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Question Number 28883 by abdo imad last updated on 31/Jan/18
find ∫_0 ^∞ (dt/((1+t^2 )^4 ))
$${find}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{4}} } \\ $$
Commented by abdo imad last updated on 02/Feb/18
I=(1/2)∫_(−∞) ^(+∞) (dt/((1+t^2 )^4 )) let put f(z)= (1/((1+z^2 )^4 )) f(z)= (1/((z−i)^4 (z+i)^4 )) the poles of f are i and −i with ordr4 ∫_(−∞) ^(+∞) f(z)dz= 2iπ Res(f,i) but Res(f,i) =lim_(z→i) (1/((4−1)!))((z−i)^4 f(z))^((3)) =lim_(z→i) (1/6)( (z+i)^(−4) )^((3)) and we have (z+i)^(−4) )^((1)) =−4(z+i)^(−5) ((z+i)^(−4) )^((2)) =20(z+i)^(−6) ((z+i)^(−4) )^((3)) =−120(z+i)^(−7) Res(f,i)= (1/6)(−120)(2i)^(−7) =((−20)/(2^7 i^7 )) but i^7 =(i^8 /i)=(1/i) Res(f,i)=((−20i)/2^7 )=−((2^2 .5i)/2^7 )=−i(5/(32)) and ∫_(−∞) ^(+∞) f(z)dz= 2iπ(−i(5/(32)))= ((10π)/(32))= ((5π)/(16)) . finally I=(1/2)∫_(−∞) ^(+∞) f(z)dz= ((5π)/(32)) .
$${I}=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{+\infty} \:\:\:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{4}} }\:\:{let}\:{put}\:{f}\left({z}\right)=\:\:\frac{\mathrm{1}}{\left(\mathrm{1}+{z}^{\mathrm{2}} \right)^{\mathrm{4}} } \\ $$$${f}\left({z}\right)=\:\frac{\mathrm{1}}{\left({z}−{i}\right)^{\mathrm{4}} \left({z}+{i}\right)^{\mathrm{4}} }\:{the}\:{poles}\:{of}\:{f}\:{are}\:{i}\:{and}\:−{i}\:{with}\:{ordr}\mathrm{4} \\ $$$$\:\int_{−\infty} ^{+\infty} \:{f}\left({z}\right){dz}=\:\mathrm{2}{i}\pi\:{Res}\left({f},{i}\right)\:{but} \\ $$$${Res}\left({f},{i}\right)\:={lim}_{{z}\rightarrow{i}} \:\:\frac{\mathrm{1}}{\left(\mathrm{4}−\mathrm{1}\right)!}\left(\left({z}−{i}\right)^{\mathrm{4}} {f}\left({z}\right)\right)^{\left(\mathrm{3}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\frac{\mathrm{1}}{\mathrm{6}}\left(\:\:\left({z}+{i}\right)^{−\mathrm{4}} \right)^{\left(\mathrm{3}\right)} \:\:{and}\:{we}\:{have} \\ $$$$\left.\left({z}+{i}\right)^{−\mathrm{4}} \right)^{\left(\mathrm{1}\right)} =−\mathrm{4}\left({z}+{i}\right)^{−\mathrm{5}} \\ $$$$\left(\left({z}+{i}\right)^{−\mathrm{4}} \right)^{\left(\mathrm{2}\right)} =\mathrm{20}\left({z}+{i}\right)^{−\mathrm{6}} \\ $$$$\left(\left({z}+{i}\right)^{−\mathrm{4}} \right)^{\left(\mathrm{3}\right)} =−\mathrm{120}\left({z}+{i}\right)^{−\mathrm{7}} \\ $$$${Res}\left({f},{i}\right)=\:\frac{\mathrm{1}}{\mathrm{6}}\left(−\mathrm{120}\right)\left(\mathrm{2}{i}\right)^{−\mathrm{7}} =\frac{−\mathrm{20}}{\mathrm{2}^{\mathrm{7}} {i}^{\mathrm{7}} }\:{but}\:{i}^{\mathrm{7}} =\frac{{i}^{\mathrm{8}} }{{i}}=\frac{\mathrm{1}}{{i}} \\ $$$${Res}\left({f},{i}\right)=\frac{−\mathrm{20}{i}}{\mathrm{2}^{\mathrm{7}} }=−\frac{\mathrm{2}^{\mathrm{2}} .\mathrm{5}{i}}{\mathrm{2}^{\mathrm{7}} }=−{i}\frac{\mathrm{5}}{\mathrm{32}}\:{and} \\ $$$$\int_{−\infty} ^{+\infty} {f}\left({z}\right){dz}=\:\mathrm{2}{i}\pi\left(−{i}\frac{\mathrm{5}}{\mathrm{32}}\right)=\:\frac{\mathrm{10}\pi}{\mathrm{32}}=\:\frac{\mathrm{5}\pi}{\mathrm{16}}\:.\:\:\:{finally} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{+\infty} {f}\left({z}\right){dz}=\:\frac{\mathrm{5}\pi}{\mathrm{32}}\:. \\ $$

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