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Question Number 136210 by Ar Brandon last updated on 19/Mar/21

Show that ∫_0 ^∞ ((lnx)/((x^2 +1)^2 ))dx=−(π/4)

$$\mathrm{Show}\:\mathrm{that}\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{lnx}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dx}=−\frac{\pi}{\mathrm{4}} \\ $$

Commented by Ar Brandon last updated on 19/Mar/21

Oh ! Thank You Sir 😃

$$\mathrm{Oh}\:!\:\mathrm{Thank}\:\mathrm{You}\:\mathrm{Sir} \\ $$😃

Commented by Ar Brandon last updated on 19/Mar/21

Ω=∫_0 ^∞ ((lnx)/((x^2 +1)^2 ))dx =(∂/∂s)∣_(s=0) ∫_0 ^∞ (x^s /((x^2 +1)^2 ))dx=^(u=x^2 ) (∂/∂s)∣_(s=0) (1/2)∫_0 ^∞ (u^((s/2)−(1/2)) /((u+1)^2 ))du =(∂/∂s)∣_(s=0) (1/2)β((s/2)+(1/2), (3/2)−(s/2))=(∂/∂s)∣_(s=0) (1/2)Γ((s/2)+(1/2))Γ((3/2)−(s/2)) =(1/2)∣_(s=0) Γ((s/2)+(1/2))Γ′((3/2)−(s/2))+Γ((3/2)−(s/2))Γ′((s/2)+(1/2)) =(1/2)∙(1/2)Γ((1/2))Γ((3/2))[ψ((1/2))−ψ((3/2))]=(π/8)[ψ((1/2))−ψ((1/2))−2]=−(π/4)

$$\Omega=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{lnx}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dx} \\ $$$$\:\:\:\:\:=\frac{\partial}{\partial\mathrm{s}}\mid_{\mathrm{s}=\mathrm{0}} \int_{\mathrm{0}} ^{\infty} \frac{\mathrm{x}^{\mathrm{s}} }{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dx}\overset{\mathrm{u}=\mathrm{x}^{\mathrm{2}} } {=}\frac{\partial}{\partial\mathrm{s}}\mid_{\mathrm{s}=\mathrm{0}} \frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{u}^{\frac{\mathrm{s}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}} }{\left(\mathrm{u}+\mathrm{1}\right)^{\mathrm{2}} }\mathrm{du} \\ $$$$\:\:\:\:\:=\frac{\partial}{\partial\mathrm{s}}\mid_{\mathrm{s}=\mathrm{0}} \frac{\mathrm{1}}{\mathrm{2}}\beta\left(\frac{\mathrm{s}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}},\:\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{s}}{\mathrm{2}}\right)=\frac{\partial}{\partial\mathrm{s}}\mid_{\mathrm{s}=\mathrm{0}} \frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{s}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{s}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\mid_{\mathrm{s}=\mathrm{0}} \Gamma\left(\frac{\mathrm{s}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma'\left(\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{s}}{\mathrm{2}}\right)+\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{s}}{\mathrm{2}}\right)\Gamma'\left(\frac{\mathrm{s}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\centerdot\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\left[\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\psi\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\right]=\frac{\pi}{\mathrm{8}}\left[\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\mathrm{2}\right]=−\frac{\pi}{\mathrm{4}} \\ $$

Commented by mnjuly1970 last updated on 19/Mar/21

error in sign of drivative of ψ((3/2)−s) ....... ′ = −ψ′((3/2)−s) ...good luck my brother mr brandon...

$$\:\:{error}\:\:{in}\:{sign}\:{of}\:{drivative} \\ $$$${of}\:\:\psi\left(\frac{\mathrm{3}}{\mathrm{2}}−{s}\right)\:.......\:\:'\:=\:−\psi'\left(\frac{\mathrm{3}}{\mathrm{2}}−{s}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:...{good}\:{luck}\:{my}\:{brother}\:{mr}\:{brandon}... \\ $$

Answered by mnjuly1970 last updated on 19/Mar/21

𝛗=^(x^2 =y) (1/4)∫_0 ^( ∞) ((y^(−(1/2)) ln(y))/((y+1)^2 ))dy f(a)=∫_0 ^( ∞) (y^(a−(1/2)) /((y+1)^2 ))dy 𝛗=((f ′(0))/4) ...✓ f(a)=∫_0 ^( ∞) (y^(a+(1/2)−1) /((y+1)^2 ))dy=β(a+(1/2),(3/2)−a) f(a)=Γ(a+(1/2)).Γ((3/2)−a) f ′(a)=Γ′(a+(1/2))Γ((3/2)−a)−Γ′((3/2)−a)Γ((1/2)+a) 𝛗=(1/4) f ′(0)=(1/4)((1/2)ψ((1/2))Γ^2 ((1/2))−(1/2)ψ((3/2))Γ^2 ((1/2))) =(π/8)(ψ((1/2))−ψ((3/2))) =(π/8)(ψ((1/2))−ψ((1/2))−2)=((−π)/4) ..✓✓

$$\:\:\:\:\boldsymbol{\phi}\overset{{x}^{\mathrm{2}} ={y}} {=}\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\:\infty} \frac{{y}^{−\frac{\mathrm{1}}{\mathrm{2}}} {ln}\left({y}\right)}{\left({y}+\mathrm{1}\right)^{\mathrm{2}} }{dy} \\ $$$$\:\:\:\:{f}\left({a}\right)=\int_{\mathrm{0}} ^{\:\infty} \frac{{y}^{{a}−\frac{\mathrm{1}}{\mathrm{2}}} }{\left({y}+\mathrm{1}\right)^{\mathrm{2}} }{dy} \\ $$$$\:\:\:\:\boldsymbol{\phi}=\frac{{f}\:'\left(\mathrm{0}\right)}{\mathrm{4}}\:...\checkmark \\ $$$$\:\:\:{f}\left({a}\right)=\int_{\mathrm{0}} ^{\:\infty} \:\frac{{y}^{{a}+\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} }{\left({y}+\mathrm{1}\right)^{\mathrm{2}} }{dy}=\beta\left({a}+\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{3}}{\mathrm{2}}−{a}\right) \\ $$$$\:\:\:\:{f}\left({a}\right)=\Gamma\left({a}+\frac{\mathrm{1}}{\mathrm{2}}\right).\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}−{a}\right) \\ $$$$\:\:{f}\:'\left({a}\right)=\Gamma'\left({a}+\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}−{a}\right)−\Gamma'\left(\frac{\mathrm{3}}{\mathrm{2}}−{a}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}+{a}\right) \\ $$$$\:\:\:\boldsymbol{\phi}=\frac{\mathrm{1}}{\mathrm{4}}\:{f}\:'\left(\mathrm{0}\right)=\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{2}}\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\psi\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right) \\ $$$$\:\:\:\:\:\:\:=\frac{\pi}{\mathrm{8}}\left(\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\psi\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\right) \\ $$$$\:\:\:\:\:\:\:=\frac{\pi}{\mathrm{8}}\left(\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\mathrm{2}\right)=\frac{−\pi}{\mathrm{4}}\:..\checkmark\checkmark \\ $$$$\:\: \\ $$

Answered by Dwaipayan Shikari last updated on 19/Mar/21

(1/4)∫_0 ^∞ ((u^(−(1/2)) log(u))/((1+u)^2 ))du=(1/4)τ′((1/2))=∫_0 ^∞ ((log(x))/((x^2 +1)^2 ))dx τ(α)=∫_0 ^∞ (u^(α−1) /((1+u)^2 ))du=((Γ(α)Γ(2−α))/(Γ(2)))=((π(1−α))/(sin(πα))) ⇒τ′(α)=∫_0 ^∞ ((u^(α−1) log(u))/((1+u)^2 ))du=−π^2 cosec(πα)cot(πα)−(π/(sin(πα))) (1/4)τ′((1/2))=−0−(π/4)=−(π/4)

$$\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} \frac{{u}^{−\frac{\mathrm{1}}{\mathrm{2}}} {log}\left({u}\right)}{\left(\mathrm{1}+{u}\right)^{\mathrm{2}} }{du}=\frac{\mathrm{1}}{\mathrm{4}}\tau'\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{log}\left({x}\right)}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$$\tau\left(\alpha\right)=\int_{\mathrm{0}} ^{\infty} \frac{{u}^{\alpha−\mathrm{1}} }{\left(\mathrm{1}+{u}\right)^{\mathrm{2}} }{du}=\frac{\Gamma\left(\alpha\right)\Gamma\left(\mathrm{2}−\alpha\right)}{\Gamma\left(\mathrm{2}\right)}=\frac{\pi\left(\mathrm{1}−\alpha\right)}{{sin}\left(\pi\alpha\right)} \\ $$$$\Rightarrow\tau'\left(\alpha\right)=\int_{\mathrm{0}} ^{\infty} \frac{{u}^{\alpha−\mathrm{1}} {log}\left({u}\right)}{\left(\mathrm{1}+{u}\right)^{\mathrm{2}} }{du}=−\pi^{\mathrm{2}} {cosec}\left(\pi\alpha\right){cot}\left(\pi\alpha\right)−\frac{\pi}{{sin}\left(\pi\alpha\right)} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\tau'\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=−\mathrm{0}−\frac{\pi}{\mathrm{4}}=−\frac{\pi}{\mathrm{4}} \\ $$

Commented by Ar Brandon last updated on 19/Mar/21

Thanks bro

$$\mathrm{Thanks}\:\mathrm{bro} \\ $$

Commented by Dwaipayan Shikari last updated on 19/Mar/21

😃

$$ \\ $$😃

Answered by Ajetunmobi last updated on 19/Mar/21

i have drop the solution before here

$$ \\ $$$$\boldsymbol{\mathrm{i}}\:\boldsymbol{\mathrm{have}}\:\boldsymbol{\mathrm{drop}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{solution}}\:\boldsymbol{\mathrm{before}}\:\boldsymbol{\mathrm{here}} \\ $$

Commented by Ar Brandon last updated on 19/Mar/21

Alright ! Before posting I used the search option to check if it was previously posted but couldn′t find it. Thank You Sir

$$\mathrm{Alright}\:! \\ $$$$\mathrm{Before}\:\mathrm{posting}\:\mathrm{I}\:\mathrm{used}\:\mathrm{the}\:\mathrm{search}\:\mathrm{option}\: \\ $$$$\mathrm{to}\:\mathrm{check}\:\mathrm{if}\:\mathrm{it}\:\mathrm{was}\:\mathrm{previously}\:\mathrm{posted}\:\mathrm{but}\: \\ $$$$\mathrm{couldn}'\mathrm{t}\:\mathrm{find}\:\mathrm{it}.\:\mathrm{Thank}\:\mathrm{You}\:\mathrm{Sir} \\ $$

Commented by Ajetunmobi last updated on 19/Mar/21

ok

$$ \\ $$$$\boldsymbol{\mathrm{ok}} \\ $$

Answered by Ajetunmobi last updated on 19/Mar/21

Answered by mathmax by abdo last updated on 19/Mar/21

let f(a)=∫_0 ^∞ ((lnx)/(x^2 +a^2 ))dx with a>0 ⇒f^′ (a)=−2a∫_0 ^∞ ((lnx)/((x^2 +a^2 )^2 ))dx ⇒ f^′ (1)=−2∫_0 ^∞ ((lnx)/((x^2 +1)^2 )) ⇒∫_0 ^∞ ((lnx)/((x^2 +1)^2 ))=−(1/2)f^′ (1) f(a) =_(x=at) ∫_0 ^∞ ((lna +lnt)/(a^2 (1+t^2 )))adt =(1/a)∫_0 ^∞ ((lna+lnt)/((1+t^2 )))dt =((lna)/a)∫_0 ^∞ (dt/(1+t^2 )) +(1/a)∫_0 ^∞ ((lnt)/(1+t^2 ))dt but ∫_0 ^∞ ((lnt)/(1+t^2 ))dt=0(proved) ⇒ f(a)=((lna)/a).(π/2) =((πlna)/(2a)) ⇒f^′ (a)=(((π/a)(2a)−2πlna)/(4a^2 )) ⇒f^′ (1)=((2π)/4)=(π/2) ⇒ ∫_0 ^∞ ((lnx)/((x^2 +1)^2 ))dx =−(1/2).(π/2)=−(π/4)

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{lnx}}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{a}^{\mathrm{2}} }\mathrm{dx}\:\mathrm{with}\:\mathrm{a}>\mathrm{0}\:\Rightarrow\mathrm{f}^{'} \left(\mathrm{a}\right)=−\mathrm{2a}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{lnx}}{\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{a}^{\mathrm{2}} \right)^{\mathrm{2}} }\mathrm{dx}\:\Rightarrow \\ $$$$\mathrm{f}^{'} \left(\mathrm{1}\right)=−\mathrm{2}\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{lnx}}{\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{lnx}}{\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{f}^{'} \left(\mathrm{1}\right) \\ $$$$\mathrm{f}\left(\mathrm{a}\right)\:=_{\mathrm{x}=\mathrm{at}} \:\:\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{lna}\:+\mathrm{lnt}}{\mathrm{a}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)}\mathrm{adt}\:=\frac{\mathrm{1}}{\mathrm{a}}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{lna}+\mathrm{lnt}}{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)}\mathrm{dt} \\ $$$$=\frac{\mathrm{lna}}{\mathrm{a}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{dt}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{a}}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{lnt}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt}\:\:\mathrm{but}\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{lnt}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt}=\mathrm{0}\left(\mathrm{proved}\right)\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{a}\right)=\frac{\mathrm{lna}}{\mathrm{a}}.\frac{\pi}{\mathrm{2}}\:=\frac{\pi\mathrm{lna}}{\mathrm{2a}}\:\Rightarrow\mathrm{f}^{'} \left(\mathrm{a}\right)=\frac{\frac{\pi}{\mathrm{a}}\left(\mathrm{2a}\right)−\mathrm{2}\pi\mathrm{lna}}{\mathrm{4a}^{\mathrm{2}} }\:\Rightarrow\mathrm{f}^{'} \left(\mathrm{1}\right)=\frac{\mathrm{2}\pi}{\mathrm{4}}=\frac{\pi}{\mathrm{2}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{lnx}}{\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dx}\:=−\frac{\mathrm{1}}{\mathrm{2}}.\frac{\pi}{\mathrm{2}}=−\frac{\pi}{\mathrm{4}} \\ $$

Commented by mathmax by abdo last updated on 19/Mar/21

folow my facebook (abdo imad)

$$\mathrm{folow}\:\mathrm{my}\:\mathrm{facebook}\:\:\left(\mathrm{abdo}\:\mathrm{imad}\right) \\ $$

Commented by Ajetunmobi last updated on 19/Mar/21

sir mathmax by abdo how can we chat privately sir i need to talk about something with you you can send me your whatsapp number or your facebook name Thanks

$$ \\ $$$$\boldsymbol{\mathrm{sir}}\:\boldsymbol{\mathrm{mathmax}}\:\boldsymbol{\mathrm{by}}\:\boldsymbol{\mathrm{abdo}} \\ $$$$\boldsymbol{\mathrm{how}}\:\boldsymbol{\mathrm{can}}\:\boldsymbol{\mathrm{we}}\:\boldsymbol{\mathrm{chat}}\:\boldsymbol{\mathrm{privately}}\:\boldsymbol{\mathrm{sir}}\:\boldsymbol{\mathrm{i}}\:\boldsymbol{\mathrm{need}}\:\boldsymbol{\mathrm{to}} \\ $$$$\boldsymbol{\mathrm{talk}}\:\boldsymbol{\mathrm{about}}\:\boldsymbol{\mathrm{something}}\:\boldsymbol{\mathrm{with}}\:\boldsymbol{\mathrm{you}}\: \\ $$$$\boldsymbol{\mathrm{you}}\:\boldsymbol{\mathrm{can}}\:\boldsymbol{\mathrm{send}}\:\boldsymbol{\mathrm{me}}\:\boldsymbol{\mathrm{your}}\:\boldsymbol{\mathrm{whatsapp}}\:\boldsymbol{\mathrm{number}} \\ $$$$\boldsymbol{\mathrm{or}}\:\boldsymbol{\mathrm{your}}\:\boldsymbol{\mathrm{facebook}}\:\boldsymbol{\mathrm{name}}\: \\ $$$$\boldsymbol{\mathrm{Thanks}} \\ $$

Commented by Ajetunmobi last updated on 20/Mar/21

there are many abdo imad there how will i recognize you sir?

$$ \\ $$$$\boldsymbol{\mathrm{there}}\:\boldsymbol{\mathrm{are}}\:\boldsymbol{\mathrm{many}}\:\boldsymbol{\mathrm{abdo}}\:\boldsymbol{\mathrm{imad}}\:\boldsymbol{\mathrm{there}}\: \\ $$$$\boldsymbol{\mathrm{how}}\:\boldsymbol{\mathrm{will}}\:\boldsymbol{\mathrm{i}}\:\boldsymbol{\mathrm{recognize}}\:\boldsymbol{\mathrm{you}}\:\boldsymbol{\mathrm{sir}}? \\ $$

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