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Question Number 162535 by mnjuly1970 last updated on 30/Dec/21

       prove that  Ω = ∫_0 ^( ∞) ((  ln ((1/x) ))/( x^( 4)  + 17x^( 2)  + 16)) dx=^?  (π/(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 😅😁

$$\mathrm{Synthax}\:\mathrm{error},\:\mathrm{Sir}.\:\mathrm{Your}\:{dx}\:\mathrm{is}\:\mathrm{missing}. \\ $$Haha 😅😁

Commented by amin96 last updated on 30/Dec/21

  very big mistake😂😂

$$ \\ $$very big mistake😂😂

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

Ω=∫_0 ^∞ ((ln((1/x)))/(x^4 +17x^2 +16))dx=−∫_0 ^∞ ((lnx)/(x^4 +17x^2 +16))dx  ϕ(z)=(((lnz)^2 )/(z^4 +17z^2 +16)) ,poles:(z^2 +16)(z^2 +1)=0⇒z_1 =4e^((π/2)i) ,z_2 =4e^(−(π/2)i) ,z_3 =e^((π/2)i) ,z_4 =e^(−(π/2)i)   Ω=−(1/2)Re(Res(ϕ, z_1 )+Res(ϕ, z_2 )+Res(ϕ, z_3 )+Res(ϕ, z_4 ))  Res (ϕ, z_1 )=lim_(z→z_1 ) ((((lnz)^2 )/((z−z_2 )(z−z_3 )(z−z_4 ))))=(((ln4+(π/2)i)^2 )/((8i)(3i)(5i)))=(i/(120))(ln^2 4−(π^2 /4)+iπln4)  Res(ϕ, z_2 )=lim_(z→z_2 ) ((((lnz)^2 )/((z−z_1 )(z−z_3 )(z−z_4 ))))=(((ln4−(π/2)i)^2 )/((−8i)(−5i)(−3i)))=−(i/(120))(ln^2 4+(π^2 /4)−iπln4)  Res(ϕ, z_3 )=lim_(z→z_3 ) ((((lnz)^2 )/((z−z_1 )(z−z_2 )(z−z_4 ))))=((((π/2)i)^2 )/((−3i)(5i)(2i)))=(i/(30))((π^2 /4))=i(π^2 /(120))  Res(ϕ,z_4 )=lim_(z→z_4 ) ((((lnz)^2 )/((z−z_1 )(z−z_2 )(z−z_3 ))))=(((−(π/2)i)^2 )/((−5i)(3i)(−2i)))=−(i/(30))((π^2 /4))=−i(π^2 /(120))  Ω=−(1/2)(−((πln4)/(120))−((πln4)/(120)))=((πln4)/(120))=(π/(60))ln(2)

$$\Omega=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}\left(\frac{\mathrm{1}}{{x}}\right)}{{x}^{\mathrm{4}} +\mathrm{17}{x}^{\mathrm{2}} +\mathrm{16}}{dx}=−\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}{x}}{{x}^{\mathrm{4}} +\mathrm{17}{x}^{\mathrm{2}} +\mathrm{16}}{dx} \\ $$$$\varphi\left({z}\right)=\frac{\left(\mathrm{ln}{z}\right)^{\mathrm{2}} }{{z}^{\mathrm{4}} +\mathrm{17}{z}^{\mathrm{2}} +\mathrm{16}}\:,\mathrm{poles}:\left({z}^{\mathrm{2}} +\mathrm{16}\right)\left({z}^{\mathrm{2}} +\mathrm{1}\right)=\mathrm{0}\Rightarrow{z}_{\mathrm{1}} =\mathrm{4}{e}^{\frac{\pi}{\mathrm{2}}{i}} ,{z}_{\mathrm{2}} =\mathrm{4}{e}^{−\frac{\pi}{\mathrm{2}}{i}} ,{z}_{\mathrm{3}} ={e}^{\frac{\pi}{\mathrm{2}}{i}} ,{z}_{\mathrm{4}} ={e}^{−\frac{\pi}{\mathrm{2}}{i}} \\ $$$$\Omega=−\frac{\mathrm{1}}{\mathrm{2}}{Re}\left({Res}\left(\varphi,\:{z}_{\mathrm{1}} \right)+{Res}\left(\varphi,\:{z}_{\mathrm{2}} \right)+{Res}\left(\varphi,\:{z}_{\mathrm{3}} \right)+{Res}\left(\varphi,\:{z}_{\mathrm{4}} \right)\right) \\ $$$${Res}\:\left(\varphi,\:{z}_{\mathrm{1}} \right)=\underset{{z}\rightarrow{z}_{\mathrm{1}} } {\mathrm{lim}}\left(\frac{\left(\mathrm{ln}{z}\right)^{\mathrm{2}} }{\left({z}−{z}_{\mathrm{2}} \right)\left({z}−{z}_{\mathrm{3}} \right)\left({z}−{z}_{\mathrm{4}} \right)}\right)=\frac{\left(\mathrm{ln4}+\frac{\pi}{\mathrm{2}}{i}\right)^{\mathrm{2}} }{\left(\mathrm{8}{i}\right)\left(\mathrm{3}{i}\right)\left(\mathrm{5}{i}\right)}=\frac{{i}}{\mathrm{120}}\left(\mathrm{ln}^{\mathrm{2}} \mathrm{4}−\frac{\pi^{\mathrm{2}} }{\mathrm{4}}+{i}\pi\mathrm{ln4}\right) \\ $$$${Res}\left(\varphi,\:{z}_{\mathrm{2}} \right)=\underset{{z}\rightarrow{z}_{\mathrm{2}} } {\mathrm{lim}}\left(\frac{\left(\mathrm{ln}{z}\right)^{\mathrm{2}} }{\left({z}−{z}_{\mathrm{1}} \right)\left({z}−{z}_{\mathrm{3}} \right)\left({z}−{z}_{\mathrm{4}} \right)}\right)=\frac{\left(\mathrm{ln4}−\frac{\pi}{\mathrm{2}}{i}\right)^{\mathrm{2}} }{\left(−\mathrm{8}{i}\right)\left(−\mathrm{5}{i}\right)\left(−\mathrm{3}{i}\right)}=−\frac{{i}}{\mathrm{120}}\left(\mathrm{ln}^{\mathrm{2}} \mathrm{4}+\frac{\pi^{\mathrm{2}} }{\mathrm{4}}−{i}\pi\mathrm{ln4}\right) \\ $$$${Res}\left(\varphi,\:{z}_{\mathrm{3}} \right)=\underset{{z}\rightarrow{z}_{\mathrm{3}} } {\mathrm{lim}}\left(\frac{\left(\mathrm{ln}{z}\right)^{\mathrm{2}} }{\left({z}−{z}_{\mathrm{1}} \right)\left({z}−{z}_{\mathrm{2}} \right)\left({z}−{z}_{\mathrm{4}} \right)}\right)=\frac{\left(\frac{\pi}{\mathrm{2}}{i}\right)^{\mathrm{2}} }{\left(−\mathrm{3}{i}\right)\left(\mathrm{5}{i}\right)\left(\mathrm{2}{i}\right)}=\frac{{i}}{\mathrm{30}}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\right)={i}\frac{\pi^{\mathrm{2}} }{\mathrm{120}} \\ $$$${Res}\left(\varphi,{z}_{\mathrm{4}} \right)=\underset{{z}\rightarrow{z}_{\mathrm{4}} } {\mathrm{lim}}\left(\frac{\left(\mathrm{ln}{z}\right)^{\mathrm{2}} }{\left({z}−{z}_{\mathrm{1}} \right)\left({z}−{z}_{\mathrm{2}} \right)\left({z}−{z}_{\mathrm{3}} \right)}\right)=\frac{\left(−\frac{\pi}{\mathrm{2}}{i}\right)^{\mathrm{2}} }{\left(−\mathrm{5}{i}\right)\left(\mathrm{3}{i}\right)\left(−\mathrm{2}{i}\right)}=−\frac{{i}}{\mathrm{30}}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\right)=−{i}\frac{\pi^{\mathrm{2}} }{\mathrm{120}} \\ $$$$\Omega=−\frac{\mathrm{1}}{\mathrm{2}}\left(−\frac{\pi\mathrm{ln4}}{\mathrm{120}}−\frac{\pi\mathrm{ln4}}{\mathrm{120}}\right)=\frac{\pi\mathrm{ln4}}{\mathrm{120}}=\frac{\pi}{\mathrm{60}}\mathrm{ln}\left(\mathrm{2}\right) \\ $$

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

=−∫_0 ^∞ −((ln(x))/((x^2 +16)(x^2 +1)))dx=−(1/(15))∫_0 ^∞ ((ln(x))/(x^2 +1))+((ln(x))/(x^2 +16))dx  =(1/(15))(−∫_0 ^∞ ((ln(x))/(x^2 +1))dx+∫((ln(4y))/(16(y^2 +1)))4y  =−(3/(60))∫_0 ^∞ ((ln(x))/(1+x^2 ))+((ln(4))/(60))∫_0 ^∞ (dx/(1+x^2 ))  ∫_0 ^∞ ((ln(x))/(1+x^2 ))dx,x→(1/x)⇒=−∫_0 ^∞ ((ln(x))/(1+x^2 ))dx  ⇒∫_0 ^∞ ((ln(x))/(1+x^2 ))dx=0  Ω=((ln(4))/(60))∫_0 ^∞ (dx/(1+x^2 ))=((ln(4))/(60))lim_(x→∞) [tan^(−1) (z)]_0 ^x   =((ln(4))/(60)).(π/2)=((πln(2))/(60))

$$=−\int_{\mathrm{0}} ^{\infty} −\frac{{ln}\left({x}\right)}{\left({x}^{\mathrm{2}} +\mathrm{16}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{dx}=−\frac{\mathrm{1}}{\mathrm{15}}\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left({x}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}+\frac{{ln}\left({x}\right)}{{x}^{\mathrm{2}} +\mathrm{16}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{15}}\left(−\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left({x}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}+\int\frac{{ln}\left(\mathrm{4}{y}\right)}{\mathrm{16}\left({y}^{\mathrm{2}} +\mathrm{1}\right)}\mathrm{4}{y}\right. \\ $$$$=−\frac{\mathrm{3}}{\mathrm{60}}\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }+\frac{{ln}\left(\mathrm{4}\right)}{\mathrm{60}}\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx},{x}\rightarrow\frac{\mathrm{1}}{{x}}\Rightarrow=−\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\mathrm{0} \\ $$$$\Omega=\frac{\mathrm{ln}\left(\mathrm{4}\right)}{\mathrm{60}}\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }=\frac{{ln}\left(\mathrm{4}\right)}{\mathrm{60}}\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left[\mathrm{tan}^{−\mathrm{1}} \left({z}\right)\right]_{\mathrm{0}} ^{{x}} \\ $$$$=\frac{{ln}\left(\mathrm{4}\right)}{\mathrm{60}}.\frac{\pi}{\mathrm{2}}=\frac{\pi{ln}\left(\mathrm{2}\right)}{\mathrm{60}} \\ $$

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