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Question Number 59893 by aliesam last updated on 15/May/19
Commented by maxmathsup by imad last updated on 15/May/19
method using the formuae ∫_0 ^∞ (t^(a−1) /(1+t))dt =(π/(sin(πa))) if 0<a<1 let I =∫_(−∞) ^(+∞) ((x^2 +4)/(x^4 +16)) ⇒I =_(x=2t) ∫_(−∞) ^(+∞) ((4t^2 +4)/(16t^4 +16)) (2)dt =(1/2) ∫_(−∞) ^(+∞) ((t^2 +1)/(t^4 +1)) dt =∫_0 ^∞ ((t^2 +1)/(t^4 +1)) dt =∫_0 ^∞ (t^2 /(t^4 +1)) dt +∫_0 ^∞ (dt/(t^4 +1)) changement t^4 =u give t =u^(1/4) ⇒∫_0 ^∞ (t^2 /(t^4 +1)) dt =∫_0 ^∞ (u^(1/2) /(1+u)) (1/4)u^((1/4)−1) du =(1/4) ∫_0 ^∞ (u^((3/4)−1) /(1+u)) du =(1/4) (π/(sin(((3π)/4)))) =(π/(4.(1/( (√2))))) =((π(√2))/4) same changement give ∫_0 ^∞ (dt/(1+t^4 )) =∫_0 ^∞ (1/4) (u^((1/4)−1) /(1+u)) du =(1/4) (π/(sin((π/4)))) =(π/(4(1/( (√2))))) =((π(√2))/4) ⇒ I =((π(√2))/4) +((π(√2))/4) = ((2π(√2))/4) =((π(√2))/2) ⇒ ★ I =(π/( (√2))) ★
$${method}\:{using}\:{the}\:{formuae}\:\int_{\mathrm{0}} ^{\infty} \frac{{t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}\:=\frac{\pi}{{sin}\left(\pi{a}\right)}\:\:{if}\:\mathrm{0}<{a}<\mathrm{1} \\ $$$${let}\:{I}\:=\int_{−\infty} ^{+\infty} \:\:\frac{{x}^{\mathrm{2}} \:+\mathrm{4}}{{x}^{\mathrm{4}} \:+\mathrm{16}}\:\Rightarrow{I}\:=_{{x}=\mathrm{2}{t}} \:\:\:\int_{−\infty} ^{+\infty} \:\:\:\frac{\mathrm{4}{t}^{\mathrm{2}} \:+\mathrm{4}}{\mathrm{16}{t}^{\mathrm{4}} \:+\mathrm{16}}\:\left(\mathrm{2}\right){dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{t}^{\mathrm{2}} \:+\mathrm{1}}{{t}^{\mathrm{4}} \:+\mathrm{1}}\:{dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\frac{{t}^{\mathrm{2}} \:+\mathrm{1}}{{t}^{\mathrm{4}} \:+\mathrm{1}}\:{dt}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{4}} \:+\mathrm{1}}\:{dt}\:+\int_{\mathrm{0}} ^{\infty} \:\frac{{dt}}{{t}^{\mathrm{4}} \:+\mathrm{1}} \\ $$$${changement}\:\:\:{t}^{\mathrm{4}} \:={u}\:\:{give}\:{t}\:={u}^{\frac{\mathrm{1}}{\mathrm{4}}} \:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{4}} \:+\mathrm{1}}\:{dt}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{u}^{\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{1}+{u}}\:\frac{\mathrm{1}}{\mathrm{4}}{u}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} \:{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{u}^{\frac{\mathrm{3}}{\mathrm{4}}−\mathrm{1}} }{\mathrm{1}+{u}}\:{du}\:=\frac{\mathrm{1}}{\mathrm{4}}\:\frac{\pi}{{sin}\left(\frac{\mathrm{3}\pi}{\mathrm{4}}\right)}\:=\frac{\pi}{\mathrm{4}.\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}\:=\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$$${same}\:{changement}\:{give}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{4}} }\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\mathrm{4}}\:\frac{{u}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} }{\mathrm{1}+{u}}\:{du}\:=\frac{\mathrm{1}}{\mathrm{4}}\:\frac{\pi}{{sin}\left(\frac{\pi}{\mathrm{4}}\right)}\:=\frac{\pi}{\mathrm{4}\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}\:=\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$$$\Rightarrow\:{I}\:=\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{4}}\:+\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{4}}\:=\:\frac{\mathrm{2}\pi\sqrt{\mathrm{2}}}{\mathrm{4}}\:=\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{2}}\:\:\Rightarrow\:\bigstar\:{I}\:=\frac{\pi}{\:\sqrt{\mathrm{2}}}\:\bigstar \\ $$
Commented by malwaan last updated on 15/May/19
great sir but ; can anyone solve it without using this formulae
$${great}\:{sir} \\ $$$${but}\:;\:{can}\:{anyone}\:{solve}\:{it} \\ $$$${without}\:{using}\:{this}\:{formulae} \\ $$
Commented by maxmathsup by imad last updated on 16/May/19
yes this integral is also solved by residus theorem i will post the method soon...
$${yes}\:{this}\:{integral}\:{is}\:{also}\:{solved}\:{by}\:{residus}\:{theorem}\:{i}\:{will}\:{post}\:{the}\:{method}\: \\ $$$${soon}… \\ $$
Commented by malwaan last updated on 16/May/19
thanks sir I am waiting ...
$${thanks}\:{sir}\: \\ $$$${I}\:{am}\:{waiting}\:… \\ $$
Commented by maxmathsup by imad last updated on 17/May/19
residus method let A =∫_(−∞) ^(+∞) ((x^2 +4)/(x^4 +16))dx ⇒A=_(x=2t) ∫_(−∞) ^(+∞) ((4t^2 +4)/(16(t^4 +1)))2dt ⇒2A =∫_(−∞) ^(+∞) ((t^2 +1)/(t^4 +1)) dt let W(z) = ((z^2 +1)/(z^4 +1)) poles of W! W(z) =((z^2 +1)/((z^2 −i)(z^2 +i))) =((z^2 +1)/((z−e^((iπ)/4) )(z+e^((iπ)/4) )(z−e^(−((iπ)/4)) )(z+e^(−((iπ)/4)) ))) so the poles of W are +^− e^((iπ)/4) and +^− e^(−((iπ)/4)) residus theorem give ∫_(−∞) ^(+∞) W(z)dz =2iπ{ Res(W,e^((iπ)/4) ) +Res(W,−e^(−((iπ)/4)) )} Res(W,e^((iπ)/4) ) =lim_(z→e^((iπ)/4) ) (z−e^((iπ)/4) )W(z) = ((i+1)/(2e^((iπ)/4) (2i))) =((1+i)/(4i )) e^(−((iπ)/4)) Res(W,−e^(−((iπ)/4)) ) =lim_(z→−e^(−((iπ)/4)) ) (z+e^(−((iπ)/4)) )W(z) = ((−i+1)/(−2e^(−((iπ)/4)) (−2i))) =(((1−i)e^((iπ)/4) )/(4i)) ⇒ ∫_(−∞) ^(+∞) W(z)dz =2iπ(((1+i)/(4i)) e^(−((iπ)/4)) +((1−i)/(4i)) e^((iπ)/4) } =(π/2) ( 2Re(1+i)e^(−((iπ)/4)) ) =π Re{ (1+i)e^(−((iπ)/4)) } but (1+i)e^(−i(π/4)) =(√2)e^((iπ)/4) e^(−((iπ)/4)) =(√2) ⇒∫_(−∞) ^(+∞) W(z)dz =π(√(2 )) ⇒ 2A =π(√2) ⇒ A =((π(√2))/2) ⇒ A =(π/( (√2))) .
$${residus}\:{method}\:{let}\:\:{A}\:=\int_{−\infty} ^{+\infty} \:\:\frac{{x}^{\mathrm{2}} \:+\mathrm{4}}{{x}^{\mathrm{4}} \:+\mathrm{16}}{dx}\:\Rightarrow{A}=_{{x}=\mathrm{2}{t}} \:\:\:\:\int_{−\infty} ^{+\infty} \:\:\:\frac{\mathrm{4}{t}^{\mathrm{2}} \:+\mathrm{4}}{\mathrm{16}\left({t}^{\mathrm{4}} \:+\mathrm{1}\right)}\mathrm{2}{dt} \\ $$$$\:\Rightarrow\mathrm{2}{A}\:=\int_{−\infty} ^{+\infty} \:\:\:\frac{{t}^{\mathrm{2}} \:+\mathrm{1}}{{t}^{\mathrm{4}} \:+\mathrm{1}}\:{dt}\:\:{let}\:{W}\left({z}\right)\:=\:\frac{{z}^{\mathrm{2}} \:+\mathrm{1}}{{z}^{\mathrm{4}} \:+\mathrm{1}}\:\:\:{poles}\:{of}\:{W}! \\ $$$${W}\left({z}\right)\:=\frac{{z}^{\mathrm{2}} \:+\mathrm{1}}{\left({z}^{\mathrm{2}} −{i}\right)\left({z}^{\mathrm{2}} \:+{i}\right)}\:=\frac{{z}^{\mathrm{2}} \:+\mathrm{1}}{\left({z}−{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)}\:\:{so}\:{the}\:{poles}\:{of}\:{W} \\ $$$${are}\:\overset{−} {+}{e}^{\frac{{i}\pi}{\mathrm{4}}} \:\:{and}\:\overset{−} {+}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:\:{residus}\:{theorem}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:{Res}\left({W},{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\:+{Res}\left({W},−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\right\} \\ $$$${Res}\left({W},{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\:={lim}_{{z}\rightarrow{e}^{\frac{{i}\pi}{\mathrm{4}}} } \:\:\:\left({z}−{e}^{\frac{{i}\pi}{\mathrm{4}}} \right){W}\left({z}\right)\:=\:\frac{{i}+\mathrm{1}}{\mathrm{2}{e}^{\frac{{i}\pi}{\mathrm{4}}} \left(\mathrm{2}{i}\right)}\:=\frac{\mathrm{1}+{i}}{\mathrm{4}{i}\:}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \\ $$$${Res}\left({W},−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\:={lim}_{{z}\rightarrow−{e}^{−\frac{{i}\pi}{\mathrm{4}}} } \:\:\:\left({z}+{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right){W}\left({z}\right)\:=\:\frac{−{i}+\mathrm{1}}{−\mathrm{2}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:\left(−\mathrm{2}{i}\right)}\:=\frac{\left(\mathrm{1}−{i}\right){e}^{\frac{{i}\pi}{\mathrm{4}}} }{\mathrm{4}{i}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left(\frac{\mathrm{1}+{i}}{\mathrm{4}{i}}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:\:+\frac{\mathrm{1}−{i}}{\mathrm{4}{i}}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \right\}\:=\frac{\pi}{\mathrm{2}}\:\left(\:\mathrm{2}{Re}\left(\mathrm{1}+{i}\right){e}^{−\frac{{i}\pi}{\mathrm{4}}} \right) \\ $$$$=\pi\:{Re}\left\{\:\left(\mathrm{1}+{i}\right){e}^{−\frac{{i}\pi}{\mathrm{4}}} \right\}\:{but}\:\:\left(\mathrm{1}+{i}\right){e}^{−{i}\frac{\pi}{\mathrm{4}}} \:=\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:=\sqrt{\mathrm{2}}\:\Rightarrow\int_{−\infty} ^{+\infty} {W}\left({z}\right){dz}\:=\pi\sqrt{\mathrm{2}\:}\:\Rightarrow \\ $$$$\mathrm{2}{A}\:=\pi\sqrt{\mathrm{2}}\:\Rightarrow\:{A}\:=\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{2}}\:\Rightarrow\:{A}\:=\frac{\pi}{\:\sqrt{\mathrm{2}}}\:. \\ $$
Commented by malwaan last updated on 18/May/19
Thank you so much sir
$${Thank}\:{you}\:{so}\:{much}\:{sir} \\ $$
Commented by maxmathsup by imad last updated on 18/May/19
you are welcome.
$${you}\:{are}\:{welcome}. \\ $$

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