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Question Number 66459 by mathmax by abdo last updated on 15/Aug/19

1) calculate by residus method ∫_0 ^∞ (dx/((1+x^2 )^3 )) 2) find the value of ∫_0 ^1 ((1+x^4 )/((1+x^2 )^3 ))dx

$$\left.\mathrm{1}\right)\:{calculate}\:{by}\:{residus}\:{method}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} } \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}+{x}^{\mathrm{4}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }{dx} \\ $$

Commented by mathmax by abdo last updated on 17/Aug/19

1)let A =∫_0 ^∞ (dx/((1+x^2 )^3 )) ⇒2A =∫_(−∞) ^(+∞) (dx/((x^2 +1)^3 )) let W(z)=(1/((z^2 +1)^3 )) ⇒W(z)=(1/((z−i)^3 (z+i)^3 )) so the poles of W are +^− i (triples) residus theoreme give ∫_(−∞) ^(+∞) W(z)dz=2iπRes(W,i) Res(W,i)=lim_(z→i) (1/((3−1)!)){(z−i)^3 W(z)}^((2)) =lim_(z→i) (1/2){(z+i)^(−3) }^((2)) =lim_(z→i) (1/2){−3(z+i)^(−4) }^((1)) =lim_(z→i) (1/2){12(z+i)^(−5) } =6(2i)^(−5) =(6/((2i)^5 )) =(6/(2^5 i)) =(3/(16i)) ⇒ ∫_(−∞) ^(+∞) W(z)dz =2iπ×(3/(16i)) =((3π)/8) ⇒A =((3π)/(16)) 2)we have ∫_0 ^∞ (dx/((1+x^2 )^3 )) =∫_0 ^1 (dx/((1+x^2 )^3 )) +∫_1 ^(+∞) (dx/((1+x^2 )^3 )) and ∫_1 ^(+∞) (dx/((1+x^2 )^3 )) =_(x=(1/t)) −∫_0 ^1 (−/((1+(1/t^2 ))^3 ))(dt/t^2 ) =∫_0 ^∞ (t^6 /(t^2 (1+t^2 )^3 ))dt =∫_0 ^∞ (t^4 /((1+t^2 )^3 ))dt ⇒∫_0 ^∞ (dx/((1+x^2 )^3 )) =∫_0 ^1 (dx/((1+x^2 )^3 )) +∫_0 ^∞ (x^4 /((1+x^2 )^3 ))dx =∫_0 ^∞ ((1+x^4 )/((1+x^2 )^3 ))dx ⇒∫_0 ^∞ ((1+x^4 )/((1+x^2 )^3 ))=((3π)/(16)) .

$$\left.\mathrm{1}\right){let}\:{A}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }\:\Rightarrow\mathrm{2}{A}\:=\int_{−\infty} ^{+\infty} \:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$${let}\:{W}\left({z}\right)=\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{3}} }\:\Rightarrow{W}\left({z}\right)=\frac{\mathrm{1}}{\left({z}−{i}\right)^{\mathrm{3}} \left({z}+{i}\right)^{\mathrm{3}} }\:{so}\:{the}\:{poles}\:{of}\:{W}\:{are} \\ $$$$\overset{−} {+}{i}\:\:\left({triples}\right)\:\:{residus}\:{theoreme}\:{give}\:\int_{−\infty} ^{+\infty} {W}\left({z}\right){dz}=\mathrm{2}{i}\pi{Res}\left({W},{i}\right) \\ $$$${Res}\left({W},{i}\right)={lim}_{{z}\rightarrow{i}} \frac{\mathrm{1}}{\left(\mathrm{3}−\mathrm{1}\right)!}\left\{\left({z}−{i}\right)^{\mathrm{3}} {W}\left({z}\right)\right\}^{\left(\mathrm{2}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\:\frac{\mathrm{1}}{\mathrm{2}}\left\{\left({z}+{i}\right)^{−\mathrm{3}} \right\}^{\left(\mathrm{2}\right)} \:={lim}_{{z}\rightarrow{i}} \:\frac{\mathrm{1}}{\mathrm{2}}\left\{−\mathrm{3}\left({z}+{i}\right)^{−\mathrm{4}} \right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{12}\left({z}+{i}\right)^{−\mathrm{5}} \right\}\:=\mathrm{6}\left(\mathrm{2}{i}\right)^{−\mathrm{5}} \:=\frac{\mathrm{6}}{\left(\mathrm{2}{i}\right)^{\mathrm{5}} }\:=\frac{\mathrm{6}}{\mathrm{2}^{\mathrm{5}} {i}}\:=\frac{\mathrm{3}}{\mathrm{16}{i}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi×\frac{\mathrm{3}}{\mathrm{16}{i}}\:=\frac{\mathrm{3}\pi}{\mathrm{8}}\:\Rightarrow{A}\:=\frac{\mathrm{3}\pi}{\mathrm{16}} \\ $$$$\left.\mathrm{2}\right){we}\:{have}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }\:+\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }\:\:{and} \\ $$$$\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }\:=_{{x}=\frac{\mathrm{1}}{{t}}} \:\:\:\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{−}{\left(\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)^{\mathrm{3}} }\frac{{dt}}{{t}^{\mathrm{2}} }\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{\mathrm{6}} }{{t}^{\mathrm{2}} \left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{3}} }{dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{{t}^{\mathrm{4}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{3}} }{dt}\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }\:+\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{\mathrm{4}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }{dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}+{x}^{\mathrm{4}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }{dx}\:\:\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}+{x}^{\mathrm{4}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }=\frac{\mathrm{3}\pi}{\mathrm{16}}\:. \\ $$

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