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Question Number 57237 by maxmathsup by imad last updated on 31/Mar/19

let f(x) =∫_0 ^(+∞) ((sin(xt^2 −1))/(t^4 +1)) dt 1) find a explicit form of f(x) 2) let g(x) =∫_0 ^∞ ((t^2 cos(xt^2 −1))/(t^4 +1)) dt find a explicit form of g(x) 3) calculate ∫_0 ^∞ ((sin(2t^2 −1))/(t^4 +1)) dt and ∫_0 ^∞ ((t^2 cos(3t^2 −1))/(t^4 +1)) dt .

$${let}\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{+\infty} \:\:\frac{{sin}\left({xt}^{\mathrm{2}} −\mathrm{1}\right)}{{t}^{\mathrm{4}} \:+\mathrm{1}}\:{dt} \\ $$$$\left.\mathrm{1}\right)\:{find}\:\:{a}\:{explicit}\:{form}\:{of}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{let}\:{g}\left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{\mathrm{2}} \:{cos}\left({xt}^{\mathrm{2}} −\mathrm{1}\right)}{{t}^{\mathrm{4}} \:+\mathrm{1}}\:{dt} \\ $$$${find}\:{a}\:{explicit}\:{form}\:{of}\:{g}\left({x}\right) \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sin}\left(\mathrm{2}{t}^{\mathrm{2}} −\mathrm{1}\right)}{{t}^{\mathrm{4}} \:+\mathrm{1}}\:{dt}\:\:{and}\:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{\mathrm{2}} \:{cos}\left(\mathrm{3}{t}^{\mathrm{2}} −\mathrm{1}\right)}{{t}^{\mathrm{4}} \:+\mathrm{1}}\:{dt}\:. \\ $$

Commented by maxmathsup by imad last updated on 05/Apr/19

1) we have 2f(x)=∫_(−∞) ^(+∞) ((sin(xt^2 −1))/(t^4 +1))dt =Im (∫_(−∞) ^(+∞) (e^(i(xt^2 −1)) /(t^4 +1))dt) let ϕ(z) =(e^(i(xz^2 −1)) /(z^4 +1)) ⇒ϕ(z) =(e^(i(xz^2 −1)) /((z^2 −i)(z^2 +i))) =(e^(i(xz^2 −1)) /((z−e^((iπ)/4) )(z+e^((iπ)/4) )(z−e^(−((iπ)/4)) )(z+e^(−((iπ)/4)) ))) so the poles of ϕ are +^− e^((iπ)/4) and +^− e^(−((iπ)/4)) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ {Res(ϕ,e^((iπ)/4) )+Res(ϕ,−e^(−((iπ)/4)) )} Res(ϕ,e^((iπ)/4) ) =lim_(z→e^((iπ)/4) ) (z−e^((iπ)/4) )ϕ(z) = (e^(i(x(i)−1)) /(2e^((iπ)/4) (2i))) =((e^(−x) e^(−i) )/(4i e^((iπ)/4) )) =(e^(−x) /(4i)) e^(−(1+(π/4))i) Res(ϕ,−e^(−((iπ)/4)) ) =lim_(z→−e^(−((iπ)/4)) ) (z+e^(−((iπ)/4)) )ϕ(z) = (e^(i(−xi −1)) /(−2 e^(−((iπ)/4)) (−2i))) = (e^x /(4i)) e^(−i) e^((iπ)/4) =(e^x /(4i)) e^((−1+(π/4))i) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (1/(4i)){ e^(−i) { e^(−x) e^(−((iπ)/4)) +e^x e^((iπ)/4) }} =(π/2) e^(−i) {e^(−x) ( (1/(√2)) −(i/(√2))) +e^x ((1/(√2)) +(i/(√2)))} =(π/2) e^(−i) { (1/(√2))(e^x +e^(−x) )+(i/(√2))(e^x −e^(−x) )} =(π/(2(√2)))(cos1 −isin1){ 2ch(x)+2ish(x)} =(π/(√2))(cos(1)−isin(1))(ch(x)+ish(x)) =(π/(√2)){cos(1)ch(x) +icos(1)sh(x)−isin(1)ch(x) +sin(1)sh(x)} ⇒ f(x) =(π/(2(√2))){cos(1)sh(x) −sin(1)ch(x)} 2) we have f^′ (x)=∫_0 ^∞ ((t^2 cos(xt^2 −1))/(t^4 +1))dt =g(x) ⇒ g(x) =(π/(2(√2))){cos(1)ch(x)−sin(1)sh(x)} .

$$\left.\mathrm{1}\right)\:{we}\:{have}\:\mathrm{2}{f}\left({x}\right)=\int_{−\infty} ^{+\infty} \:\:\:\frac{{sin}\left({xt}^{\mathrm{2}} −\mathrm{1}\right)}{{t}^{\mathrm{4}} \:+\mathrm{1}}{dt}\:={Im}\:\left(\int_{−\infty} ^{+\infty} \:\:\frac{{e}^{{i}\left({xt}^{\mathrm{2}} −\mathrm{1}\right)} }{{t}^{\mathrm{4}} \:+\mathrm{1}}{dt}\right) \\ $$$${let}\:\varphi\left({z}\right)\:=\frac{{e}^{{i}\left({xz}^{\mathrm{2}} −\mathrm{1}\right)} }{{z}^{\mathrm{4}} \:+\mathrm{1}}\:\Rightarrow\varphi\left({z}\right)\:=\frac{{e}^{{i}\left({xz}^{\mathrm{2}} −\mathrm{1}\right)} }{\left({z}^{\mathrm{2}} −{i}\right)\left({z}^{\mathrm{2}} \:+{i}\right)}\:=\frac{{e}^{{i}\left({xz}^{\mathrm{2}} −\mathrm{1}\right)} }{\left({z}−{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)} \\ $$$${so}\:{the}\:{poles}\:{of}\:\varphi\:{are}\:\overset{−} {+}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \:{and}\:\overset{−} {+}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:\:\:{residus}\:{theorem}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\left\{{Res}\left(\varphi,{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)+{Res}\left(\varphi,−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\right\} \\ $$$${Res}\left(\varphi,{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\:={lim}_{{z}\rightarrow{e}^{\frac{{i}\pi}{\mathrm{4}}} } \:\:\:\left({z}−{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\varphi\left({z}\right)\:=\:\frac{{e}^{{i}\left({x}\left({i}\right)−\mathrm{1}\right)} }{\mathrm{2}{e}^{\frac{{i}\pi}{\mathrm{4}}} \left(\mathrm{2}{i}\right)}\:=\frac{{e}^{−{x}} \:{e}^{−{i}} }{\mathrm{4}{i}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} }\:=\frac{{e}^{−{x}} }{\mathrm{4}{i}}\:{e}^{−\left(\mathrm{1}+\frac{\pi}{\mathrm{4}}\right){i}} \\ $$$${Res}\left(\varphi,−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\:={lim}_{{z}\rightarrow−{e}^{−\frac{{i}\pi}{\mathrm{4}}} } \:\:\:\left({z}+{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\varphi\left({z}\right)\:=\:\frac{{e}^{{i}\left(−{xi}\:−\mathrm{1}\right)} }{−\mathrm{2}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \left(−\mathrm{2}{i}\right)} \\ $$$$=\:\frac{{e}^{{x}} }{\mathrm{4}{i}}\:\:{e}^{−{i}} \:{e}^{\frac{{i}\pi}{\mathrm{4}}} \:=\frac{{e}^{{x}} }{\mathrm{4}{i}}\:{e}^{\left(−\mathrm{1}+\frac{\pi}{\mathrm{4}}\right){i}} \:\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\frac{\mathrm{1}}{\mathrm{4}{i}}\left\{\:{e}^{−{i}} \:\left\{\:{e}^{−{x}} {e}^{−\frac{{i}\pi}{\mathrm{4}}} \:+{e}^{{x}} {e}^{\frac{{i}\pi}{\mathrm{4}}} \right\}\right\} \\ $$$$=\frac{\pi}{\mathrm{2}}\:\:{e}^{−{i}} \:\:\left\{{e}^{−{x}} \left(\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:−\frac{{i}}{\sqrt{\mathrm{2}}}\right)\:+{e}^{{x}} \left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:+\frac{{i}}{\sqrt{\mathrm{2}}}\right)\right\} \\ $$$$=\frac{\pi}{\mathrm{2}}\:{e}^{−{i}} \left\{\:\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\left({e}^{{x}} \:+{e}^{−{x}} \right)+\frac{{i}}{\sqrt{\mathrm{2}}}\left({e}^{{x}} \:−{e}^{−{x}} \right)\right\} \\ $$$$=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\left({cos}\mathrm{1}\:−{isin}\mathrm{1}\right)\left\{\:\mathrm{2}{ch}\left({x}\right)+\mathrm{2}{ish}\left({x}\right)\right\} \\ $$$$=\frac{\pi}{\sqrt{\mathrm{2}}}\left({cos}\left(\mathrm{1}\right)−{isin}\left(\mathrm{1}\right)\right)\left({ch}\left({x}\right)+{ish}\left({x}\right)\right) \\ $$$$=\frac{\pi}{\sqrt{\mathrm{2}}}\left\{{cos}\left(\mathrm{1}\right){ch}\left({x}\right)\:+{icos}\left(\mathrm{1}\right){sh}\left({x}\right)−{isin}\left(\mathrm{1}\right){ch}\left({x}\right)\:+{sin}\left(\mathrm{1}\right){sh}\left({x}\right)\right\}\:\Rightarrow \\ $$$${f}\left({x}\right)\:=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\left\{{cos}\left(\mathrm{1}\right){sh}\left({x}\right)\:−{sin}\left(\mathrm{1}\right){ch}\left({x}\right)\right\}\: \\ $$$$\left.\mathrm{2}\right)\:{we}\:{have}\:{f}^{'} \left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{{t}^{\mathrm{2}} {cos}\left({xt}^{\mathrm{2}} −\mathrm{1}\right)}{{t}^{\mathrm{4}} \:+\mathrm{1}}{dt}\:={g}\left({x}\right)\:\Rightarrow \\ $$$${g}\left({x}\right)\:=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\left\{{cos}\left(\mathrm{1}\right){ch}\left({x}\right)−{sin}\left(\mathrm{1}\right){sh}\left({x}\right)\right\}\:. \\ $$$$ \\ $$

Commented by maxmathsup by imad last updated on 05/Apr/19

3) ∫_0 ^∞ ((sin(2t^2 −1))/(t^4 +1)) dt =f(2) =(π/(2(√2))){cos(1)sh(2)−sin(1)ch(2)} ∫_0 ^∞ ((t^2 sin(3t^2 −1))/(t^4 +1)) dt =g(3) =(π/(2(√2))){cos(1)ch(3)−sin(1)sh(3)} .

$$\left.\mathrm{3}\right)\:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{sin}\left(\mathrm{2}{t}^{\mathrm{2}} −\mathrm{1}\right)}{{t}^{\mathrm{4}} \:+\mathrm{1}}\:{dt}\:={f}\left(\mathrm{2}\right)\:=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\left\{{cos}\left(\mathrm{1}\right){sh}\left(\mathrm{2}\right)−{sin}\left(\mathrm{1}\right){ch}\left(\mathrm{2}\right)\right\} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{\mathrm{2}} \:{sin}\left(\mathrm{3}{t}^{\mathrm{2}} −\mathrm{1}\right)}{{t}^{\mathrm{4}} \:+\mathrm{1}}\:{dt}\:={g}\left(\mathrm{3}\right)\:=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\left\{{cos}\left(\mathrm{1}\right){ch}\left(\mathrm{3}\right)−{sin}\left(\mathrm{1}\right){sh}\left(\mathrm{3}\right)\right\}\:. \\ $$

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