: $$ \int _ {- \infty } ^ { \infty } ( \frac{x}{2+ 2^{x} + 2^{-x} } )^{2} dx= \frac{ \zeta (2)-1}{3} . \frac{1}{ ln^{3} 2} $$

Question

مطلوب است محاسبه انتگرال معین زیر: $$ \int _ {- \infty } ^ { \infty } ( \frac{x}{2+ 2^{x} + 2^{-x} } )^{2} dx= \frac{ \zeta (2)-1}{3} . \frac{1}{ ln^{3} 2} $$

Comments

$$ \int _ {-  \infty } ^ { \infty }  ( \frac{x}{ a+a^{x}+ a^{-x}  } )^{2} dx \wedge  \int _ {- \infty } ^ { \infty }  ( \frac{x}{a+ e^{ ln a^{x} } + e^{ln a^{-x} } } )^{2} dx \wedge  \int _ {- \infty } ^ { \infty }  ( \frac{x}{a+ e^{xlna} + e^{-xlna} } )^{2}  \Longrightarrow u=xlna \Longrightarrow du=lna.dx \Longrightarrow = \int _ {- \infty } ^ { \infty }  \frac{ ( \frac{u}{lna} )^{2} }{ ( a+ e^{u} + e^{-u} )^{2} } . \frac{du}{lna} = \frac{1}{ ln^{3} a}  \int _ {- \infty } ^ { \infty }  \frac{ u^{2} }{ (a+ e^{u} + e^{-u} )^{2} } du \wedge t= e^{u}  \Longrightarrow dt= e^{u} du \Longrightarrow  \frac{dt}{t} =du \Longrightarrow = \frac{1}{ ln^{3} a}  \int _0^ \infty  \frac{t ln^{2}t }{ (a+t+ \frac{1}{t} )^{2} } . \frac{dt}{t}= \frac{a}{ ln^{3}a }   \int _0^ \infty  \frac{t ln^{2} t}{ ( t^{2} +at+1)^{2} } dt \Longrightarrow a=2:= \frac{1}{ ln^{3} a}  \int _0^ \infty  \frac{t ln^{2}t }{ (t+1)^{4} } dt$$$$ \Longrightarrow  \int _0^ \infty  \frac{ x^{s-1} }{ (1+x)^{4} } dx=I(s) \wedge t= \frac{1}{1+x}  \Longrightarrow x+1= \frac{1}{t}  \Longrightarrow x= \frac{1}{t} -1 \Longrightarrow I(s)= \int _1^0 ( \frac{1}{t} -1)^{s-1} . t^{4} . \frac{-dt}{ t^{2} } = \int _0^1 t^{2}  (1-t)^{s-1}. t^{1-s} dt = \int _0^1 t^{3-s}  (1-t)^{s-1} dt= \beta (4-s,s) \Longrightarrow  \frac{\partial}{\partial s} I(s)= \int _0^ \infty  \frac{ x^{s-1} lnx}{ ( 1+x)^{4} } dx= \frac{\partial^2}{\partial s^2} I(s)= \int _0^ \infty  \frac{ x^{s-1}  ln^{2} x}{ ( 1+x)^{4} } dx \Longrightarrow  \frac{\partial^2}{\partial s^2}  \beta (4-s,s)= \int _0^ \infty  \frac{ x^{s-1}  ln^{2}x }{ ( 1+x)^{4} } dx= \frac{\partial^2}{\partial s^2} [ \frac{ \Gamma (4-s) \Gamma (s)}{ \Gamma (4)} ]= \frac{\partial}{\partial s} .[ \frac{1}{ \Gamma (4)} [- \Gamma (s) \Gamma '( 4-s)+ \Gamma (4-s ) \Gamma '(s)]]= \frac{1}{ \Gamma (4)}  \frac{\partial}{\partial s} [ \Gamma '(s) \Gamma (4-s)- \Gamma (s) \Gamma '(4-s)]$$

Answer 1

$$ \Longrightarrow \int _0^ \infty \frac{ x^{s-1} }{ (1+x)^{4} } dx=I(s) \wedge t= \frac{1}{1+x} \Longrightarrow x+1= \frac{1}{t} \Longrightarrow x= \frac{1}{t} -1 \Longrightarrow I(s)= \int _1^0 ( \frac{1}{t} -1)^{s-1} . t^{4} . \frac{-dt}{ t^{2} } = \int _0^1 t^{2} (1-t)^{s-1}. t^{1-s} dt = \int _0^1 t^{3-s} (1-t)^{s-1} dt= \beta (4-s,s) \Longrightarrow \frac{\partial}{\partial s} I(s)= \int _0^ \infty \frac{ x^{s-1} lnx}{ ( 1+x)^{4} } dx= \frac{\partial^2}{\partial s^2} I(s)= \int _0^ \infty \frac{ x^{s-1} ln^{2} x}{ ( 1+x)^{4} } dx \Longrightarrow \frac{\partial^2}{\partial s^2} \beta (4-s,s)= \int _0^ \infty \frac{ x^{s-1} ln^{2}x }{ ( 1+x)^{4} } dx= \frac{\partial^2}{\partial s^2} [ \frac{ \Gamma (4-s) \Gamma (s)}{ \Gamma (4)} ]= \frac{\partial}{\partial s} .[ \frac{1}{ \Gamma (4)} [- \Gamma (s) \Gamma '( 4-s)+ \Gamma (4-s ) \Gamma '(s)]]= \frac{1}{ \Gamma (4)} \frac{\partial}{\partial s} [ \Gamma '(s) \Gamma (4-s)- \Gamma (s) \Gamma '(4-s)]$$$$= \frac{1}{ \Gamma (4)} [ \Gamma 'x(s) \Gamma (4-s)- \Gamma '(s) \Gamma '(4-s)- \Gamma '(s) \Gamma '(4-s)+ \Gamma (s) \Gamma ''(4-s) ] \wedge s=2, \frac{1}{ \Gamma (4)} [ \Gamma ''(2) \Gamma (2)- \Gamma '(2) \Gamma '(2)- \Gamma '(2) \Gamma '(2)+ \Gamma (2) \Gamma ''(2) ]= \frac{2}{ \Gamma (4)} [ \Gamma ''(2)- ( \Gamma '(2))^{2} ]= \frac{1}{3} [ \Gamma ''(2)- (1- \gamma )^{2} ] \wedge \psi (x)= \frac{ \Gamma '(x)}{ \Gamma (x)} =- \gamma - \frac{1}{x} + \sum _ {k=1} ^ \infty \frac{1}{k} - \frac{1}{k+x} \Longrightarrow \psi '(x)= \frac{1}{ x^{2} } + \sum _ {k=1} ^ \infty \frac{1}{ (k+x)^{2} } \Longrightarrow \psi '(2)= \frac{1}{ 2^{2} } + \sum _ {k=1} ^ \infty \frac{1}{ (x+2)^{2} }= \frac{1}{ 2^{2} } + \frac{1}{ 3^{2} } + \frac{1}{ 4^{2} } +...=( \frac{ \pi ^{2} }{6} -1) \wedge \Gamma ''(2)=-1+ \frac{ \pi ^{2} }{6} + ( 1- \gamma )^{2} \wedge \Longrightarrow ?= \frac{1}{3} [-1+ \frac{ \pi ^{2} }{6} + (1- \gamma )^{2}- ( 1- \gamma )^{2}]= \frac{1}{3} [ \zeta (2)-1 ] $$