let f(x) =∫_0 ^∞ ((sin(t^2 ))/((x^2 +t^2 )^2 ))dt with x>0 1)determine a explicit form for f(x) 2) find also g(x) =∫_0 ^∞ ((sin(t^2 ))/((x^2 +t^2 )^3 ))dt 3) give f^((n)) (x) at form of integral and calculate f^((n)) (1). 4) find the valueof ∫_0 ^∞ ((sin(t^2 ))/((1+t^2 )^2 )) dt and ∫


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Question Number 67744 by mathmax by abdo last updated on 31/Aug/19

$${let}\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sin}\left({t}^{\mathrm{2}} \right)}{\left({x}^{\mathrm{2}} \:+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt}\:\:{with}\:{x}>\mathrm{0} \\ $$ $$\left.\mathrm{1}\right){determine}\:{a}\:{explicit}\:{form}\:{for}\:{f}\left({x}\right) \\ $$ $$\left.\mathrm{2}\right)\:{find}\:\:{also}\:{g}\left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sin}\left({t}^{\mathrm{2}} \right)}{\left({x}^{\mathrm{2}} \:+{t}^{\mathrm{2}} \right)^{\mathrm{3}} }{dt} \\ $$ $$\left.\mathrm{3}\right)\:{give}\:{f}^{\left({n}\right)} \left({x}\right)\:{at}\:{form}\:{of}\:{integral}\:{and}\:{calculate}\:{f}^{\left({n}\right)} \left(\mathrm{1}\right). \\ $$ $$\left.\mathrm{4}\right)\:{find}\:{the}\:{valueof}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sin}\left({t}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dt}\:{and}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sin}\left({t}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{3}} }{dt} \\ $$
Commented bymathmax by abdo last updated on 31/Aug/19
$1) we have f(x)=∫_0 ^∞ ((sin(t^2 ))/((x^2 +t^2 )^2 ))dt cha7gement t=xu give f(x) =∫_0 ^∞ ((sin(x^2 u^2 ))/(x^4 (1+u^2 )^2 )) xdu =(1/x^3 ) ∫_0 ^∞ ((sin(x^2 u^2 ))/((u^2 +1)^2 ))du =(1/(2x^3 ))∫_(−∞) ^(+∞) ((sin(x^2 u^2 ))/((u^2 +1)^2 ))du =(1/(2x^3 )) Im(∫_(−∞) ^(+∞) (e^(ix^2 u^2 ) /((u^2 +1)^2 ))du)let W(z)=(e^(ix^2 z^2 ) /((z^2 +1)^2 )) ⇒W(z) =(e^(ix^2 z^2 ) /((z−i)^2 (z+i)^2 )) so the poles of W are i and −i(doubles) residus theorem give ∫_(−∞) ^(+∞) W(z)dz =2iπ Res(W,i) and Res(W,i) =lim_(z→i) (1/((2−1)!)){(z−i)^2 W(z)}^((1)) =lim_(z→i) { (e^(ix^2 z^2 ) /((z+i)^2 ))}^((1)) =lim_(z→i) ((2ix^2 z e^(ix^2 z^2 ) (z+i)^2 −2(z+i)e^(ix^2 z^2 ) )/((z+i)^4 )) =lim_(z→i) ((2ix^2 z e^(ix^2 z^2 ) (z+i)−2 e^(ix^2 z^2 ) )/((z+i)^3 )) =lim_(z→i) ((2ix^2 z(z+i)−2)/((z+i)^3 )) e^(ix^2 z^2 ) =((((2i)(−2x^2 )−2)e^(−ix^2 ) )/((2i)^3 )) =(((−4ix^2 −2) e^(−ix^2 ) )/(−8i)) =(((2ix^2 −1)e^(−ix^2 ) )/(4i)) ⇒ ∫_(−∞) ^(+∞) W(z)dz =2iπ×(((2ix^2 −1)e^(−ix^2 ) )/(4i)) =(π/2)(2ix^2 −1)e^(−ix^2 ) =(π/2)(2ix^2 −1)(cos(x^2 )−isin(x^2 )) =(π/2){2ix^2 cos(x^2 ) +2x^2 sin(x^2 )−cos(x^2 )+isin(x^2 )} =(π/2){2x^2 sin(x^2 )−cos(x^2 ) +i(2x^2 cos(x^2 )+sin(x^2 )} ⇒ f(x)=((2x^2 cos(x^2 )+sin(x^2 ))/(2x^3 )) .$
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sin}\left({t}^{\mathrm{2}} \right)}{\left({x}^{\mathrm{2}} +{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt}\:{cha}\mathrm{7}{gement}\:\:{t}={xu}\:{give} \\ $$ $${f}\left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sin}\left({x}^{\mathrm{2}} {u}^{\mathrm{2}} \right)}{{x}^{\mathrm{4}} \left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{xdu}\:=\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{sin}\left({x}^{\mathrm{2}} {u}^{\mathrm{2}} \right)}{\left({u}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{du} \\ $$ $$=\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{3}} }\int_{−\infty} ^{+\infty} \:\frac{{sin}\left({x}^{\mathrm{2}} {u}^{\mathrm{2}} \right)}{\left({u}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{du}\:\:=\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{3}} }\:{Im}\left(\int_{−\infty} ^{+\infty} \:\frac{{e}^{{ix}^{\mathrm{2}} {u}^{\mathrm{2}} } }{\left({u}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{du}\right){let} \\ $$ $${W}\left({z}\right)=\frac{{e}^{{ix}^{\mathrm{2}} {z}^{\mathrm{2}} } }{\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow{W}\left({z}\right)\:=\frac{{e}^{{ix}^{\mathrm{2}} {z}^{\mathrm{2}} } }{\left({z}−{i}\right)^{\mathrm{2}} \left({z}+{i}\right)^{\mathrm{2}} }\:\:{so}\:{the}\:{poles}\:{of}\:{W}\:{are}\:{i} \\ $$ $${and}\:−{i}\left({doubles}\right)\:{residus}\:{theorem}\:{give} \\ $$ $$\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left({W},{i}\right)\:{and} \\ $$ $${Res}\left({W},{i}\right)\:={lim}_{{z}\rightarrow{i}} \:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\left({z}−{i}\right)^{\mathrm{2}} {W}\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$ $$={lim}_{{z}\rightarrow{i}} \:\:\:\left\{\:\frac{{e}^{{ix}^{\mathrm{2}} {z}^{\mathrm{2}} } }{\left({z}+{i}\right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \:={lim}_{{z}\rightarrow{i}} \:\:\frac{\mathrm{2}{ix}^{\mathrm{2}} {z}\:{e}^{{ix}^{\mathrm{2}} {z}^{\mathrm{2}} } \left({z}+{i}\right)^{\mathrm{2}} −\mathrm{2}\left({z}+{i}\right){e}^{{ix}^{\mathrm{2}} {z}^{\mathrm{2}} } }{\left({z}+{i}\right)^{\mathrm{4}} } \\ $$ $$={lim}_{{z}\rightarrow{i}} \:\:\frac{\mathrm{2}{ix}^{\mathrm{2}} {z}\:{e}^{{ix}^{\mathrm{2}} {z}^{\mathrm{2}} } \left({z}+{i}\right)−\mathrm{2}\:{e}^{{ix}^{\mathrm{2}} {z}^{\mathrm{2}} } }{\left({z}+{i}\right)^{\mathrm{3}} } \\ $$ $$={lim}_{{z}\rightarrow{i}} \:\frac{\mathrm{2}{ix}^{\mathrm{2}} {z}\left({z}+{i}\right)−\mathrm{2}}{\left({z}+{i}\right)^{\mathrm{3}} }\:{e}^{{ix}^{\mathrm{2}} {z}^{\mathrm{2}} } \:=\frac{\left(\left(\mathrm{2}{i}\right)\left(−\mathrm{2}{x}^{\mathrm{2}} \right)−\mathrm{2}\right){e}^{−{ix}^{\mathrm{2}} } }{\left(\mathrm{2}{i}\right)^{\mathrm{3}} } \\ $$ $$=\frac{\left(−\mathrm{4}{ix}^{\mathrm{2}} −\mathrm{2}\right)\:{e}^{−{ix}^{\mathrm{2}} } }{−\mathrm{8}{i}}\:=\frac{\left(\mathrm{2}{ix}^{\mathrm{2}} −\mathrm{1}\right){e}^{−{ix}^{\mathrm{2}} } }{\mathrm{4}{i}}\:\Rightarrow \\ $$ $$\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi×\frac{\left(\mathrm{2}{ix}^{\mathrm{2}} −\mathrm{1}\right){e}^{−{ix}^{\mathrm{2}} } }{\mathrm{4}{i}}\:=\frac{\pi}{\mathrm{2}}\left(\mathrm{2}{ix}^{\mathrm{2}} −\mathrm{1}\right){e}^{−{ix}^{\mathrm{2}} } \\ $$ $$=\frac{\pi}{\mathrm{2}}\left(\mathrm{2}{ix}^{\mathrm{2}} −\mathrm{1}\right)\left({cos}\left({x}^{\mathrm{2}} \right)−{isin}\left({x}^{\mathrm{2}} \right)\right) \\ $$ $$=\frac{\pi}{\mathrm{2}}\left\{\mathrm{2}{ix}^{\mathrm{2}} {cos}\left({x}^{\mathrm{2}} \right)\:+\mathrm{2}{x}^{\mathrm{2}} {sin}\left({x}^{\mathrm{2}} \right)−{cos}\left({x}^{\mathrm{2}} \right)+{isin}\left({x}^{\mathrm{2}} \right)\right\} \\ $$ $$=\frac{\pi}{\mathrm{2}}\left\{\mathrm{2}{x}^{\mathrm{2}} {sin}\left({x}^{\mathrm{2}} \right)−{cos}\left({x}^{\mathrm{2}} \right)\:+{i}\left(\mathrm{2}{x}^{\mathrm{2}} {cos}\left({x}^{\mathrm{2}} \right)+{sin}\left({x}^{\mathrm{2}} \right)\right\}\:\Rightarrow\right. \\ $$ $${f}\left({x}\right)=\frac{\mathrm{2}{x}^{\mathrm{2}} {cos}\left({x}^{\mathrm{2}} \right)+{sin}\left({x}^{\mathrm{2}} \right)}{\mathrm{2}{x}^{\mathrm{3}} }\:. \\ $$ $$ \\ $$
Commented bymathmax by abdo last updated on 31/Aug/19
$sorry f(x) =(π/(4x^3 )){2x^2 cos(x^2 )+sin(x^2 )}$
$${sorry}\:{f}\left({x}\right)\:=\frac{\pi}{\mathrm{4}{x}^{\mathrm{3}} }\left\{\mathrm{2}{x}^{\mathrm{2}} {cos}\left({x}^{\mathrm{2}} \right)+{sin}\left({x}^{\mathrm{2}} \right)\right\} \\ $$
Commented bymathmax by abdo last updated on 31/Aug/19
$2) we have f^′ (x) =−∫_0 ^∞ ((−2(2x)(x^2 +t^2 ))/((x^2 +t^2 )^4 ))sin(t^2 )dt =4x∫_0 ^∞ ((sin(t^2 ))/((x^2 +t^2 )^3 )) =4x g(x) ⇒g(x) =(1/(4x))f^′ (x) we have f(x)=(π/(4x^3 )){2x^2 cos(x^2 )+sin(x^2 )} ⇒ f^′ (x) =(π/4)(((−3x^2 )/x^6 )){2x^2 cos(x^2 )+sin(x^2 )} +(π/(4x^3 )){ 4xcos(x^2 )−4x^3 sin(x^2 )+2xcos(x^2 )}=...$
$$\left.\mathrm{2}\right)\:{we}\:{have}\:{f}^{'} \left({x}\right)\:=−\int_{\mathrm{0}} ^{\infty} \:\frac{−\mathrm{2}\left(\mathrm{2}{x}\right)\left({x}^{\mathrm{2}} +{t}^{\mathrm{2}} \right)}{\left({x}^{\mathrm{2}} \:+{t}^{\mathrm{2}} \right)^{\mathrm{4}} }{sin}\left({t}^{\mathrm{2}} \right){dt}\:=\mathrm{4}{x}\int_{\mathrm{0}} ^{\infty} \:\frac{{sin}\left({t}^{\mathrm{2}} \right)}{\left({x}^{\mathrm{2}} \:+{t}^{\mathrm{2}} \right)^{\mathrm{3}} } \\ $$ $$=\mathrm{4}{x}\:{g}\left({x}\right)\:\Rightarrow{g}\left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{4}{x}}{f}^{'} \left({x}\right)\:\:{we}\:{have}\: \\ $$ $${f}\left({x}\right)=\frac{\pi}{\mathrm{4}{x}^{\mathrm{3}} }\left\{\mathrm{2}{x}^{\mathrm{2}} {cos}\left({x}^{\mathrm{2}} \right)+{sin}\left({x}^{\mathrm{2}} \right)\right\}\:\Rightarrow \\ $$ $${f}^{'} \left({x}\right)\:=\frac{\pi}{\mathrm{4}}\left(\frac{−\mathrm{3}{x}^{\mathrm{2}} }{{x}^{\mathrm{6}} }\right)\left\{\mathrm{2}{x}^{\mathrm{2}} {cos}\left({x}^{\mathrm{2}} \right)+{sin}\left({x}^{\mathrm{2}} \right)\right\} \\ $$ $$+\frac{\pi}{\mathrm{4}{x}^{\mathrm{3}} }\left\{\:\mathrm{4}{xcos}\left({x}^{\mathrm{2}} \right)−\mathrm{4}{x}^{\mathrm{3}} {sin}\left({x}^{\mathrm{2}} \right)+\mathrm{2}{xcos}\left({x}^{\mathrm{2}} \right)\right\}=... \\ $$
Commented bymathmax by abdo last updated on 31/Aug/19
$4) ∫_0 ^∞ ((sin(t^2 ))/((1+t^2 )^2 ))dt =f(1)=(π/4){2cos(1) +sin(1)} =(π/2)cos(1)+(π/4)sin(1). ∫_0 ^∞ ((sin(t^2 ))/((1+t^2 )^3 ))dt =g(1) rest to calculste g(1)...$
$$\left.\mathrm{4}\right)\:\int_{\mathrm{0}} ^{\infty} \:\frac{{sin}\left({t}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt}\:={f}\left(\mathrm{1}\right)=\frac{\pi}{\mathrm{4}}\left\{\mathrm{2}{cos}\left(\mathrm{1}\right)\:+{sin}\left(\mathrm{1}\right)\right\} \\ $$ $$=\frac{\pi}{\mathrm{2}}{cos}\left(\mathrm{1}\right)+\frac{\pi}{\mathrm{4}}{sin}\left(\mathrm{1}\right). \\ $$ $$\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sin}\left({t}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{3}} }{dt}\:={g}\left(\mathrm{1}\right)\:{rest}\:{to}\:{calculste}\:{g}\left(\mathrm{1}\right)... \\ $$
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