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Question Number 116097 by mathmax by abdo last updated on 30/Sep/20
calculate ∫_0 ^∞ ((lnx)/(x^4 +1))dx
$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{lnx}}{\mathrm{x}^{\mathrm{4}} \:+\mathrm{1}}\mathrm{dx}\: \\ $$
Answered by mnjuly1970 last updated on 01/Oct/20
solution: Ω=∫_0 ^( ∞) ((ln(x))/(1+x^4 )) =??? x^4 =t ⇒ x=t^(1/4) ⇒dx=(1/4)t^((−3)/4) Ω =(1/(16))∫_0 ^( ∞) ((ln(t)t^(−(3/4)) )/(1+t))dt f(λ)=∫_(0 ) ^( ∞) (t^(−λ) /(1+t))dt ⇒ Ω =((−1)/(16)) f ′((3/4)) but:: f(λ) = Γ(λ)Γ(1−λ) ; 0<λ<1 f(λ)=(π/(sin(λπ))) ⇒ f ′(λ)=[− ((π^2 cos(πλ))/(sin^2 (πλ)))]_(λ=(3/4)) f ′((3/4))=π^2 ∗(((√2)/2)/(1/2)) =π^2 (√2) Ω:=− (1/(16))f ′((3/4))=−((π^2 (√2))/(16)) ✓ ...m.n.july.1970...
$$\:\:{solution}:\:\:\:\Omega=\int_{\mathrm{0}} ^{\:\infty} \frac{{ln}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{4}} }\:=??? \\ $$$${x}^{\mathrm{4}} ={t}\:\Rightarrow\:{x}={t}^{\frac{\mathrm{1}}{\mathrm{4}}} \Rightarrow{dx}=\frac{\mathrm{1}}{\mathrm{4}}{t}^{\frac{−\mathrm{3}}{\mathrm{4}}} \\ $$$$\:\Omega\:=\frac{\mathrm{1}}{\mathrm{16}}\int_{\mathrm{0}} ^{\:\infty} \frac{{ln}\left({t}\right){t}^{−\frac{\mathrm{3}}{\mathrm{4}}} }{\mathrm{1}+{t}}{dt} \\ $$$$\:{f}\left(\lambda\right)=\int_{\mathrm{0}\:\:} ^{\:\infty} \frac{{t}^{−\lambda} }{\mathrm{1}+{t}}{dt}\:\Rightarrow\:\:\Omega\:=\frac{−\mathrm{1}}{\mathrm{16}}\:{f}\:'\left(\frac{\mathrm{3}}{\mathrm{4}}\right) \\ $$$${but}::\:\:{f}\left(\lambda\right)\:=\:\Gamma\left(\lambda\right)\Gamma\left(\mathrm{1}−\lambda\right)\:\:;\:\:\:\mathrm{0}<\lambda<\mathrm{1}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{f}\left(\lambda\right)=\frac{\pi}{{sin}\left(\lambda\pi\right)}\:\Rightarrow\:{f}\:'\left(\lambda\right)=\left[−\:\frac{\pi^{\mathrm{2}} {cos}\left(\pi\lambda\right)}{{sin}^{\mathrm{2}} \left(\pi\lambda\right)}\right]_{\lambda=\frac{\mathrm{3}}{\mathrm{4}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{f}\:'\left(\frac{\mathrm{3}}{\mathrm{4}}\right)=\pi^{\mathrm{2}} \:\ast\frac{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}{\frac{\mathrm{1}}{\mathrm{2}}}\:=\pi^{\mathrm{2}} \:\sqrt{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\Omega:=−\:\frac{\mathrm{1}}{\mathrm{16}}{f}\:'\left(\frac{\mathrm{3}}{\mathrm{4}}\right)=−\frac{\pi^{\mathrm{2}} \:\sqrt{\mathrm{2}}}{\mathrm{16}}\:\:\checkmark \\ $$$$\:\:\:\:\:…{m}.{n}.{july}.\mathrm{1970}… \\ $$$$\:\:\: \\ $$$$\: \\ $$$$\: \\ $$$$ \\ $$$$ \\ $$
Commented by 1549442205PVT last updated on 01/Oct/20
third line →^(?) fourth line( lnt missed?)
$$\mathrm{third}\:\mathrm{line}\overset{?} {\:\rightarrow}\mathrm{fourth}\:\mathrm{line}\left(\:\mathrm{lnt}\:\mathrm{missed}?\right) \\ $$
Commented by mnjuly1970 last updated on 01/Oct/20
method: feynman′s trick please look at again.
$$\:{method}:\:{feynman}'{s}\:{trick} \\ $$$${please}\:{look}\:{at}\:{again}. \\ $$$$ \\ $$$$\:\:\: \\ $$
Commented by 1549442205PVT last updated on 01/Oct/20
Thank Sir,i understanded.(a^x )′=a^x lna and put t=(1/x)−1⇒f(λ)=∫_0 ^1 x^(−λ) (1−x)^(1−λ) =β(λ,1−λ)=Γ(λ)Γ(1−λ)=(π/(sin(λπ))) Great Sir thanh you very much
$$\mathrm{Thank}\:\mathrm{Sir},\mathrm{i}\:\mathrm{understanded}.\left(\mathrm{a}^{\mathrm{x}} \right)'=\mathrm{a}^{\mathrm{x}} \mathrm{lna} \\ $$$$\mathrm{and}\:\mathrm{put}\:\mathrm{t}=\frac{\mathrm{1}}{\mathrm{x}}−\mathrm{1}\Rightarrow\mathrm{f}\left(\lambda\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}^{−\lambda} \left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{1}−\lambda} =\beta\left(\lambda,\mathrm{1}−\lambda\right)=\Gamma\left(\lambda\right)\Gamma\left(\mathrm{1}−\lambda\right)=\frac{\pi}{\mathrm{sin}\left(\lambda\pi\right)} \\ $$$$\mathrm{Great}\:\mathrm{Sir}\:\mathrm{thanh}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much} \\ $$
Commented by mnjuly1970 last updated on 01/Oct/20
you are welcome.
$${you}\:{are}\:{welcome}. \\ $$
Commented by Tawa11 last updated on 06/Sep/21
grest sir
$$\mathrm{grest}\:\mathrm{sir} \\ $$
Answered by mathdave last updated on 01/Oct/20
solution let I=∫_0 ^1 ((lnx)/(x^4 +1))dx+∫_1 ^∞ ((lnx)/(1+x^4 ))dx=A+B putting x=(1/x) into B I=∫_0 ^1 ((lnx)/(1+x^4 ))dx+∫_1 ^0 ((ln((1/x)))/(1+(1/x^4 )))×−(1/x^2 )dx=∫_0 ^1 ((lnx)/(1+x^4 ))dx−∫_0 ^1 ((x^2 lnx)/(1+x^4 ))dx I=(−)^n Σ_(n=0) ^∞ ∫_0 ^1 x^(4n) lnxdx−(−1)^n Σ_(n=0) ^∞ ∫_0 ^1 x^2 .x^(4n) lnxdx I=(−1)^n Σ_(n=0) ^∞ (∂/∂a)∣_(a=1) ∫_0 ^1 x^(4n) .x^(a−1) dx−(−1)^n Σ_(n=0) ^∞ (∂/∂a)∣_(a=1) ∫_0 ^1 x^(4n+2) .x^(a−1) dx I=Σ_(n=0) ^∞ (∂/∂a)∣_(a=1) (((−1)^n )/((4n+a)))−Σ_(n=0) ^∞ (∂/∂a)∣_(a=1) (((−1)^n )/((4n+2+a))) I=−Σ_(n=0) ^∞ (((−1)^n )/((4n+1)^2 ))+Σ_(n=0) ^∞ (((−1)^n )/((4n+3)^2 ))=−Σ_(n=−∞) ^∞ (((−1)^n )/((4n+1)^2 )) note Σ_(n=−∞) ^∞ (((−1)^n )/((ax+1)^n ))=−(π/a^n )lim_(z→−(1/a)) (1/((n−1)!))•(d^(n−1) /dz^(n−1) )(cosec(πz)) I=−Σ_(n=−∞) ^∞ (1/((4n+1)^2 ))=−(−(π/4^2 ))lim_(z→−(1/4)) (1/((2−1)!))•(d/dz)(cosec(πz)) I=(π/(16))lim_(z→−(1/4)) (1/(1!))(−πcosec(πz)cot(πz)) I=−(π^2 /(16))cosec(−(π/4))cot(−(π/4))=−(π^2 /(16))(−(√2))(−1)=−(π^2 /(16))(√2) ∵∫_0 ^∞ ((lnx)/(1+x^4 ))dx=−(π^2 /(8(√2))) by mathdave(01/10/2020)
$${solution} \\ $$$${let}\:{I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}{x}}{{x}^{\mathrm{4}} +\mathrm{1}}{dx}+\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{ln}{x}}{\mathrm{1}+{x}^{\mathrm{4}} }{dx}={A}+{B} \\ $$$${putting}\:\:{x}=\frac{\mathrm{1}}{{x}}\:\:\:{into}\:{B} \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}{x}}{\mathrm{1}+{x}^{\mathrm{4}} }{dx}+\int_{\mathrm{1}} ^{\mathrm{0}} \frac{\mathrm{ln}\left(\frac{\mathrm{1}}{{x}}\right)}{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{4}} }}×−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}{x}}{\mathrm{1}+{x}^{\mathrm{4}} }{dx}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} \mathrm{ln}{x}}{\mathrm{1}+{x}^{\mathrm{4}} }{dx} \\ $$$${I}=\left(−\right)^{{n}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{4}{n}} \mathrm{ln}{xdx}−\left(−\mathrm{1}\right)^{{n}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}} .{x}^{\mathrm{4}{n}} \mathrm{ln}{xdx} \\ $$$${I}=\left(−\mathrm{1}\right)^{{n}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\partial}{\partial{a}}\mid_{{a}=\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{4}{n}} .{x}^{{a}−\mathrm{1}} {dx}−\left(−\mathrm{1}\right)^{{n}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\partial}{\partial{a}}\mid_{{a}=\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{4}{n}+\mathrm{2}} .{x}^{{a}−\mathrm{1}} {dx} \\ $$$${I}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\partial}{\partial{a}}\mid_{{a}=\mathrm{1}} \frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{4}{n}+{a}\right)}−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\partial}{\partial{a}}\mid_{{a}=\mathrm{1}} \frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{4}{n}+\mathrm{2}+{a}\right)} \\ $$$${I}=−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{4}{n}+\mathrm{1}\right)^{\mathrm{2}} }+\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{4}{n}+\mathrm{3}\right)^{\mathrm{2}} }=−\underset{{n}=−\infty} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{4}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${note}\:\underset{{n}=−\infty} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({ax}+\mathrm{1}\right)^{{n}} }=−\frac{\pi}{{a}^{{n}} }\underset{{z}\rightarrow−\frac{\mathrm{1}}{{a}}} {\mathrm{lim}}\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}\bullet\frac{{d}^{{n}−\mathrm{1}} }{{dz}^{{n}−\mathrm{1}} }\left(\mathrm{cosec}\left(\pi{z}\right)\right) \\ $$$${I}=−\underset{{n}=−\infty} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{4}{n}+\mathrm{1}\right)^{\mathrm{2}} }=−\left(−\frac{\pi}{\mathrm{4}^{\mathrm{2}} }\right)\underset{{z}\rightarrow−\frac{\mathrm{1}}{\mathrm{4}}} {\mathrm{lim}}\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\bullet\frac{{d}}{{dz}}\left({cosec}\left(\pi{z}\right)\right) \\ $$$${I}=\frac{\pi}{\mathrm{16}}\underset{{z}\rightarrow−\frac{\mathrm{1}}{\mathrm{4}}} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{1}!}\left(−\pi\mathrm{cosec}\left(\pi{z}\right)\mathrm{cot}\left(\pi{z}\right)\right) \\ $$$${I}=−\frac{\pi^{\mathrm{2}} }{\mathrm{16}}\mathrm{cosec}\left(−\frac{\pi}{\mathrm{4}}\right)\mathrm{cot}\left(−\frac{\pi}{\mathrm{4}}\right)=−\frac{\pi^{\mathrm{2}} }{\mathrm{16}}\left(−\sqrt{\mathrm{2}}\right)\left(−\mathrm{1}\right)=−\frac{\pi^{\mathrm{2}} }{\mathrm{16}}\sqrt{\mathrm{2}} \\ $$$$\because\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}{x}}{\mathrm{1}+{x}^{\mathrm{4}} }{dx}=−\frac{\pi^{\mathrm{2}} }{\mathrm{8}\sqrt{\mathrm{2}}} \\ $$$${by}\:{mathdave}\left(\mathrm{01}/\mathrm{10}/\mathrm{2020}\right) \\ $$
Commented by Tawa11 last updated on 06/Sep/21
great sir
$$\mathrm{great}\:\mathrm{sir} \\ $$
Answered by Bird last updated on 01/Oct/20
let f(a) =∫_0 ^∞ (t^(a−1) /(1+t)) dt with 0<a<1 we have f(a) =(π/(sin(πa))) and f(a) =∫_0 ^∞ (e^((a−1)lnt) /(1+t))dt ⇒ f^′ (a) =∫_0 ^∞ ((lnt t^(a−1) )/(1+t))dt=∫_0 ^∞ ((t^(a−1) lnt)/(1+t))dt ∫_0 ^∞ ((lnx)/(x^4 +1))dx =_(x=t^(1/4) ) (1/4) ∫_0 ^∞ ((lnt)/(1+t)).(1/4)t^((1/4)−1) dt =(1/(16)) ∫_0 ^∞ ((t^((1/4)−1) lnt)/(1+t))dt =(1/(16))f^′ ((1/4)) but f^′ (a)=−((π^2 cos(πa))/(sin^2 (πa))) ⇒f^′ ((1/4))=−π^2 ×((1/( (√2)))/(1/2)) =((−2π^2 )/( (√2))) =−(√2)π^(2 ) ⇒ ∫_0 ^∞ ((lnx)/(1+x^4 ))dx =−((π^2 (√2))/(16))
$${let}\:{f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}\:{dt}\:\:{with}\:\mathrm{0}<{a}<\mathrm{1} \\ $$$${we}\:{have}\:{f}\left({a}\right)\:=\frac{\pi}{{sin}\left(\pi{a}\right)} \\ $$$${and}\:{f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{\left({a}−\mathrm{1}\right){lnt}} }{\mathrm{1}+{t}}{dt}\:\Rightarrow \\ $$$${f}^{'} \left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \frac{{lnt}\:{t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}=\int_{\mathrm{0}} ^{\infty} \:\frac{{t}^{{a}−\mathrm{1}} {lnt}}{\mathrm{1}+{t}}{dt} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{lnx}}{{x}^{\mathrm{4}} +\mathrm{1}}{dx}\:=_{{x}={t}^{\frac{\mathrm{1}}{\mathrm{4}}} } \frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{lnt}}{\mathrm{1}+{t}}.\frac{\mathrm{1}}{\mathrm{4}}{t}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} {dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{16}}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{t}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} \:{lnt}}{\mathrm{1}+{t}}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{16}}{f}^{'} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)\:{but}\:{f}^{'} \left({a}\right)=−\frac{\pi^{\mathrm{2}} {cos}\left(\pi{a}\right)}{{sin}^{\mathrm{2}} \left(\pi{a}\right)} \\ $$$$\Rightarrow{f}^{'} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)=−\pi^{\mathrm{2}} ×\frac{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}{\frac{\mathrm{1}}{\mathrm{2}}}\:=\frac{−\mathrm{2}\pi^{\mathrm{2}} }{\:\sqrt{\mathrm{2}}} \\ $$$$=−\sqrt{\mathrm{2}}\pi^{\mathrm{2}\:} \:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{{lnx}}{\mathrm{1}+{x}^{\mathrm{4}} }{dx}\:=−\frac{\pi^{\mathrm{2}} \sqrt{\mathrm{2}}}{\mathrm{16}} \\ $$
Commented by mnjuly1970 last updated on 01/Oct/20
excellent sir..
$${excellent}\:{sir}.. \\ $$
Commented by Bird last updated on 01/Oct/20
you are welcome
$${you}\:{are}\:{welcome} \\ $$

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