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Question Number 162535 by mnjuly1970 last updated on 30/Dec/21
prove that Ξ© = ∫_0 ^( ∞) (( ln ((1/x) ))/( x^( 4) + 17x^( 2) + 16)) dx=^? (Ο€/(60)) ln(2)
$$ \\ $$$$\:\:\:\:\:\mathrm{prove}\:\mathrm{that} \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\:\infty} \frac{\:\:\mathrm{ln}\:\left(\frac{\mathrm{1}}{{x}}\:\right)}{\:{x}^{\:\mathrm{4}} \:+\:\mathrm{17}{x}^{\:\mathrm{2}} \:+\:\mathrm{16}}\:{dx}\overset{?} {=}\:\frac{\pi}{\mathrm{60}}\:\mathrm{ln}\left(\mathrm{2}\right) \\ $$$$ \\ $$
Commented by Ar Brandon last updated on 30/Dec/21
Synthax error, Sir. Your dx is missing. Haha πŸ˜…πŸ˜
$$\mathrm{Synthax}\:\mathrm{error},\:\mathrm{Sir}.\:\mathrm{Your}\:{dx}\:\mathrm{is}\:\mathrm{missing}. \\ $$Haha πŸ˜…πŸ˜
Commented by amin96 last updated on 30/Dec/21
very big mistakeπŸ˜‚πŸ˜‚
$$ \\ $$very big mistakeπŸ˜‚πŸ˜‚
Commented by mnjuly1970 last updated on 30/Dec/21
yes you are right sir ...
$$\:{yes}\:{you}\:{are}\:{right}\:{sir}\:\:… \\ $$
Answered by Ar Brandon last updated on 24/Mar/22
Ξ©=∫_0 ^∞ ((ln((1/x)))/(x^4 +17x^2 +16))dx=βˆ’βˆ«_0 ^∞ ((lnx)/(x^4 +17x^2 +16))dx Ο•(z)=(((lnz)^2 )/(z^4 +17z^2 +16)) ,poles:(z^2 +16)(z^2 +1)=0β‡’z_1 =4e^((Ο€/2)i) ,z_2 =4e^(βˆ’(Ο€/2)i) ,z_3 =e^((Ο€/2)i) ,z_4 =e^(βˆ’(Ο€/2)i) Ξ©=βˆ’(1/2)Re(Res(Ο•, z_1 )+Res(Ο•, z_2 )+Res(Ο•, z_3 )+Res(Ο•, z_4 )) Res (Ο•, z_1 )=lim_(zβ†’z_1 ) ((((lnz)^2 )/((zβˆ’z_2 )(zβˆ’z_3 )(zβˆ’z_4 ))))=(((ln4+(Ο€/2)i)^2 )/((8i)(3i)(5i)))=(i/(120))(ln^2 4βˆ’(Ο€^2 /4)+iΟ€ln4) Res(Ο•, z_2 )=lim_(zβ†’z_2 ) ((((lnz)^2 )/((zβˆ’z_1 )(zβˆ’z_3 )(zβˆ’z_4 ))))=(((ln4βˆ’(Ο€/2)i)^2 )/((βˆ’8i)(βˆ’5i)(βˆ’3i)))=βˆ’(i/(120))(ln^2 4+(Ο€^2 /4)βˆ’iΟ€ln4) Res(Ο•, z_3 )=lim_(zβ†’z_3 ) ((((lnz)^2 )/((zβˆ’z_1 )(zβˆ’z_2 )(zβˆ’z_4 ))))=((((Ο€/2)i)^2 )/((βˆ’3i)(5i)(2i)))=(i/(30))((Ο€^2 /4))=i(Ο€^2 /(120)) Res(Ο•,z_4 )=lim_(zβ†’z_4 ) ((((lnz)^2 )/((zβˆ’z_1 )(zβˆ’z_2 )(zβˆ’z_3 ))))=(((βˆ’(Ο€/2)i)^2 )/((βˆ’5i)(3i)(βˆ’2i)))=βˆ’(i/(30))((Ο€^2 /4))=βˆ’i(Ο€^2 /(120)) Ξ©=βˆ’(1/2)(βˆ’((Ο€ln4)/(120))βˆ’((Ο€ln4)/(120)))=((Ο€ln4)/(120))=(Ο€/(60))ln(2)
$$\Omega=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}\left(\frac{\mathrm{1}}{{x}}\right)}{{x}^{\mathrm{4}} +\mathrm{17}{x}^{\mathrm{2}} +\mathrm{16}}{dx}=βˆ’\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}{x}}{{x}^{\mathrm{4}} +\mathrm{17}{x}^{\mathrm{2}} +\mathrm{16}}{dx} \\ $$$$\varphi\left({z}\right)=\frac{\left(\mathrm{ln}{z}\right)^{\mathrm{2}} }{{z}^{\mathrm{4}} +\mathrm{17}{z}^{\mathrm{2}} +\mathrm{16}}\:,\mathrm{poles}:\left({z}^{\mathrm{2}} +\mathrm{16}\right)\left({z}^{\mathrm{2}} +\mathrm{1}\right)=\mathrm{0}\Rightarrow{z}_{\mathrm{1}} =\mathrm{4}{e}^{\frac{\pi}{\mathrm{2}}{i}} ,{z}_{\mathrm{2}} =\mathrm{4}{e}^{βˆ’\frac{\pi}{\mathrm{2}}{i}} ,{z}_{\mathrm{3}} ={e}^{\frac{\pi}{\mathrm{2}}{i}} ,{z}_{\mathrm{4}} ={e}^{βˆ’\frac{\pi}{\mathrm{2}}{i}} \\ $$$$\Omega=βˆ’\frac{\mathrm{1}}{\mathrm{2}}{Re}\left({Res}\left(\varphi,\:{z}_{\mathrm{1}} \right)+{Res}\left(\varphi,\:{z}_{\mathrm{2}} \right)+{Res}\left(\varphi,\:{z}_{\mathrm{3}} \right)+{Res}\left(\varphi,\:{z}_{\mathrm{4}} \right)\right) \\ $$$${Res}\:\left(\varphi,\:{z}_{\mathrm{1}} \right)=\underset{{z}\rightarrow{z}_{\mathrm{1}} } {\mathrm{lim}}\left(\frac{\left(\mathrm{ln}{z}\right)^{\mathrm{2}} }{\left({z}βˆ’{z}_{\mathrm{2}} \right)\left({z}βˆ’{z}_{\mathrm{3}} \right)\left({z}βˆ’{z}_{\mathrm{4}} \right)}\right)=\frac{\left(\mathrm{ln4}+\frac{\pi}{\mathrm{2}}{i}\right)^{\mathrm{2}} }{\left(\mathrm{8}{i}\right)\left(\mathrm{3}{i}\right)\left(\mathrm{5}{i}\right)}=\frac{{i}}{\mathrm{120}}\left(\mathrm{ln}^{\mathrm{2}} \mathrm{4}βˆ’\frac{\pi^{\mathrm{2}} }{\mathrm{4}}+{i}\pi\mathrm{ln4}\right) \\ $$$${Res}\left(\varphi,\:{z}_{\mathrm{2}} \right)=\underset{{z}\rightarrow{z}_{\mathrm{2}} } {\mathrm{lim}}\left(\frac{\left(\mathrm{ln}{z}\right)^{\mathrm{2}} }{\left({z}βˆ’{z}_{\mathrm{1}} \right)\left({z}βˆ’{z}_{\mathrm{3}} \right)\left({z}βˆ’{z}_{\mathrm{4}} \right)}\right)=\frac{\left(\mathrm{ln4}βˆ’\frac{\pi}{\mathrm{2}}{i}\right)^{\mathrm{2}} }{\left(βˆ’\mathrm{8}{i}\right)\left(βˆ’\mathrm{5}{i}\right)\left(βˆ’\mathrm{3}{i}\right)}=βˆ’\frac{{i}}{\mathrm{120}}\left(\mathrm{ln}^{\mathrm{2}} \mathrm{4}+\frac{\pi^{\mathrm{2}} }{\mathrm{4}}βˆ’{i}\pi\mathrm{ln4}\right) \\ $$$${Res}\left(\varphi,\:{z}_{\mathrm{3}} \right)=\underset{{z}\rightarrow{z}_{\mathrm{3}} } {\mathrm{lim}}\left(\frac{\left(\mathrm{ln}{z}\right)^{\mathrm{2}} }{\left({z}βˆ’{z}_{\mathrm{1}} \right)\left({z}βˆ’{z}_{\mathrm{2}} \right)\left({z}βˆ’{z}_{\mathrm{4}} \right)}\right)=\frac{\left(\frac{\pi}{\mathrm{2}}{i}\right)^{\mathrm{2}} }{\left(βˆ’\mathrm{3}{i}\right)\left(\mathrm{5}{i}\right)\left(\mathrm{2}{i}\right)}=\frac{{i}}{\mathrm{30}}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\right)={i}\frac{\pi^{\mathrm{2}} }{\mathrm{120}} \\ $$$${Res}\left(\varphi,{z}_{\mathrm{4}} \right)=\underset{{z}\rightarrow{z}_{\mathrm{4}} } {\mathrm{lim}}\left(\frac{\left(\mathrm{ln}{z}\right)^{\mathrm{2}} }{\left({z}βˆ’{z}_{\mathrm{1}} \right)\left({z}βˆ’{z}_{\mathrm{2}} \right)\left({z}βˆ’{z}_{\mathrm{3}} \right)}\right)=\frac{\left(βˆ’\frac{\pi}{\mathrm{2}}{i}\right)^{\mathrm{2}} }{\left(βˆ’\mathrm{5}{i}\right)\left(\mathrm{3}{i}\right)\left(βˆ’\mathrm{2}{i}\right)}=βˆ’\frac{{i}}{\mathrm{30}}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\right)=βˆ’{i}\frac{\pi^{\mathrm{2}} }{\mathrm{120}} \\ $$$$\Omega=βˆ’\frac{\mathrm{1}}{\mathrm{2}}\left(βˆ’\frac{\pi\mathrm{ln4}}{\mathrm{120}}βˆ’\frac{\pi\mathrm{ln4}}{\mathrm{120}}\right)=\frac{\pi\mathrm{ln4}}{\mathrm{120}}=\frac{\pi}{\mathrm{60}}\mathrm{ln}\left(\mathrm{2}\right) \\ $$
Commented by mnjuly1970 last updated on 30/Dec/21
very nice solution ..residues theory.. thanks alot sir brandon..
$$\:\:\:{very}\:{nice}\:{solution}\:..{residues}\:{theory}.. \\ $$$$\:\:\:\:{thanks}\:{alot}\:{sir}\:{brandon}.. \\ $$
Answered by mindispower last updated on 30/Dec/21
=βˆ’βˆ«_0 ^∞ βˆ’((ln(x))/((x^2 +16)(x^2 +1)))dx=βˆ’(1/(15))∫_0 ^∞ ((ln(x))/(x^2 +1))+((ln(x))/(x^2 +16))dx =(1/(15))(βˆ’βˆ«_0 ^∞ ((ln(x))/(x^2 +1))dx+∫((ln(4y))/(16(y^2 +1)))4y =βˆ’(3/(60))∫_0 ^∞ ((ln(x))/(1+x^2 ))+((ln(4))/(60))∫_0 ^∞ (dx/(1+x^2 )) ∫_0 ^∞ ((ln(x))/(1+x^2 ))dx,xβ†’(1/x)β‡’=βˆ’βˆ«_0 ^∞ ((ln(x))/(1+x^2 ))dx β‡’βˆ«_0 ^∞ ((ln(x))/(1+x^2 ))dx=0 Ξ©=((ln(4))/(60))∫_0 ^∞ (dx/(1+x^2 ))=((ln(4))/(60))lim_(xβ†’βˆž) [tan^(βˆ’1) (z)]_0 ^x =((ln(4))/(60)).(Ο€/2)=((Ο€ln(2))/(60))
$$=βˆ’\int_{\mathrm{0}} ^{\infty} βˆ’\frac{{ln}\left({x}\right)}{\left({x}^{\mathrm{2}} +\mathrm{16}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{dx}=βˆ’\frac{\mathrm{1}}{\mathrm{15}}\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left({x}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}+\frac{{ln}\left({x}\right)}{{x}^{\mathrm{2}} +\mathrm{16}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{15}}\left(βˆ’\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left({x}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}+\int\frac{{ln}\left(\mathrm{4}{y}\right)}{\mathrm{16}\left({y}^{\mathrm{2}} +\mathrm{1}\right)}\mathrm{4}{y}\right. \\ $$$$=βˆ’\frac{\mathrm{3}}{\mathrm{60}}\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }+\frac{{ln}\left(\mathrm{4}\right)}{\mathrm{60}}\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx},{x}\rightarrow\frac{\mathrm{1}}{{x}}\Rightarrow=βˆ’\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\mathrm{0} \\ $$$$\Omega=\frac{\mathrm{ln}\left(\mathrm{4}\right)}{\mathrm{60}}\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }=\frac{{ln}\left(\mathrm{4}\right)}{\mathrm{60}}\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left[\mathrm{tan}^{βˆ’\mathrm{1}} \left({z}\right)\right]_{\mathrm{0}} ^{{x}} \\ $$$$=\frac{{ln}\left(\mathrm{4}\right)}{\mathrm{60}}.\frac{\pi}{\mathrm{2}}=\frac{\pi{ln}\left(\mathrm{2}\right)}{\mathrm{60}} \\ $$

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