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Question Number 108750 by mathmax by abdo last updated on 19/Aug/20
calculste ∫_0 ^∞ ((ln(x))/(x^2 −x+1))dx
$$\mathrm{calculste}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{ln}\left(\mathrm{x}\right)}{\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}}\mathrm{dx} \\ $$
Answered by mnjuly1970 last updated on 19/Aug/20
sol....: put : x=(1/t)⇒Ω= ∫_0 ^( ∞) ((ln(x))/(x^2 −x+1)) dx=−Ω 2Ω=0 ⇒Ω=∫_0 ^( ∞) ((ln(x))/(1−x+x^2 )) dx =0 .....M.N.....
$$\:\mathrm{sol}….:\:\mathrm{put}\::\:{x}=\frac{\mathrm{1}}{{t}}\Rightarrow\Omega=\:\int_{\mathrm{0}} ^{\:\infty} \frac{{ln}\left({x}\right)}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}\:{dx}=−\Omega \\ $$$$\mathrm{2}\Omega=\mathrm{0}\:\Rightarrow\Omega=\int_{\mathrm{0}} ^{\:\infty} \frac{{ln}\left({x}\right)}{\mathrm{1}−{x}+{x}^{\mathrm{2}} }\:{dx}\:=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…..\mathscr{M}.\mathscr{N}….. \\ $$$$ \\ $$
Answered by mathmax by abdo last updated on 19/Aug/20
if q is a fraction with no real poles we have ∫_0 ^∞ q(x)lnx dx =−(1/2)Re(Σ_i Re( q(z)ln^2 z,a_i ) let use this we have q(z)ln^2 z =((ln^2 z)/(z^2 −z+1)) =w(z) poles? z^2 −z+1 =0 →Δ =−3 ⇒z_1 =((1+i(√3))/2) =e^((i2π)/3) and z_2 =((1−i(√3))/2) =e^(−((i2π)/3)) w(z) =((ln^2 z)/((z−z_1 )(z−z_2 ))) ⇒Res(w,z_1 ) =((ln^2 (z_1 ))/(z_1 −z_2 )) =((ln^2 (e^((i2π)/3) ))/(i(√3))) =(1/(i(√3)))(((i2π)/3))^2 =((−4π^2 )/(3i(√3))) also Res(w,z_2 ) =((ln^2 (z_2 ))/(z_2 −z_1 )) =((4π^2 )/(3i(√3))) ⇒ Σ Re(q(z)ln^2 (z),a_i )=0 ⇒∫_0 ^∞ ((ln(x))/(x^2 −x+1))dx =0
$$\mathrm{if}\:\mathrm{q}\:\mathrm{is}\:\mathrm{a}\:\mathrm{fraction}\:\mathrm{with}\:\mathrm{no}\:\mathrm{real}\:\mathrm{poles}\:\mathrm{we}\:\mathrm{have} \\ $$$$\int_{\mathrm{0}} ^{\infty} \mathrm{q}\left(\mathrm{x}\right)\mathrm{lnx}\:\mathrm{dx}\:=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{Re}\left(\sum_{\mathrm{i}} \:\mathrm{Re}\left(\:\mathrm{q}\left(\mathrm{z}\right)\mathrm{ln}^{\mathrm{2}} \mathrm{z},\mathrm{a}_{\mathrm{i}} \right)\:\mathrm{let}\:\mathrm{use}\:\mathrm{this}\right. \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{q}\left(\mathrm{z}\right)\mathrm{ln}^{\mathrm{2}} \mathrm{z}\:=\frac{\mathrm{ln}^{\mathrm{2}} \mathrm{z}}{\mathrm{z}^{\mathrm{2}} −\mathrm{z}+\mathrm{1}}\:=\mathrm{w}\left(\mathrm{z}\right)\:\mathrm{poles}? \\ $$$$\mathrm{z}^{\mathrm{2}} −\mathrm{z}+\mathrm{1}\:=\mathrm{0}\:\rightarrow\Delta\:=−\mathrm{3}\:\Rightarrow\mathrm{z}_{\mathrm{1}} =\frac{\mathrm{1}+\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:=\mathrm{e}^{\frac{\mathrm{i2}\pi}{\mathrm{3}}} \:\mathrm{and}\:\mathrm{z}_{\mathrm{2}} =\frac{\mathrm{1}−\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:=\mathrm{e}^{−\frac{\mathrm{i2}\pi}{\mathrm{3}}} \\ $$$$\mathrm{w}\left(\mathrm{z}\right)\:=\frac{\mathrm{ln}^{\mathrm{2}} \mathrm{z}}{\left(\mathrm{z}−\mathrm{z}_{\mathrm{1}} \right)\left(\mathrm{z}−\mathrm{z}_{\mathrm{2}} \right)}\:\Rightarrow\mathrm{Res}\left(\mathrm{w},\mathrm{z}_{\mathrm{1}} \right)\:=\frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{z}_{\mathrm{1}} \right)}{\mathrm{z}_{\mathrm{1}} −\mathrm{z}_{\mathrm{2}} }\:=\frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{e}^{\frac{\mathrm{i2}\pi}{\mathrm{3}}} \right)}{\mathrm{i}\sqrt{\mathrm{3}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{i}\sqrt{\mathrm{3}}}\left(\frac{\mathrm{i2}\pi}{\mathrm{3}}\right)^{\mathrm{2}} \:=\frac{−\mathrm{4}\pi^{\mathrm{2}} }{\mathrm{3i}\sqrt{\mathrm{3}}}\:\:\mathrm{also}\:\mathrm{Res}\left(\mathrm{w},\mathrm{z}_{\mathrm{2}} \right)\:=\frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{z}_{\mathrm{2}} \right)}{\mathrm{z}_{\mathrm{2}} −\mathrm{z}_{\mathrm{1}} }\:=\frac{\mathrm{4}\pi^{\mathrm{2}} }{\mathrm{3i}\sqrt{\mathrm{3}}}\:\Rightarrow \\ $$$$\Sigma\:\mathrm{Re}\left(\mathrm{q}\left(\mathrm{z}\right)\mathrm{ln}^{\mathrm{2}} \left(\mathrm{z}\right),\mathrm{a}_{\mathrm{i}} \right)=\mathrm{0}\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{ln}\left(\mathrm{x}\right)}{\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}}\mathrm{dx}\:=\mathrm{0} \\ $$$$ \\ $$

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