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Question Number 56699 by maxmathsup by imad last updated on 21/Mar/19
let f_n (a)=∫_(−∞) ^∞ ((sin(x^n ))/(x^2 +a^2 )) dx with a positif real not 0 and n from N 1) find a explicit form of f(a) 2) calculate g_n (a) =∫_(−∞) ^(+∞) ((sin(x^n ))/((x^2 +a^2 )^2 ))dx 3) calculate ∫_(−∞) ^(+∞) ((sin(x^3 ))/(x^2 +4)) dx and ∫_(−∞) ^(+∞) ((sin(x^2 ))/(x^2 +9))dx 4) calculate ∫_(−∞) ^(+∞) ((sin(x^3 ))/((x^2 +4)^2 ))dx .
$${let}\:{f}_{{n}} \left({a}\right)=\int_{−\infty} ^{\infty} \:\:\frac{{sin}\left({x}^{{n}} \right)}{{x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} }\:{dx}\:\:\:{with}\:{a}\:{positif}\:{real}\:{not}\:\mathrm{0}\:\:{and}\:{n}\:{from}\:{N} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{explicit}\:{form}\:{of}\:{f}\left({a}\right) \\ $$$$\left.\mathrm{2}\right)\:\:{calculate}\:{g}_{{n}} \left({a}\right)\:=\int_{−\infty} ^{+\infty} \:\:\:\frac{{sin}\left({x}^{{n}} \right)}{\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{sin}\left({x}^{\mathrm{3}} \right)}{{x}^{\mathrm{2}} \:+\mathrm{4}}\:{dx}\:{and}\:\int_{−\infty} ^{+\infty} \:\:\frac{{sin}\left({x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} \:+\mathrm{9}}{dx} \\ $$$$\left.\mathrm{4}\right)\:{calculate}\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{sin}\left({x}^{\mathrm{3}} \right)}{\left({x}^{\mathrm{2}} \:+\mathrm{4}\right)^{\mathrm{2}} }{dx}\:. \\ $$
Commented by maxmathsup by imad last updated on 23/Mar/19
1)we have f_n (a) =Im( ∫_(−∞) ^(+∞) (e^(ix^n ) /(x^2 +a^2 ))dx) let consider the complex function ϕ(z) =(e^(iz^n ) /(z^2 +a^2 )) ⇒ϕ(z)=(e^(iz^n ) /((z−ia)(z+ia))) so the poles of ϕ are ia and −ia residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,ia) Res(ϕ,ia) =lim_(z→ia) (z−ia)ϕ(z)= (e^(i(ia)^n ) /(2ia)) =(e^(i^(n+1) a^n ) /(2ia)) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (e^(i^(n+1) a^n ) /(2ia)) =(π/a) e^(i^(n+1) a^n ) if n is even n=2p ⇒∫_(−∞) ^(+∞) ϕ(z)dz =(π/a) e^(i^(2p+1) a^n ) =(π/a) e^(i(−1)^p a^(2p) ) =(π/a){cos((−1)^p a^(2p) +isin((−1)^p a^(2p) } ⇒f_n (a) =(π/a) sin((−1)^p a^(2p) ) (n=2p) if n is odd ⇒n=2p+1 ⇒∫_(−∞) ^(+∞) ϕ(z)dz =(π/a) e^(i^(2p+2) a^(2p+1) ) =(π/a) e^((−1)^(p+1) a^(2p+1) ) ⇒ f_n (a) =0 2) we have by derivation f_n ^′ (a) = ∫_(−∞) ^(+∞) (∂/∂a)(((sin(x^n ))/(x^2 +a^2 )))dx =−∫_(−∞) ^(+∞) ((2a sin(x^n ))/((x^2 +a^2 )^2 )) dx =−2a g_n (a) ⇒g_n (a)=−(1/(2a)) f_n ^′ (a) g_(2p) (a) =−(1/(2a)) f_(2p) ^′ (a) =−(1/(2a)) {−(π/a^2 ) sin{(−1)^p a^(2p) }+(π/a)(2p)(−1)^p a^(2p−1) cos{(−1)^p a^(2p) } =(π/(2a^3 )) sin{(−1)^p a^(2p) } +((pπ)/a^2 ) (−1)^(p+1) a^(2p−1) cos{(−1)^p a^(2p) } g_(2p+1) (a) =0 due to f_(2p+1) (a)=0 . 3) ∫_(−∞) ^(+∞) ((sin(x^3 ))/(x^2 +4)) dx =f_3 (2)=0 because 3 is odd. ∫_(−∞) ^(+∞) ((sin(x^2 ))/(x^2 +9)) =f_2 (3) =(π/3) sin(−9) =−(π/3) sin(9) .
$$\left.\mathrm{1}\right){we}\:{have}\:{f}_{{n}} \left({a}\right)\:={Im}\left(\:\int_{−\infty} ^{+\infty} \:\:\frac{{e}^{{ix}^{{n}} } }{{x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} }{dx}\right)\:\:{let}\:{consider}\:{the}\:{complex}\:{function} \\ $$$$\varphi\left({z}\right)\:=\frac{{e}^{{iz}^{{n}} } }{{z}^{\mathrm{2}} \:+{a}^{\mathrm{2}} }\:\Rightarrow\varphi\left({z}\right)=\frac{{e}^{{iz}^{{n}} } }{\left({z}−{ia}\right)\left({z}+{ia}\right)}\:{so}\:{the}\:{poles}\:{of}\:\varphi\:{are}\:{ia}\:{and}\:−{ia} \\ $$$${residus}\:{theorem}\:{give}\:\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{ia}\right) \\ $$$${Res}\left(\varphi,{ia}\right)\:={lim}_{{z}\rightarrow{ia}} \left({z}−{ia}\right)\varphi\left({z}\right)=\:\frac{{e}^{{i}\left({ia}\right)^{{n}} } }{\mathrm{2}{ia}}\:=\frac{{e}^{{i}^{{n}+\mathrm{1}} {a}^{{n}} } }{\mathrm{2}{ia}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\frac{{e}^{{i}^{{n}+\mathrm{1}} {a}^{{n}} } }{\mathrm{2}{ia}}\:=\frac{\pi}{{a}}\:{e}^{{i}^{{n}+\mathrm{1}} {a}^{{n}} } \\ $$$${if}\:{n}\:{is}\:{even}\:{n}=\mathrm{2}{p}\:\Rightarrow\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\frac{\pi}{{a}}\:{e}^{{i}^{\mathrm{2}{p}+\mathrm{1}} {a}^{{n}} } =\frac{\pi}{{a}}\:{e}^{{i}\left(−\mathrm{1}\right)^{{p}} {a}^{\mathrm{2}{p}} } \\ $$$$=\frac{\pi}{{a}}\left\{{cos}\left(\left(−\mathrm{1}\right)^{{p}} \:{a}^{\mathrm{2}{p}} \:+{isin}\left(\left(−\mathrm{1}\right)^{{p}} {a}^{\mathrm{2}{p}} \right\}\:\Rightarrow{f}_{{n}} \left({a}\right)\:=\frac{\pi}{{a}}\:{sin}\left(\left(−\mathrm{1}\right)^{{p}} {a}^{\mathrm{2}{p}} \right)\:\:\left({n}=\mathrm{2}{p}\right)\right.\right. \\ $$$${if}\:{n}\:{is}\:{odd}\:\:\Rightarrow{n}=\mathrm{2}{p}+\mathrm{1}\:\Rightarrow\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\frac{\pi}{{a}}\:{e}^{{i}^{\mathrm{2}{p}+\mathrm{2}} {a}^{\mathrm{2}{p}+\mathrm{1}} } =\frac{\pi}{{a}}\:{e}^{\left(−\mathrm{1}\right)^{{p}+\mathrm{1}} {a}^{\mathrm{2}{p}+\mathrm{1}} } \:\Rightarrow \\ $$$${f}_{{n}} \left({a}\right)\:=\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\:{we}\:{have}\:{by}\:{derivation}\:{f}_{{n}} ^{'} \left({a}\right)\:=\:\int_{−\infty} ^{+\infty} \:\frac{\partial}{\partial{a}}\left(\frac{{sin}\left({x}^{{n}} \right)}{{x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} }\right){dx} \\ $$$$=−\int_{−\infty} ^{+\infty} \:\:\frac{\mathrm{2}{a}\:{sin}\left({x}^{{n}} \right)}{\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dx}\:=−\mathrm{2}{a}\:{g}_{{n}} \left({a}\right)\:\Rightarrow{g}_{{n}} \left({a}\right)=−\frac{\mathrm{1}}{\mathrm{2}{a}}\:{f}_{{n}} ^{'} \left({a}\right) \\ $$$${g}_{\mathrm{2}{p}} \left({a}\right)\:=−\frac{\mathrm{1}}{\mathrm{2}{a}}\:{f}_{\mathrm{2}{p}} ^{'} \left({a}\right)\:=−\frac{\mathrm{1}}{\mathrm{2}{a}}\:\left\{−\frac{\pi}{{a}^{\mathrm{2}} }\:{sin}\left\{\left(−\mathrm{1}\right)^{{p}} {a}^{\mathrm{2}{p}} \right\}+\frac{\pi}{{a}}\left(\mathrm{2}{p}\right)\left(−\mathrm{1}\right)^{{p}} {a}^{\mathrm{2}{p}−\mathrm{1}} {cos}\left\{\left(−\mathrm{1}\right)^{{p}} {a}^{\mathrm{2}{p}} \right\}\right. \\ $$$$=\frac{\pi}{\mathrm{2}{a}^{\mathrm{3}} }\:{sin}\left\{\left(−\mathrm{1}\right)^{{p}} \:{a}^{\mathrm{2}{p}} \right\}\:+\frac{{p}\pi}{{a}^{\mathrm{2}} }\:\left(−\mathrm{1}\right)^{{p}+\mathrm{1}} \:{a}^{\mathrm{2}{p}−\mathrm{1}} \:{cos}\left\{\left(−\mathrm{1}\right)^{{p}} \:{a}^{\mathrm{2}{p}} \right\} \\ $$$${g}_{\mathrm{2}{p}+\mathrm{1}} \left({a}\right)\:=\mathrm{0}\:\:{due}\:{to}\:{f}_{\mathrm{2}{p}+\mathrm{1}} \left({a}\right)=\mathrm{0}\:. \\ $$$$\left.\mathrm{3}\right)\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{sin}\left({x}^{\mathrm{3}} \right)}{{x}^{\mathrm{2}} \:+\mathrm{4}}\:{dx}\:={f}_{\mathrm{3}} \left(\mathrm{2}\right)=\mathrm{0}\:{because}\:\mathrm{3}\:{is}\:{odd}. \\ $$$$\int_{−\infty} ^{+\infty} \:\:\:\frac{{sin}\left({x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} \:+\mathrm{9}}\:={f}_{\mathrm{2}} \left(\mathrm{3}\right)\:=\frac{\pi}{\mathrm{3}}\:{sin}\left(−\mathrm{9}\right)\:=−\frac{\pi}{\mathrm{3}}\:{sin}\left(\mathrm{9}\right)\:. \\ $$
Commented by maxmathsup by imad last updated on 23/Mar/19
4) ∫_(−∞) ^(+∞) ((sin(x^3 ))/((x^2 +4)^2 ))dx =g_3 (2)=0 because 3 is odd.
$$\left.\mathrm{4}\right)\:\int_{−\infty} ^{+\infty} \:\:\frac{{sin}\left({x}^{\mathrm{3}} \right)}{\left({x}^{\mathrm{2}} \:+\mathrm{4}\right)^{\mathrm{2}} }{dx}\:={g}_{\mathrm{3}} \left(\mathrm{2}\right)=\mathrm{0}\:{because}\:\mathrm{3}\:{is}\:{odd}. \\ $$

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