} }{1+x} dx+ \frac{1}{2} \int _0^ \infty \frac{ x^{k} (1-x)}{1+ x^{2} } dx= \frac{1}{2} \int_0^ \infty \frac{ x^{k} }{1+x} dx+ \frac{1}{2} \int _0^ \infty \frac{ x^{k} }{1+ x^{2} } dx- \frac{1}{2} \int _0^ \infty \frac{ x^{k+1} }{1+ x^{2} } dx= \frac{1}{2} \int _0^ \infty \frac{ x^{k} }{1+x} dx+ \frac{1}{2} \int _0^ \infty \frac{ ( x^{2} )^{ \frac{k-1}{2} } }{1+ x^{2} } dx- \frac{1}{2} \int _0^ \infty \frac{ ( x^{2} )^{ \frac{k}{2} } -x }{1+ x^{2} } dx \wedge y= x^{2} \Longrightarrow dy=2xdx \Longrightarrow \frac{1}{2} dy=xdx \Longrightarrow = \frac{1}{2} \int _0^ \infty \frac{ x^{k} }{1+x} dx+ \frac{1}{4} \int _0^ \infty \frac{ y^{ \frac{k-1}{2} } }{1+y} dy- \frac{1}{4} \int _0^ \infty \frac{ y^{ \frac{k}{2} } }{1+y} dy or= \frac{1}{2} \int _0^ \infty \frac{ x^{k} }{1+x} dx+ \frac{1}{4} \int _0^ \infty \frac{ x^{ \frac{k-1}{2} } }{1+x} dx- \frac{1}{4} \int _0^ \infty \frac{ x^{ \frac{k}{2} } }{1+x} dx$$$$from: \beta (m,n)= \int _0^ \infty \frac{ x^{m-1} }{ (1+x)^{m+n} } dx= \frac{ \Gamma (m) \Gamma (n)}{ \Gamma (m+n)} \Longrightarrow = \frac{1}{2} \Gamma (1+k) \Gamma (-k)+ \frac{1}{4} \Gamma ( \frac{k+1}{2} ) \Gamma ( \frac{1-k}{2} )- \frac{1}{4} \Gamma (1+ \frac{k}{2} ) \Gamma (- \frac{k}{2} ) from \Gamma (a) \Gamma (1-a)= \frac{ \pi }{sin( \pi a)} \Longrightarrow \Gamma (-a) \Gamma (1+a)=- \frac{ \pi }{sin( \pi a)} \wedge \Gamma ( \frac{a+1}{2} ) \Gamma ( \frac{1-a}{2} )= \frac{ \pi }{sin[ \frac{ \pi }{2} (a+1)]} = \frac{ \pi }{cos( \frac{ \pi a}{2} )} \wedge \Gamma ( \frac{-a}{2} ) \Gamma (1+ \frac{a}{2} )= \frac{ -\pi }{sin( \frac{ \pi a}{2} )} \Longrightarrow = \frac{- \pi }{2} . \frac{1}{sin( \pi k)} + \frac{ \pi }{4} . \frac{1}{cos( \frac{ \pi }{2} k)} + \frac{ \pi }{4} . \frac{1}{sin( \frac{ \pi k}{2} )} = \frac { \pi }{4} . \frac{1}{sin( \frac{ \pi k}{2} )cos( \frac{ \pi k}{2} )} + \frac{ \pi }{4} . \frac{sin( \frac{ \pi k}{2} )}{sin( \frac{ \pi k}{2} )cos( \frac{ \pi k}{2} )} + \frac{ \pi }{4} . \frac{cos( \frac{ \pi k}{2} )}{sin( \frac{ \pi k}{2} )cos( \frac{ \pi k}{2} )} $$$$ \frac{ -\pi }{2} . \frac{1}{sin( \pi k)} + \frac{ \pi }{4} . \frac{1}{cos( \frac{ \pi k}{2} )} + \frac{ \pi }{4} . \frac{cos( \frac{ \pi k}{2} )}{sin( \frac{ \pi k}{2} )cos( \frac{ \pi k }{2} )} = \frac{- \pi }{2} . \frac{1}{sin( \pi k)} + \frac{ \pi }{2} . \frac{sin( \frac{ \pi k }{2} )}{sin( \pi k)} + \frac{ \pi }{2} . \frac{cos( \frac{ \pi k}{2} )}{sin( \pi k)} = \frac{- \pi }{2} . \frac{1}{sin( \pi k)} + \frac{ \pi }{2} . \frac{sin( \frac{ \pi k}{2} )}{sin( \pi k)} + \frac{ \pi }{2} . \frac{cos( \frac{ \pi k}{2} )}{sin( \pi k)} = \frac{ \pi }{2} . \frac{(sin( \frac{ \pi k}{2} )+cos( \frac{ \pi k}{2} )-1)}{sin( \pi k)} = \frac{ \pi }{2} . \frac{sin \pi k}{(sin \pi k)[1+sin( \frac{ \pi k}{2} )+cos( \frac{ \pi k}{2} )]} = \frac{ \pi }{2} . \frac{1}{1+sin( \frac{ \pi k}{2} )+cos( \frac{ \pi k}{2} )} \wedge -1 < k < 2 \Longrightarrow \int _0^ \infty \frac{ x^{k} }{(1+x)(1+ x^{2} )} dx= \frac{ \pi }{2} . \frac{1}{1+sin( \frac{ \pi k}{2} )+cos( \frac{ \pi k}{2} )} = \frac{\partial}{\partial k} \int _0^ \infty \frac{ x^{k} }{(1+x)(1+ x^{2} )} dx= \frac{ \pi }{2} \frac{\partial}{\partial k} [(1+sin( \frac{ \pi k}{2} )+cos ( \frac{ \pi k}{2} ))]^{-1} \Longrightarrow \int _0^ \infty \frac{ x^{k} lnx}{(1+x)(1+ x^{2} )} dx= \frac{- \pi }{2} . \frac{ \pi }{2} . \frac{cos( \frac{ \pi k}{2} )-sin( \frac{ \pi k}{2} )}{ [1+sin ( \frac{ \pi k}{2} )+cos ( \frac{ \pi k}{2} )]^{2} } = \frac{ \pi ^{2} }{4} . \frac{cos( \frac{ \pi k}{2} )-sin( \frac{ \pi k}{2} )}{ (1+sin ( \frac{ \pi k}{2} )+cos ( \frac{ \pi k}{2} ))^{2} } $$$$ \int _0^ \infty \frac{ x^{k} ln^{2}x }{(1+x)(1+ x^{2} )} dx= -\frac{ \pi ^{2} }{4} \frac{\partial}{\partial k} [ \frac{cos( \frac{ \pi k}{2} )-sin( \frac{ \pi k}{2} )}{ (1+sin ( \frac{ \pi k}{2} )+cos ( \frac{ \pi k}{2} ))^{2} } ]=- \frac{ \pi ^{2} }{4} [ \frac{ \frac{- \pi }{2} (sin( \frac{ \pi k}{2} )+cos( \frac{ \pi k}{2} ))( 1+sin( \frac{ \pi k}{2} )+cos ( \frac{ \pi k}{2} ))^{2} - \pi (cos ( \frac{ \pi k}{2} )-sin ( \frac{ \pi k}{2} ))^{2} (1+sin( \frac{ \pi k}{2} )+cos( \frac{ \pi k}{2} ))}{ (1+sin ( \frac{ \pi k}{2} )+cos ( \frac{ \pi k}{2} ))^{4} } ]= \frac{ \pi ^{3} }{4} [ \frac{ \frac{1}{2}(sin ( \frac{ \pi k}{2} )+cos( \frac{ \pi k}{2} ))}{ (1+sin ( \frac{ \pi k}{2} )+cos ( \frac{ \pi k}{2} ))^{2} }+ \frac{ (cos( \frac{ \pi k}{2} )-sin( \frac{ \pi k}{2} ))^{2} }{ (1+sin( \frac{ \pi k}{2} )+cos( \frac{ \pi k}{2} ))^{3} } ]= \frac{ \pi ^{3} }{4} [ \frac{1}{2} . \frac{(sin( \frac{ \pi k}{2} )+cos( \frac{ \pi k}{2} ))}{ (1+sin( \frac{ \pi k}{2} )+cos ( \frac{ \pi k}{2} ))^{2} } + \frac{1-sin \pi k}{ (1+sin ( \frac{ \pi k}{2} )+cos ( \frac{ \pi k}{2} ))^{3} } ] \wedge \int _0^ \infty \frac{ x^{k} ln^{3} x}{(1+x)(1+ x^{2} )} dx= \frac{ \pi ^{3} }{8} \frac{\partial}{\partial k} [ \frac{sin( \frac{ \pi k}{2} )+cos( \frac{ \pi k}{2} )}{ (1+sin( \frac{ \pi k}{2} )+cos( \frac{ \pi k}{2} ))^{2} } ]+ \frac{ \pi ^{3} }{4} \frac{\partial}{\partial k} [ \frac{1-sin \pi k}{ (1+sin ( \frac{ \pi k}{2} )+cos ( \frac{ \pi k}{2} ))^{3} } ]= \frac{ \pi ^{3} }{8} [ \frac{ \pi }{2} ( \frac{ (cos( \frac{ \pi k}{2} )-sin( \frac{ \pi k}{2} )) (1+sin( \frac{ \pi k}{2} )+cos ( \frac{ \pi k}{2} ))^{2} )- \pi (sin( \frac{ \pi k}{2} )+cos( \frac{ \pi k}{2} )(cos( \frac{ \pi k}{2} )-sin( \frac{ \pi k}{2} ))(1+sin( \frac{ \pi k}{2} )+cos( \frac{ \pi k}{2} ))] }{ (1+sin ( \frac{ \pi k}{2} )+cos ( \frac{ \pi k}{2} ))^{6} } )+ \frac{ \pi ^{3} }{4} [ \frac{- \pi cos \pi k (1+sin ( \frac{ \pi k}{2} )+cos ( \frac{ \pi k}{2} ))^{3} - \frac{3 \pi }{2} (1-sin \pi k)^{2} (1+sin( \frac{ \pi k}{2} )+cos ( \frac{ \pi k}{2} ))^{2} }{ (1+sin ( \frac{ \pi k}{2} )+cos ( \frac{ \pi k}{2} ))^{6} } ]$$$$ \frac{ \pi ^{4} }{8} [ \frac{1}{2} \frac{(cos( \frac{ \pi k}{2} )-sin( \frac{ \pi k}{2} ))}{ (1+sin ( \frac{ \pi k}{2} )+cos ( \frac{ \pi k}{2} ))^{2} } - \frac{cos( \frac{ \pi k}{2} )}{ (1+sin ( \frac{ \pi k}{2} )+cos ( \frac{ \pi k}{2} ))^{3} } ]- \frac{ \pi ^{4} }{4} [ \frac{cos \pi k}{ (1+sin ( \frac{ \pi k}{2} )+cos ( \frac{ \pi k}{2} ))^{3} } + \frac{3}{2} \frac{ (1-sin \pi k)^{2} }{ (1+sin ( \frac{ \pi k}{2} )+cos ( \frac{ \pi k}{2} ))^{4} } ] \Longrightarrow \int _0^ \infty \frac{ x^{k} ln^{2} x}{(1+x)(1+ x^{2} )} dx= \frac{ \pi ^{4} }{16} \frac{[cos( \frac{ \pi k}{2} )-sin( \frac{ \pi k}{2} )]}{ (1+sin ( \frac{ \pi k}{2} )+cos ( \frac{ \pi k}{2} ))^{2} } - \frac{3 \pi ^{4} }{8} \frac{cos \pi k}{ (1+sin ( \frac{ \pi k}{2} )+cos ( \frac{ \pi k}{2} ))^{3} } - \frac{3 \pi ^{4} }{8} \frac{ (1-sin \pi k)^{2} }{ (1+sin ( \frac{ \pi k}{2} )+cos ( \frac{ \pi k}{2} ))^{4} } \wedge \int _0^ \infty \frac{ x^{k} ln^{2}x }{(1+x)(1+ x^{2} )} dx= \frac{ \pi ^{3} }{8} . \frac{sin( \frac{ \pi k}{2} )+cos( \frac{ \pi k}{2} )}{ (1+sin ( \frac{ \pi k}{2} )+cos ( \frac{ \pi k}{2} ))^{2} } + \frac{ \pi ^{3} }{6} \frac{(1-sin \pi k)}{ (1+sin ( \frac{ \pi k}{2} )+cos ( \frac{ \pi k}{2} ))^{3} } $$$$ \Longrightarrow k=0: \int _0^ \infty \frac{ ln^{3}x }{(1+x)(1+ x^{2} )} dx= \frac{ \pi ^{4} }{16} . \frac{1}{4} - \frac{3 \pi ^{4} }{8} . \frac{1}{8} - \frac{3 \pi ^{4} }{8} . \frac{1}{16} = \frac{ \pi ^{4} }{64} - \frac{ 3 \pi ^{4} }{64} - \frac{3 \pi ^{4} }{128} = -\frac{7 \pi ^{4} }{128} \wedge k=0: \int _0^ \infty \frac{ ln^{2} x}{(1+x)(1+ x^{2} )} dx= \frac{ \pi ^{3}}{8} . \frac{1}{4} + \frac{ \pi ^{3} }{4} . \frac{1}{8} = \frac{ \pi ^{3} }{16} \Longrightarrow \int _0^ \infty \frac{a}{b} dx= \frac{a}{b} . \frac{a}{b} \Longrightarrow k=0, \int _0^ \infty \frac{lnx}{(1+x)(1+ x^{2} )} dx=- \frac{ x^{2} }{4} . \frac{1}{4} =- \frac{ x^{2} }{16} \Longrightarrow ?= -\frac{ \pi ^{2} }{16} + \frac{ \pi ^{3} }{16} - \frac{7 \pi ^{4} }{128} =- \frac{8 \pi ^{2} }{128} + \frac{8 \pi ^{3} }{128} - \frac{7 \pi ^{4} }{128} = -\frac{ \pi ^{2} }{128} (7 \pi ^{2} -8 \pi +8)$$


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کانال تلگرام محفل ریاضی
امروز : تاریخ شمسی اینجا نمایش داده می‌شود
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