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Question Number 136555 by mnjuly1970 last updated on 23/Mar/21
....nice ...... calculus.... prove that 𝛗=∫_0 ^( ∞) ((ln(1+x^2 ))/((1+x^2 )^2 ))dx=(π/2)ln(2)−(π/4) ::::::::
$$\:\:\:\:\:\:\:\:\:….{nice}\:\:\:\:……\:\:\:{calculus}…. \\ $$$$\:\:\:{prove}\:{that} \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\infty} \frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}=\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right)−\frac{\pi}{\mathrm{4}} \\ $$$$\:\:\:\:\::::::::: \\ $$
Answered by mnjuly1970 last updated on 23/Mar/21
f(a)=∫_0 ^( ∞) (dx/((1+x^2 )^a )) ...(a>(1/2)) f(a)=^(x^2 =y) (1/2)∫_0 ^( ∞) (y^((−1)/2) /((1+y)^a ))dy=β((1/2),a−(1/2)) f(a)=(1/2)((((√π) Γ(a−(1/2)))/(Γ(a)))) goal:: 𝛗=−f ′(2) f ′(a)=((√π)/2) (((Γ′(a−(1/2)).Γ(a)−Γ′(a).Γ(a−(1/2)))/(Γ^2 (a)))) f ′(2)=((√π)/2)(((ψ((3/2))Γ((3/2))−ψ(2)Γ((3/2) ))/1^2 )) =((√π)/2)(((√π)/2)(2−γ−2ln(2)−(1−γ))) ∴ −𝛗=(π/4)(2−γ−2ln(2)−1+γ) =(π/4)(1−2ln(2))=(π/4)−(π/2)ln(2)..✓ ∴ 𝛗=(π/2)ln(2)−(π/4) .... m.n
$$\:\:\:{f}\left({a}\right)=\int_{\mathrm{0}} ^{\:\infty} \frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{a}} }\:\:\:\:…\left({a}>\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:{f}\left({a}\right)\overset{{x}^{\mathrm{2}} ={y}} {=}\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\infty} \frac{{y}^{\frac{−\mathrm{1}}{\mathrm{2}}} }{\left(\mathrm{1}+{y}\right)^{{a}} }{dy}=\beta\left(\frac{\mathrm{1}}{\mathrm{2}},{a}−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:{f}\left({a}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\sqrt{\pi}\:\Gamma\left({a}−\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left({a}\right)}\right) \\ $$$$\:\:\:\:\:{goal}::\:\:\boldsymbol{\phi}=−{f}\:'\left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{f}\:'\left({a}\right)=\frac{\sqrt{\pi}}{\mathrm{2}}\:\left(\frac{\Gamma'\left({a}−\frac{\mathrm{1}}{\mathrm{2}}\right).\Gamma\left({a}\right)−\Gamma'\left({a}\right).\Gamma\left({a}−\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma^{\mathrm{2}} \left({a}\right)}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:{f}\:'\left(\mathrm{2}\right)=\frac{\sqrt{\pi}}{\mathrm{2}}\left(\frac{\psi\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)−\psi\left(\mathrm{2}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\:\right)}{\mathrm{1}^{\mathrm{2}} }\right) \\ $$$$\:\:\:\:=\frac{\sqrt{\pi}}{\mathrm{2}}\left(\frac{\sqrt{\pi}}{\mathrm{2}}\left(\mathrm{2}−\gamma−\mathrm{2}{ln}\left(\mathrm{2}\right)−\left(\mathrm{1}−\gamma\right)\right)\right) \\ $$$$\:\:\:\:\:\:\:\therefore\:\:\:−\boldsymbol{\phi}=\frac{\pi}{\mathrm{4}}\left(\mathrm{2}−\gamma−\mathrm{2}{ln}\left(\mathrm{2}\right)−\mathrm{1}+\gamma\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\pi}{\mathrm{4}}\left(\mathrm{1}−\mathrm{2}{ln}\left(\mathrm{2}\right)\right)=\frac{\pi}{\mathrm{4}}−\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right)..\checkmark \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\therefore\:\:\boldsymbol{\phi}=\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right)−\frac{\pi}{\mathrm{4}}\:…. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:{m}.{n} \\ $$
Answered by mindispower last updated on 23/Mar/21
=∫_0 ^(π/2) ((ln((1/(cos^2 (x)))))/((1+tg^2 (x))^2 ))dtg(x) =∫_0 ^(π/2) −2cos^2 (x)ln(cos(x))dx..A β(a,b)=2∫_0 ^(π/2) cos^(2a−1) (x).sin^(2b−1) (x)dx A=−(1/2)∂^a β(a,b)∣(a,b)=((3/2),(1/2)) =−(1/2)β(a,b)(Ψ(a)−Ψ(a+b)) =−(1/2).((Γ((1/2))Γ((3/2)))/(Γ(2)))(Ψ((3/2))−Ψ(2)) =−(1/2).(1/2).π(2+Ψ((1/2))−(1−γ)) =−(π/4)(1−2ln(2))=((πln(2))/4)−(π/4)
$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{ln}\left(\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \left({x}\right)}\right)}{\left(\mathrm{1}+{tg}^{\mathrm{2}} \left({x}\right)\right)^{\mathrm{2}} }{dtg}\left({x}\right) \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} −\mathrm{2}{cos}^{\mathrm{2}} \left({x}\right){ln}\left({cos}\left({x}\right)\right){dx}..{A} \\ $$$$\beta\left({a},{b}\right)=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cos}^{\mathrm{2}{a}−\mathrm{1}} \left({x}\right).{sin}^{\mathrm{2}{b}−\mathrm{1}} \left({x}\right){dx} \\ $$$${A}=−\frac{\mathrm{1}}{\mathrm{2}}\partial^{{a}} \beta\left({a},{b}\right)\mid\left({a},{b}\right)=\left(\frac{\mathrm{3}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\beta\left({a},{b}\right)\left(\Psi\left({a}\right)−\Psi\left({a}+{b}\right)\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}.\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}{\Gamma\left(\mathrm{2}\right)}\left(\Psi\left(\frac{\mathrm{3}}{\mathrm{2}}\right)−\Psi\left(\mathrm{2}\right)\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{1}}{\mathrm{2}}.\pi\left(\mathrm{2}+\Psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\left(\mathrm{1}−\gamma\right)\right) \\ $$$$=−\frac{\pi}{\mathrm{4}}\left(\mathrm{1}−\mathrm{2}{ln}\left(\mathrm{2}\right)\right)=\frac{\pi{ln}\left(\mathrm{2}\right)}{\mathrm{4}}−\frac{\pi}{\mathrm{4}} \\ $$$$ \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 23/Mar/21
thanks alot....
$${thanks}\:{alot}…. \\ $$
Answered by Dwaipayan Shikari last updated on 23/Mar/21
ϑ(α)=∫_0 ^∞ (((1+x^2 )^α )/((1+x^2 )^2 ))dx=∫_0 ^∞ (1/((1+x^2 )^(2−α) ))dx =(1/2)∫_0 ^∞ (u^((1/2)−1) /((1+u)^((1/2)+(3/2)−α) ))du=(1/2).((Γ((1/2))Γ((3/2)−α))/(Γ(2−α))) ϑ′(α)=((−(√π) Γ(2−α)Γ′((3/2)−α)+(√π) Γ(2−α)ψ(2−α)Γ((3/2)−α))/(2Γ^2 (2−α))) ϑ′(0)=((−(π/2)ψ((3/2))+(π/2)ψ(2))/2)=(π/4)(−γ+1+γ−2log(2)) =(π/4)−(π/2)log(2) −ϑ′(0)=∫_0 ^∞ ((log(x^2 +1))/((x^2 +1)^2 ))dx=(π/2)log(2)−(π/4)
$$\vartheta\left(\alpha\right)=\int_{\mathrm{0}} ^{\infty} \frac{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\alpha} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}−\alpha} }{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{{u}^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} }{\left(\mathrm{1}+{u}\right)^{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{2}}−\alpha} }{du}=\frac{\mathrm{1}}{\mathrm{2}}.\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}−\alpha\right)}{\Gamma\left(\mathrm{2}−\alpha\right)} \\ $$$$\vartheta'\left(\alpha\right)=\frac{−\sqrt{\pi}\:\Gamma\left(\mathrm{2}−\alpha\right)\Gamma'\left(\frac{\mathrm{3}}{\mathrm{2}}−\alpha\right)+\sqrt{\pi}\:\Gamma\left(\mathrm{2}−\alpha\right)\psi\left(\mathrm{2}−\alpha\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}−\alpha\right)}{\mathrm{2}\Gamma^{\mathrm{2}} \left(\mathrm{2}−\alpha\right)} \\ $$$$\vartheta'\left(\mathrm{0}\right)=\frac{−\frac{\pi}{\mathrm{2}}\psi\left(\frac{\mathrm{3}}{\mathrm{2}}\right)+\frac{\pi}{\mathrm{2}}\psi\left(\mathrm{2}\right)}{\mathrm{2}}=\frac{\pi}{\mathrm{4}}\left(−\gamma+\mathrm{1}+\gamma−\mathrm{2}{log}\left(\mathrm{2}\right)\right) \\ $$$$=\frac{\pi}{\mathrm{4}}−\frac{\pi}{\mathrm{2}}{log}\left(\mathrm{2}\right) \\ $$$$−\vartheta'\left(\mathrm{0}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{log}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{dx}=\frac{\pi}{\mathrm{2}}{log}\left(\mathrm{2}\right)−\frac{\pi}{\mathrm{4}} \\ $$
Commented by mnjuly1970 last updated on 23/Mar/21
thanking...
$${thanking}… \\ $$

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