$$ \int _0^ \infty \frac{Arctanx}{x(1+x)(1+ x^{2} )} dx= \frac{3 \pi }{8} ln2- \frac{ \pi ^{2} }{16} + \frac{G}{2} \wedge G= \beta (2)$$

Question

ثابت کنید که: $$ \int _0^ \infty \frac{Arctanx}{x(1+x)(1+ x^{2} )} dx= \frac{3 \pi }{8} ln2- \frac{ \pi ^{2} }{16} + \frac{G}{2} \wedge G= \beta (2)$$

Comments

$$t=Arctan x \wedge x=0 \leadsto t=0 \wedge x \rightarrow  \infty  \wedge t= \frac{ \pi }{2}  \Longrightarrow I= \int _0^ \frac{ \pi }{2}  \frac{t}{tant(1+tant)} dt \wedge  \frac{1}{tant(1+tant)} = \frac{A}{tant} + \frac{B}{1+tant} \Longrightarrow A=1 \wedge B=-1 \Longrightarrow I_1= \int _0^ \frac{ \pi }{2}  \frac{t}{tant} dt \wedge I_2= -\int _0^ \frac{ \pi }{2}  \frac{t}{1+tant} dt$$$$I_1= \int _0^  \frac{ \pi }{2}  \frac{t}{tant}  dt= \int _0^ \frac{ \pi }{2} t \frac{cost}{sint} dt= \underbrace{tlogsint]_0^ \frac{ \pi }{2} =0} - \int _0^ \frac{ \pi }{2} log(sint)dt=- \int _0^ \frac{ \pi }{2} log(sint)dt \wedge I(a,b)= \int _0^ \frac{ \pi }{2}  cos^{2a-1}(t)  sin^{2b-1} (t)dt= \frac{1}{2}  \frac{ \Gamma (a) \Gamma (b)}{ \Gamma (a+b)}  \Longrightarrow  \int _0^ \frac{ \pi }{2}  cos^{2a-1}(t)  sin^{2b-1} (t)log(sint)dt= \frac{1}{2}  \frac{\partial}{\partial b}  \frac{ \Gamma (a) \Gamma (b)}{ \Gamma (a+b)} = \frac{1}{4}  \frac{ \Gamma (a) \Gamma (b)}{ \Gamma (a+b)} [ \psi (b)- \psi (a+b)]dt$$$$ \longrightarrow a=b= \frac{1}{2} : \int _0^ \frac{ \pi }{2} log(sint)dt= \frac{1}{4}  \frac{ \overbrace{ \Gamma ( \frac{1}{2} ) \Gamma ( \frac{1}{2} )} = \sqrt{ \pi }  \sqrt{ \pi } }{ \underbrace{ \Gamma (1)}=1 } [ \psi ( \frac{1}{2} )- \psi (1)] \leadsto  \psi (m+ \frac{1}{2} )=2 \psi (2m)- \psi(m) -2ln2 \wedge m= \frac{1}{2} : \psi ( \frac{1}{2} )=- \gamma -2ln2 \wedge  \psi (1)=- \gamma  \Longrightarrow  \int _0^ \frac{ \pi }{2} log(sint)dt=- \frac{ \pi }{2} ln2 \Longrightarrow I_1=- \int _0^ \frac{ \pi }{2} log(sint)dt= \frac{ \pi }{2} log2 $$$$I_2= \int _0^ \frac{ \pi }{2}  \frac{t}{1+tant} dt= \int_0^ \frac{ \pi }{2}  \frac{tcost}{sint+cost} dt \Longrightarrow IBP:I_2=t \int  \frac{cost}{sint+cost} dt- \int [ \frac{d}{dt}  \int  \frac{cost}{sint+cost} dt]dt= \frac{1}{2} t[t+log(sint+cost)]_0^ \frac{ \pi }{2} - \frac{1}{2}  \int _0^ \frac{ \pi }{2} [t+log(sint+cost)]dt= \frac{  \pi}{4} [ \frac{ \pi }{2} ]- \frac{1}{2}  \int _0^ \frac{ \pi }{2} tdt- \frac{1}{2}  \int _0^ \frac{ \pi }{2} log(sint+cost)dt= \frac{  \pi ^{2} }{8} - \frac{1}{4} [ \frac{  \pi ^{2} }{4} ]- \frac{1}{2}  \int _0^ \frac{ \pi }{2} log( \sqrt{2} cos(t- \frac{ \pi }{4} ))dt= \frac {  \pi ^{2} }{16} - \frac{1}{2}  \int _0^ \frac{ \pi }{2} log( \sqrt{2} )dt- \frac{1}{2}  \int _0^ \frac{ \pi }{2} log(cos(t- \frac{ \pi }{4} ))dt= \frac{  \pi ^{2} }{16} - \underbrace{ \frac{1}{4} log(2) \frac{ \pi }{2}= \frac{ \pi }{8} log(2) } - \frac{1}{2}  \int _0^ \frac{ \pi }{2} log[cos(t- \frac{ \pi }{4} )]dt$$$$I= \int _0^ \frac{ \pi }{2} log[cos (t- \frac{ \pi }{4} )]dt \wedge t- \frac{ \pi }{4} =u  \Longrightarrow  dt=du \wedge t=0   \wedge u= \frac{ \pi }{4}  \wedge t= \frac{ \pi }{2}  \Longrightarrow u= \frac{ \pi }{4}  \Longrightarrow I= \int _ {- \frac{ \pi }{4} } ^ { \frac{ \pi }{4} } log(cos u)du=2 \int _0^ \frac{ \pi }{4} log(cos u)du now we consider I'= \int _0^ \frac{ \pi }{4} log(cos u)du \wedge J= \int _0^ \frac{ \pi }{4} log(sinu)du \Longrightarrow I'+J= \int _0^ \frac{ \pi }{4} log(cos usinu)du= \int _0^ \frac{ \pi }{4} log ( \frac{1}{2} sin2u)du= \int _0^ \frac{ \pi }{4} log(sin2u)du- \int _0^ \frac{ \pi }{4} log(2)du= \underbrace{ \frac{1 }{2}  \int _0^ \frac{ \pi }{2} log(sinu) \leadsto 2u \longrightarrow u} du- \frac{ \pi }{4} log(2)=- \frac{ \pi }{2} log(2)$$$$I'-J= \int _0^ \frac{ \pi }{4} log ( \frac{sinu}{cosu} )du= \int _0^ \frac{ \pi }{4} log (tan u)du \wedge let tanu=w \longrightarrow du= \frac{dw}{1+ w^{2} }  \wedge when u=0,w=0 \wedge u= \frac{ \pi }{4} ,w=1 \Longrightarrow I'-J= \int _0^1 \frac{logw}{1+ w^{2} } dw \Longrightarrow we know: \beta (z)= \frac{ (-1)^{z-1} }{ \Gamma (z)}  \int _0^1 \frac{ log^{z-1} x}{1+ x^{2} } dx \Longrightarrow I'-J= \frac{ \Gamma (2)}{ (-1)^{1} }  \beta (2)=- \beta (2)=G:catalans constant \Longrightarrow J= -\frac{1}{2} (G+ \frac{ \pi }{2} log(2)),I'= \frac{1}{2} (G- \frac{ \pi }{2} log(2)) \Longrightarrow I=2I'=G- \frac{ \pi }{2} log (2) \Longrightarrow I_2= \frac{  \pi ^{2} }{16} - \frac{ \pi }{8} log (2)+ \frac{ \pi }{4} log(2)- \frac{1}{2} G.$$$$ \Longrightarrow  \frac{ \pi }{2} log(2)-[ \frac{  \pi ^{2} }{16} + \frac{ \pi }{8} log(2)- \frac{1}{2} G]= \frac{3 \pi }{8} log(2)- \frac{  \pi ^{2} }{16} + \frac{G}{2} $$

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مخرخ کسر تابع زیر انتگرال واضح نیست.

Answer 1

$$t=Arctan x \wedge x=0 \leadsto t=0 \wedge x \rightarrow \infty \wedge t= \frac{ \pi }{2} \Longrightarrow I= \int _0^ \frac{ \pi }{2} \frac{t}{tant(1+tant)} dt \wedge \frac{1}{tant(1+tant)} = \frac{A}{tant} + \frac{B}{1+tant} \Longrightarrow A=1 \wedge B=-1 \Longrightarrow I_1= \int _0^ \frac{ \pi }{2} \frac{t}{tant} dt \wedge I_2= -\int _0^ \frac{ \pi }{2} \frac{t}{1+tant} dt$$$$I_1= \int _0^ \frac{ \pi }{2} \frac{t}{tant} dt= \int _0^ \frac{ \pi }{2} t \frac{cost}{sint} dt= \underbrace{tlogsint]_0^ \frac{ \pi }{2} =0} - \int _0^ \frac{ \pi }{2} log(sint)dt=- \int _0^ \frac{ \pi }{2} log(sint)dt \wedge I(a,b)= \int _0^ \frac{ \pi }{2} cos^{2a-1}(t) sin^{2b-1} (t)dt= \frac{1}{2} \frac{ \Gamma (a) \Gamma (b)}{ \Gamma (a+b)} \Longrightarrow \int _0^ \frac{ \pi }{2} cos^{2a-1}(t) sin^{2b-1} (t)log(sint)dt= \frac{1}{2} \frac{\partial}{\partial b} \frac{ \Gamma (a) \Gamma (b)}{ \Gamma (a+b)} = \frac{1}{4} \frac{ \Gamma (a) \Gamma (b)}{ \Gamma (a+b)} [ \psi (b)- \psi (a+b)]dt$$$$ \longrightarrow a=b= \frac{1}{2} : \int _0^ \frac{ \pi }{2} log(sint)dt= \frac{1}{4} \frac{ \overbrace{ \Gamma ( \frac{1}{2} ) \Gamma ( \frac{1}{2} )} = \sqrt{ \pi } \sqrt{ \pi } }{ \underbrace{ \Gamma (1)}=1 } [ \psi ( \frac{1}{2} )- \psi (1)] \leadsto \psi (m+ \frac{1}{2} )=2 \psi (2m)- \psi(m) -2ln2 \wedge m= \frac{1}{2} : \psi ( \frac{1}{2} )=- \gamma -2ln2 \wedge \psi (1)=- \gamma \Longrightarrow \int _0^ \frac{ \pi }{2} log(sint)dt=- \frac{ \pi }{2} ln2 \Longrightarrow I_1=- \int _0^ \frac{ \pi }{2} log(sint)dt= \frac{ \pi }{2} log2 $$$$I_2= \int _0^ \frac{ \pi }{2} \frac{t}{1+tant} dt= \int_0^ \frac{ \pi }{2} \frac{tcost}{sint+cost} dt \Longrightarrow IBP:I_2=t \int \frac{cost}{sint+cost} dt- \int [ \frac{d}{dt} \int \frac{cost}{sint+cost} dt]dt= \frac{1}{2} t[t+log(sint+cost)]_0^ \frac{ \pi }{2} - \frac{1}{2} \int _0^ \frac{ \pi }{2} [t+log(sint+cost)]dt= \frac{ \pi}{4} [ \frac{ \pi }{2} ]- \frac{1}{2} \int _0^ \frac{ \pi }{2} tdt- \frac{1}{2} \int _0^ \frac{ \pi }{2} log(sint+cost)dt= \frac{ \pi ^{2} }{8} - \frac{1}{4} [ \frac{ \pi ^{2} }{4} ]- \frac{1}{2} \int _0^ \frac{ \pi }{2} log( \sqrt{2} cos(t- \frac{ \pi }{4} ))dt= \frac { \pi ^{2} }{16} - \frac{1}{2} \int _0^ \frac{ \pi }{2} log( \sqrt{2} )dt- \frac{1}{2} \int _0^ \frac{ \pi }{2} log(cos(t- \frac{ \pi }{4} ))dt= \frac{ \pi ^{2} }{16} - \underbrace{ \frac{1}{4} log(2) \frac{ \pi }{2}= \frac{ \pi }{8} log(2) } - \frac{1}{2} \int _0^ \frac{ \pi }{2} log[cos(t- \frac{ \pi }{4} )]dt$$$$I= \int _0^ \frac{ \pi }{2} log[cos (t- \frac{ \pi }{4} )]dt \wedge t- \frac{ \pi }{4} =u \Longrightarrow dt=du \wedge t=0 \wedge u= \frac{ \pi }{4} \wedge t= \frac{ \pi }{2} \Longrightarrow u= \frac{ \pi }{4} \Longrightarrow I= \int _ {- \frac{ \pi }{4} } ^ { \frac{ \pi }{4} } log(cos u)du=2 \int _0^ \frac{ \pi }{4} log(cos u)du now we consider I'= \int _0^ \frac{ \pi }{4} log(cos u)du \wedge J= \int _0^ \frac{ \pi }{4} log(sinu)du \Longrightarrow I'+J= \int _0^ \frac{ \pi }{4} log(cos usinu)du= \int _0^ \frac{ \pi }{4} log ( \frac{1}{2} sin2u)du= \int _0^ \frac{ \pi }{4} log(sin2u)du- \int _0^ \frac{ \pi }{4} log(2)du= \underbrace{ \frac{1 }{2} \int _0^ \frac{ \pi }{2} log(sinu) \leadsto 2u \longrightarrow u} du- \frac{ \pi }{4} log(2)=- \frac{ \pi }{2} log(2)$$$$I'-J= \int _0^ \frac{ \pi }{4} log ( \frac{sinu}{cosu} )du= \int _0^ \frac{ \pi }{4} log (tan u)du \wedge let tanu=w \longrightarrow du= \frac{dw}{1+ w^{2} } \wedge when u=0,w=0 \wedge u= \frac{ \pi }{4} ,w=1 \Longrightarrow I'-J= \int _0^1 \frac{logw}{1+ w^{2} } dw \Longrightarrow we know: \beta (z)= \frac{ (-1)^{z-1} }{ \Gamma (z)} \int _0^1 \frac{ log^{z-1} x}{1+ x^{2} } dx \Longrightarrow I'-J= \frac{ \Gamma (2)}{ (-1)^{1} } \beta (2)=- \beta (2)=G:catalans constant \Longrightarrow J= -\frac{1}{2} (G+ \frac{ \pi }{2} log(2)),I'= \frac{1}{2} (G- \frac{ \pi }{2} log(2)) \Longrightarrow I=2I'=G- \frac{ \pi }{2} log (2) \Longrightarrow I_2= \frac{ \pi ^{2} }{16} - \frac{ \pi }{8} log (2)+ \frac{ \pi }{4} log(2)- \frac{1}{2} G.$$$$ \Longrightarrow \frac{ \pi }{2} log(2)-[ \frac{ \pi ^{2} }{16} + \frac{ \pi }{8} log(2)- \frac{1}{2} G]= \frac{3 \pi }{8} log(2)- \frac{ \pi ^{2} }{16} + \frac{G}{2} $$