Menu Close

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-




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

Leave a Reply

Your email address will not be published. Required fields are marked *