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Question Number 42680 by prof Abdo imad last updated on 31/Aug/18

calculale A_n (α) = ∫_(−∞) ^(+∞) ((cos(αx^n ))/(1+x^2 )) dx with n integr natural.

$${calculale}\:\:{A}_{{n}} \left(\alpha\right)\:=\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{cos}\left(\alpha{x}^{{n}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx}\:{with} \\ $$$${n}\:{integr}\:{natural}. \\ $$$$ \\ $$

Commented by prof Abdo imad last updated on 31/Aug/18

α real.

$$\alpha\:{real}. \\ $$

Commented by maxmathsup by imad last updated on 01/Sep/18

we have A_n (α) =Re( ∫_(−∞) ^(+∞) (e^(iαx^n ) /(1+x^2 )) dx) let consider the comlex function ϕ(z) =(e^(iαz^n ) /(1+z^2 )) ⇒ϕ(z) = (e^(iαz^n ) /((z−i)(z+i))) the poles of ϕ are +^− i ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) but Res(ϕ,i)= (e^(iαi^n ) /(2i)) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (e^(iαi^n ) /(2i)) =π e^(iα(−1)^(n/2) ) =π{ cos((−1)^(n/2) α)+isin((−1)^(n/2) α)} ⇒ A_n = π cos{(−1)^(n/2) α} ⇒ A_(2n) = π cos{(−1)^n α} and A_(2n+1) =π cos{i(−1)^n α} but cosz =((e^(iz) +e^(−iz) )/2) ⇒cos(ix)=((e^(−x) +e^x )/2) =ch(x) (x∈R) ⇒ A_(2n+1) =π ch{(−1)^n α} .

$${we}\:{have}\:{A}_{{n}} \left(\alpha\right)\:={Re}\left(\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{e}^{{i}\alpha{x}^{{n}} } }{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx}\right)\:{let}\:{consider}\:{the}\:{comlex}\:{function} \\ $$$$\varphi\left({z}\right)\:=\frac{{e}^{{i}\alpha{z}^{{n}} } }{\mathrm{1}+{z}^{\mathrm{2}} }\:\:\:\:\Rightarrow\varphi\left({z}\right)\:=\:\frac{{e}^{{i}\alpha{z}^{{n}} } }{\left({z}−{i}\right)\left({z}+{i}\right)}\:{the}\:{poles}\:{of}\:\varphi\:{are}\:\overset{−} {+}{i} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{i}\right)\:{but}\:\:\:{Res}\left(\varphi,{i}\right)=\:\frac{{e}^{{i}\alpha{i}^{{n}} } }{\mathrm{2}{i}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\frac{{e}^{{i}\alpha{i}^{{n}} } }{\mathrm{2}{i}}\:=\pi\:{e}^{{i}\alpha\left(−\mathrm{1}\right)^{\frac{{n}}{\mathrm{2}}} } \:=\pi\left\{\:{cos}\left(\left(−\mathrm{1}\right)^{\frac{{n}}{\mathrm{2}}} \alpha\right)+{isin}\left(\left(−\mathrm{1}\right)^{\frac{{n}}{\mathrm{2}}} \alpha\right)\right\}\:\Rightarrow \\ $$$${A}_{{n}} =\:\pi\:{cos}\left\{\left(−\mathrm{1}\right)^{\frac{{n}}{\mathrm{2}}} \alpha\right\}\:\Rightarrow\:{A}_{\mathrm{2}{n}} =\:\pi\:\:{cos}\left\{\left(−\mathrm{1}\right)^{{n}} \alpha\right\}\:{and} \\ $$$${A}_{\mathrm{2}{n}+\mathrm{1}} =\pi\:{cos}\left\{{i}\left(−\mathrm{1}\right)^{{n}} \alpha\right\}\:\:\:{but}\:\:{cosz}\:=\frac{{e}^{{iz}} \:+{e}^{−{iz}} }{\mathrm{2}}\:\Rightarrow{cos}\left({ix}\right)=\frac{{e}^{−{x}} \:+{e}^{{x}} }{\mathrm{2}}\:={ch}\left({x}\right) \\ $$$$\left({x}\in{R}\right)\:\Rightarrow\:{A}_{\mathrm{2}{n}+\mathrm{1}} =\pi\:\:{ch}\left\{\left(−\mathrm{1}\right)^{{n}} \alpha\right\}\:. \\ $$

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