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Question Number 69803 by Abdo msup. last updated on 28/Sep/19
1)find f(α) =∫_0 ^∞ ((cos(αx))/((x^4 +1)^2 ))dx with α real 2) find the value of ∫_0 ^∞ ((cos(2x))/((x^4 +1)^2 ))dx 3) find nature of the serie Σf(n)
$$\left.\mathrm{1}\right){find}\:\:\:{f}\left(\alpha\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{cos}\left(\alpha{x}\right)}{\left({x}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{2}} }{dx}\:\:{with}\:\alpha\:{real} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left(\mathrm{2}{x}\right)}{\left({x}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$$\left.\mathrm{3}\right)\:{find}\:{nature}\:{of}\:{the}\:{serie}\:\Sigma{f}\left({n}\right) \\ $$
Commented by mathmax by abdo last updated on 29/Sep/19
1) f(α) =∫_0 ^∞ ((cos(αx))/((x^4 +1)^2 ))dx ⇒2f(α) =∫_(−∞) ^(+∞) ((cos(αx))/((x^4 +1)^2 ))dx =Re(∫_(−∞) ^(+∞) (e^(iαx) /((x^4 +1)^2 ))dx) let W(z) =(e^(iαz) /((z^4 +1)^2 )) poles of W? W(z) =(e^(iαz) /((z^2 −i)^2 (z^2 +i)^2 )) =(e^(iαz) /((z−e^((iπ)/4) )^2 (z+e^((iπ)/4) )^2 (z−e^(−((iπ)/4)) )^2 (z+e^(−((iπ)/4)) )^2 )) so the poles of W are +^− e^((iπ)/4) and +^− e^(−((iπ)/4)) ∫_(−∞) ^(+∞) W(z)dz =2iπ { Res(W,e^((iπ)/4) ) +Res(W,−e^(−((iπ)/4)) )} Res(W,e^((iπ)/4) ) =lim_(z→e^((iπ)/4) ) (1/((2−1)!)){ (z−e^((iπ)/4) )^2 W(z)}^((1)) =lim_(z→e^((iπ)/4) ) {(e^(iαz) /((z+e^((iπ)/4) )^2 (z^2 +i)^2 ))}^((1)) =lim_(z→e^((iπ)/4) ) ((iα e^(iαz) (z+e^((iπ)/4) )^2 (z^2 +i)^2 −e^(iαz) {(z+e^((iπ)/4) )^2 (z^2 +i)^2 }^((1)) )/((z+e^((iπ)/4) )^4 (z^2 +i)^4 )) but (d/dz){(z+e^((iπ)/4) )^2 (z^2 +i)^2 }=2(z+e^((iπ)/4) )(z^2 +i)^2
$$\left.\mathrm{1}\right)\:{f}\left(\alpha\right)\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{cos}\left(\alpha{x}\right)}{\left({x}^{\mathrm{4}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dx}\:\Rightarrow\mathrm{2}{f}\left(\alpha\right)\:=\int_{−\infty} ^{+\infty} \:\frac{{cos}\left(\alpha{x}\right)}{\left({x}^{\mathrm{4}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$$={Re}\left(\int_{−\infty} ^{+\infty} \:\frac{{e}^{{i}\alpha{x}} }{\left({x}^{\mathrm{4}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dx}\right)\:\:{let}\:{W}\left({z}\right)\:=\frac{{e}^{{i}\alpha{z}} }{\left({z}^{\mathrm{4}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:\:\:{poles}\:{of}\:{W}? \\ $$$${W}\left({z}\right)\:=\frac{{e}^{{i}\alpha{z}} }{\left({z}^{\mathrm{2}} −{i}\right)^{\mathrm{2}} \left({z}^{\mathrm{2}} +{i}\right)^{\mathrm{2}} }\:=\frac{{e}^{{i}\alpha{z}} }{\left({z}−{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} \left({z}+{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} \left({z}−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} \left({z}+{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} } \\ $$$${so}\:{the}\:{poles}\:{of}\:{W}\:{are}\:\overset{−} {+}{e}^{\frac{{i}\pi}{\mathrm{4}}} \:\:{and}\:\overset{−} {+}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \\ $$$$\int_{−\infty} ^{+\infty} \:\:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\left\{\:{Res}\left({W},{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\:+{Res}\left({W},−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\right\} \\ $$$${Res}\left({W},{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\:={lim}_{{z}\rightarrow{e}^{\frac{{i}\pi}{\mathrm{4}}} } \:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\:\left({z}−{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} {W}\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{e}^{\frac{{i}\pi}{\mathrm{4}}} } \:\:\:\left\{\frac{{e}^{{i}\alpha{z}} }{\left({z}+{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} \left({z}^{\mathrm{2}} \:+{i}\right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{e}^{\frac{{i}\pi}{\mathrm{4}}} } \:\:\:\:\frac{{i}\alpha\:{e}^{{i}\alpha{z}} \left({z}+{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} \left({z}^{\mathrm{2}} \:+{i}\right)^{\mathrm{2}} \:−{e}^{{i}\alpha{z}} \left\{\left({z}+{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} \left({z}^{\mathrm{2}} \:+{i}\right)^{\mathrm{2}} \right\}^{\left(\mathrm{1}\right)} }{\left({z}+{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{4}} \left({z}^{\mathrm{2}} \:+{i}\right)^{\mathrm{4}} } \\ $$$${but}\:\frac{{d}}{{dz}}\left\{\left({z}+{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} \left({z}^{\mathrm{2}} \:+{i}\right)^{\mathrm{2}} \right\}=\mathrm{2}\left({z}+{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}^{\mathrm{2}} \:+{i}\right)^{\mathrm{2}} \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 29/Sep/19
(d/dz){(z+e^((iπ)/4) )^2 (z^2 +i)^2 } =2(z+e^((iπ)/4) )(z^2 +i)+4z(z^2 +i)(z+e^((iπ)/4) )^2 Res(W,e^((iπ)/4) )=lim_(z→e^((iπ)/4) ) ((iαe^(iαz) (z+e^((iπ)/4) )^2 (z^2 +i)^2 −e^(iαz) {2(z+e^((iπ)/4) )(z^2 +i)+4z(z^2 +i)(z+e^((iπ)/4) )^2 )/((z+e^((iπ)/4) )^4 (z^2 +i)^4 )) ...be continued...
$$\frac{{d}}{{dz}}\left\{\left({z}+{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} \left({z}^{\mathrm{2}} \:+{i}\right)^{\mathrm{2}} \right\}\:=\mathrm{2}\left({z}+{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}^{\mathrm{2}} \:+{i}\right)+\mathrm{4}{z}\left({z}^{\mathrm{2}} \:+{i}\right)\left({z}+{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} \\ $$$${Res}\left({W},{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)={lim}_{{z}\rightarrow{e}^{\frac{{i}\pi}{\mathrm{4}}} } \:\:\frac{{i}\alpha{e}^{{i}\alpha{z}} \left({z}+{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} \left({z}^{\mathrm{2}} \:+{i}\right)^{\mathrm{2}} −{e}^{{i}\alpha{z}} \left\{\mathrm{2}\left({z}+{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}^{\mathrm{2}} +{i}\right)+\mathrm{4}{z}\left({z}^{\mathrm{2}} +{i}\right)\left({z}+{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} \right.}{\left({z}+{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{4}} \left({z}^{\mathrm{2}} \:+{i}\right)^{\mathrm{4}} } \\ $$$$…{be}\:{continued}… \\ $$

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