Question-76146

Question Number 76146 by john santuy last updated on 24/Dec/19

Commented by john santuy last updated on 24/Dec/19

$${how}\:{to}\:{proof}? \\ $$

Commented by mr W last updated on 25/Dec/19

value given in question ((π/( (√6)))) is correct, sir. i think you had a mistake in your calculation. it should be A =−∫_1 ^(+∞) (√(−1+(√t)))((−4)/((t+3)^2 ))dt please recheck.

$${value}\:{given}\:{in}\:{question}\:\left(\frac{\pi}{\:\sqrt{\mathrm{6}}}\right)\:{is} \\ $$$${correct},\:{sir}.\: \\ $$$${i}\:{think}\:{you}\:{had}\:{a}\:{mistake}\:{in}\:{your} \\ $$$${calculation}.\:\:{it}\:{should}\:{be} \\ $$$${A}\:=−\int_{\mathrm{1}} ^{+\infty} \sqrt{−\mathrm{1}+\sqrt{{t}}}\frac{−\mathrm{4}}{\left({t}+\mathrm{3}\right)^{\mathrm{2}} }{dt}\: \\ $$$${please}\:{recheck}. \\ $$

Commented by mr W last updated on 25/Dec/19

i have pointed the error in your answer. it′s in the third line: A =−∫_1 ^(+∞) (√(−1+t))((−4)/((t+3)^2 ))dt =...... it should be A =−∫_1 ^(+∞) (√(−1+(√t)))((−4)/((t+3)^2 ))dt =......

$${i}\:{have}\:{pointed}\:{the}\:{error}\:{in}\:{your} \\ $$$${answer}.\:{it}'{s}\:{in}\:{the}\:{third}\:{line}: \\ $$$${A}\:=−\int_{\mathrm{1}} ^{+\infty} \sqrt{−\mathrm{1}+\boldsymbol{{t}}}\frac{−\mathrm{4}}{\left({t}+\mathrm{3}\right)^{\mathrm{2}} }{dt}\:=…… \\ $$$${it}\:{should}\:{be} \\ $$$${A}\:=−\int_{\mathrm{1}} ^{+\infty} \sqrt{−\mathrm{1}+\sqrt{{t}}}\frac{−\mathrm{4}}{\left({t}+\mathrm{3}\right)^{\mathrm{2}} }{dt}\:=…… \\ $$

Commented by abdomathmax last updated on 25/Dec/19

$${yes}\:{you}\:{are}\:{right}\:{sir}\:…{i}\:{will}\:{coreect}\:{the}\:{answer}… \\ $$

Commented by mathmax by abdo last updated on 26/Dec/19

let I =∫_0 ^1 (√(−1+(√((4/x)−3))))dx cha7gement (√((4/x)−3))=t give (4/x) −3 =t^2 ⇒4−3x=t^2 x ⇒4=(t^2 +3)x ⇒x=(4/(t^2 +3)) ⇒ dx =((−8t)/((t^2 +3)^2 )) ⇒ I =−∫_1 ^(+∞) (√(t−1))×((−8t)/((t^2 +3)^2 ))dt =8 ∫_1 ^(+∞) ((t(√(t−1)))/((t^(2 ) +3)^2 ))dt =_((√(t−1))=u) 8 ∫_o ^(+∞) (((1+u^2 )u)/(((1+u^2 )^2 +3)^2 ))(2u)du =16 ∫_0 ^∞ ((u^2 (1+u^2 ))/((u^4 +2u^2 +4)^2 ))du =8 ∫_(−∞) ^(+∞) ((u^4 +u^2 )/((u^4 +2u^2 +4)^2 ))du letϕ(z)=((z^4 +z^2 )/((z^4 +2z^2 +4)^2 )) poles of ϕ? z^4 +2z^2 +4 =0 ⇒t^2 +2t +4=0 with t=z^2 Δ^′ =1−4 =−3 ⇒t_1 =−1+(√3)=2(−(1/2)+((√3)/2)) =2e^((i2π)/3) t_1 =−1−(√3)=2 e^(−((i2π)/3)) ⇒z^4 +2z^2 +4 =(z^2 −2e^((i2π)/3) )(z^2 −2e^(−((i2π)/3)) ) =(z−(√2)e^((iπ)/3) )(z+(√2)e^((iπ)/3) )(z−(√2)e^(−((iπ)/3)) )(z+(√2)e^(−((iπ)/3)) )⇒ ϕ(z) =((z^4 +z^2 )/((z−(√2)e^((iπ)/3) )^2 (z+(√2)e^((iπ)/3) )^2 (z−(√2)e^(−((iπ)/3)) )(z+(√2)e^(−((iπ)/3)) ))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ {Res(ϕ,(√2)e^((iπ)/3) )+Res(ϕ,−(√2)e^(−((iπ)/3)) )} Res(ϕ,(√2)e^((iπ)/3) ) =lim_(z→(√2)e^((iπ)/3) ) (1/((2−1)!)){(z−(√2)e^((iπ)/3) )^2 ϕ(z)}^((1)) =lim_(z→(√2)e^((iπ)/3) ) {((z^4 +z^2 )/((z+(√2)e^((iπ)/3) )^2 (z−(√2)e^(−((iπ)/3)) )^2 (z+(√2)e^(−((iπ)/3)) )^2 ))}^((1)) ...be continued...

$${let}\:{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{−\mathrm{1}+\sqrt{\frac{\mathrm{4}}{{x}}−\mathrm{3}}}{dx}\:\:{cha}\mathrm{7}{gement}\:\sqrt{\frac{\mathrm{4}}{{x}}−\mathrm{3}}={t}\:{give} \\ $$$$\frac{\mathrm{4}}{{x}}\:−\mathrm{3}\:={t}^{\mathrm{2}} \:\Rightarrow\mathrm{4}−\mathrm{3}{x}={t}^{\mathrm{2}} {x}\:\Rightarrow\mathrm{4}=\left({t}^{\mathrm{2}} \:+\mathrm{3}\right){x}\:\Rightarrow{x}=\frac{\mathrm{4}}{{t}^{\mathrm{2}} \:+\mathrm{3}}\:\Rightarrow \\ $$$${dx}\:=\frac{−\mathrm{8}{t}}{\left({t}^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{2}} }\:\Rightarrow\:{I}\:=−\int_{\mathrm{1}} ^{+\infty} \sqrt{{t}−\mathrm{1}}×\frac{−\mathrm{8}{t}}{\left({t}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} }{dt} \\ $$$$=\mathrm{8}\:\int_{\mathrm{1}} ^{+\infty} \:\frac{{t}\sqrt{{t}−\mathrm{1}}}{\left({t}^{\mathrm{2}\:} +\mathrm{3}\right)^{\mathrm{2}} }{dt}\:=_{\sqrt{{t}−\mathrm{1}}={u}} \:\mathrm{8}\:\int_{{o}} ^{+\infty} \frac{\left(\mathrm{1}+{u}^{\mathrm{2}} \right){u}}{\left(\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} }\left(\mathrm{2}{u}\right){du} \\ $$$$=\mathrm{16}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{u}^{\mathrm{2}} \left(\mathrm{1}+{u}^{\mathrm{2}} \right)}{\left({u}^{\mathrm{4}} \:+\mathrm{2}{u}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} }{du}\:=\mathrm{8}\:\int_{−\infty} ^{+\infty} \:\frac{{u}^{\mathrm{4}} +{u}^{\mathrm{2}} }{\left({u}^{\mathrm{4}} \:+\mathrm{2}{u}^{\mathrm{2}} \:+\mathrm{4}\right)^{\mathrm{2}} }{du} \\ $$$${let}\varphi\left({z}\right)=\frac{{z}^{\mathrm{4}} \:+{z}^{\mathrm{2}} }{\left({z}^{\mathrm{4}} \:+\mathrm{2}{z}^{\mathrm{2}} \:+\mathrm{4}\right)^{\mathrm{2}} }\:\:{poles}\:{of}\:\varphi? \\ $$$${z}^{\mathrm{4}} \:+\mathrm{2}{z}^{\mathrm{2}} \:+\mathrm{4}\:=\mathrm{0}\:\Rightarrow{t}^{\mathrm{2}} \:+\mathrm{2}{t}\:+\mathrm{4}=\mathrm{0}\:\:{with}\:{t}={z}^{\mathrm{2}} \\ $$$$\Delta^{'} =\mathrm{1}−\mathrm{4}\:=−\mathrm{3}\:\Rightarrow{t}_{\mathrm{1}} =−\mathrm{1}+\sqrt{\mathrm{3}}=\mathrm{2}\left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\:=\mathrm{2}{e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \\ $$$${t}_{\mathrm{1}} =−\mathrm{1}−\sqrt{\mathrm{3}}=\mathrm{2}\:{e}^{−\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \:\Rightarrow{z}^{\mathrm{4}} \:+\mathrm{2}{z}^{\mathrm{2}} \:+\mathrm{4}\:=\left({z}^{\mathrm{2}} −\mathrm{2}{e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \right)\left({z}^{\mathrm{2}} −\mathrm{2}{e}^{−\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \right) \\ $$$$=\left({z}−\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{3}}} \right)\left({z}+\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{3}}} \right)\left({z}−\sqrt{\mathrm{2}}{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)\left({z}+\sqrt{\mathrm{2}}{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)\Rightarrow \\ $$$$\varphi\left({z}\right)\:=\frac{{z}^{\mathrm{4}} \:+{z}^{\mathrm{2}} }{\left({z}−\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} \left({z}+\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} \left({z}−\sqrt{\mathrm{2}}{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)\left({z}+\sqrt{\mathrm{2}}{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\left\{{Res}\left(\varphi,\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{3}}} \right)+{Res}\left(\varphi,−\sqrt{\mathrm{2}}{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)\right\} \\ $$$${Res}\left(\varphi,\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{3}}} \right)\:={lim}_{{z}\rightarrow\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{3}}} } \:\:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\left({z}−\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} \varphi\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{3}}} } \:\:\:\:\left\{\frac{{z}^{\mathrm{4}} \:+{z}^{\mathrm{2}} }{\left({z}+\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} \left({z}−\sqrt{\mathrm{2}}{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} \left({z}+\sqrt{\mathrm{2}}{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$…{be}\:{continued}… \\ $$

Answered by mr W last updated on 24/Dec/19

y=(√(−1+(√((4/x)−3)))) ⇒x=(4/(3+(1+y^2 )^2 )) ∫_0 ^1 ydx=∫_0 ^∞ xdy=∫_0 ^∞ (4/(3+(1+y^2 )^2 ))dy =(1/(2(√2)))[ln ((2+y(y+(√2)))/(2+y(y−(√2))))]_0 ^∞ +(1/( (√6)))[tan^(−1) (((√2)y+1)/( (√3)))+tan^(−1) (((√2)y−1)/( (√3)))]_0 ^∞ =(1/(2(√2)))(0)+(1/( (√6)))((π/2)+(π/2)−((π/6)−(π/6))) =(π/( (√6)))

$${y}=\sqrt{−\mathrm{1}+\sqrt{\frac{\mathrm{4}}{{x}}−\mathrm{3}}} \\ $$$$\Rightarrow{x}=\frac{\mathrm{4}}{\mathrm{3}+\left(\mathrm{1}+{y}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {ydx}=\int_{\mathrm{0}} ^{\infty} {xdy}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{4}}{\mathrm{3}+\left(\mathrm{1}+{y}^{\mathrm{2}} \right)^{\mathrm{2}} }{dy} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\left[\mathrm{ln}\:\frac{\mathrm{2}+{y}\left({y}+\sqrt{\mathrm{2}}\right)}{\mathrm{2}+{y}\left({y}−\sqrt{\mathrm{2}}\right)}\right]_{\mathrm{0}} ^{\infty} +\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}}\left[\mathrm{tan}^{−\mathrm{1}} \frac{\sqrt{\mathrm{2}}{y}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}+\mathrm{tan}^{−\mathrm{1}} \frac{\sqrt{\mathrm{2}}{y}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right]_{\mathrm{0}} ^{\infty} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\left(\mathrm{0}\right)+\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}}\left(\frac{\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{2}}−\left(\frac{\pi}{\mathrm{6}}−\frac{\pi}{\mathrm{6}}\right)\right) \\ $$$$=\frac{\pi}{\:\sqrt{\mathrm{6}}} \\ $$

Commented by john santuy last updated on 24/Dec/19