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Question Number 120025 by mathmax by abdo last updated on 28/Oct/20
calculate f^′ (x)  1) f(x) =∫_0 ^∞   ((cos(xt))/(t^2 +x^2 ))dt  2)f(x)=∫_0 ^∞   ((sin(xt^2 +(√2)))/(t^2 +x^2  +3))dt
$$\mathrm{calculate}\:\mathrm{f}^{'} \left(\mathrm{x}\right) \\ $$$$\left.\mathrm{1}\right)\:\mathrm{f}\left(\mathrm{x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{cos}\left(\mathrm{xt}\right)}{\mathrm{t}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} }\mathrm{dt} \\ $$$$\left.\mathrm{2}\right)\mathrm{f}\left(\mathrm{x}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{sin}\left(\mathrm{xt}^{\mathrm{2}} +\sqrt{\mathrm{2}}\right)}{\mathrm{t}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} \:+\mathrm{3}}\mathrm{dt} \\ $$
Answered by Bird last updated on 29/Oct/20
1)f(x)=_(t=xu)   ∫_0 ^∞  ((cos(x^2 u))/(x^2 (u^2 +1)))xdu  =(1/x)∫_0 ^∞   ((cos(x^2 u))/(u^2 +1))du   (we tske x>0)  =(1/(2x)) Re(∫_(−∞) ^∞   (e^(ix^2 u) /(u^2 +1))du)  ∫_(−∞) ^∞  (e^(ix^2 z) /(z^2 +1))dz =2iπ Res(f,i)  f(z)=(e^(ix^2 z) /((z−i)(z+i))) ⇒Res(f,i)=(e^(−x^2 ) /(2i))  ⇒∫_(−∞) ^(+∞)  (e^(ix^2 z) /(z^2 +1))dz =2iπ.(e^(−x^2 ) /(2i))  =π e^(−x^2 )  ⇒f(x) =(π/(2x))e^(−x^2 )  ⇒  f^′ (x) =(π/2){−(1/x^2 )e^(−x^2 ) +(1/x)(−2x)e^(−x^2 ) }  =(π/2){−(1/x^2 )e^(−x^2 ) −2e^(−x^2 ) }  =−(π/2){(1/x^2 )+2}e^(−x^2 )   =−π×((1+2x^2 )/(2x^2 )) e^(−x^2 )
$$\left.\mathrm{1}\right){f}\left({x}\right)=_{{t}={xu}} \:\:\int_{\mathrm{0}} ^{\infty} \:\frac{{cos}\left({x}^{\mathrm{2}} {u}\right)}{{x}^{\mathrm{2}} \left({u}^{\mathrm{2}} +\mathrm{1}\right)}{xdu} \\ $$$$=\frac{\mathrm{1}}{{x}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left({x}^{\mathrm{2}} {u}\right)}{{u}^{\mathrm{2}} +\mathrm{1}}{du}\:\:\:\left({we}\:{tske}\:{x}>\mathrm{0}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{x}}\:{Re}\left(\int_{−\infty} ^{\infty} \:\:\frac{{e}^{{ix}^{\mathrm{2}} {u}} }{{u}^{\mathrm{2}} +\mathrm{1}}{du}\right) \\ $$$$\int_{−\infty} ^{\infty} \:\frac{{e}^{{ix}^{\mathrm{2}} {z}} }{{z}^{\mathrm{2}} +\mathrm{1}}{dz}\:=\mathrm{2}{i}\pi\:{Res}\left({f},{i}\right) \\ $$$${f}\left({z}\right)=\frac{{e}^{{ix}^{\mathrm{2}} {z}} }{\left({z}−{i}\right)\left({z}+{i}\right)}\:\Rightarrow{Res}\left({f},{i}\right)=\frac{{e}^{−{x}^{\mathrm{2}} } }{\mathrm{2}{i}} \\ $$$$\Rightarrow\int_{−\infty} ^{+\infty} \:\frac{{e}^{{ix}^{\mathrm{2}} {z}} }{{z}^{\mathrm{2}} +\mathrm{1}}{dz}\:=\mathrm{2}{i}\pi.\frac{{e}^{−{x}^{\mathrm{2}} } }{\mathrm{2}{i}} \\ $$$$=\pi\:{e}^{−{x}^{\mathrm{2}} } \:\Rightarrow{f}\left({x}\right)\:=\frac{\pi}{\mathrm{2}{x}}{e}^{−{x}^{\mathrm{2}} } \:\Rightarrow \\ $$$${f}^{'} \left({x}\right)\:=\frac{\pi}{\mathrm{2}}\left\{−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }{e}^{−{x}^{\mathrm{2}} } +\frac{\mathrm{1}}{{x}}\left(−\mathrm{2}{x}\right){e}^{−{x}^{\mathrm{2}} } \right\} \\ $$$$=\frac{\pi}{\mathrm{2}}\left\{−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }{e}^{−{x}^{\mathrm{2}} } −\mathrm{2}{e}^{−{x}^{\mathrm{2}} } \right\} \\ $$$$=−\frac{\pi}{\mathrm{2}}\left\{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\mathrm{2}\right\}{e}^{−{x}^{\mathrm{2}} } \\ $$$$=−\pi×\frac{\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{2}{x}^{\mathrm{2}} }\:{e}^{−{x}^{\mathrm{2}} } \\ $$
Answered by Bird last updated on 29/Oct/20
2)f(x)=∫_0 ^∞  ((sin(xt^2 +(√2)))/(t^2 +x^2 +3))dt ⇒  f(x)=(1/2)∫_(−∞) ^(+∞)  ((sin(xt^2 +(√2)))/(t^2 +x^2  +3))dt  =(1/2)Im(∫_(−∞) ^(+∞)  (e^(i(xt^2 +(√2))) /(t^2 +x^2 +3))dt)  g(z)=(e^(i(xz^2 +(√2))) /(z^2 +x^2 +3)) ⇒  g(z) =(e^(i(xz^2 +(√2))) /((z−i(√(x^2 +3)))(z+i(√(x^2 +3)))))  ∫_(−∞) ^(+∞) g(x)dz =2iπ Res(g,i(√(x^2 +3)))  =2iπ×(e^(i(x(−x^2 −3)+(√2))) /(2i(√(x^2 +3))))  =(π/( (√(x^2 +3)))).e^(i(−x^3 −3x+(√2)))   =(π/( (√(x^2 +3)))){cos(−x^3 −3x+(√2))  +isin(−x^3 −3x+(√2))} ⇒  f(x)=(π/(2(√(x^2 +3))))sin(−x^3 −3x+(√2))  =−(π/(2(√(x^2 +3))))sin(x^3 +3x−(√2))  f(x)=−(π/2)(x^2 +3)^(−(1/2)) sin(x^3 +3x−(√2)) ⇒  f^′ (x)=−(π/2){−x(x^2 +3)^(−(3/2)) sin(x^3 +3x−(√2))+(x^2 +3)^(−(1/2)) (3x^2 +3)cos(x^3 +3x−(√2))}  f^′ (x) =−(π/2){−(x/((x^2 +3)(√(x^2 +3))))sin(x^3 +3x−(√2))+((3x^2 +3)/( (√(x^2 +3))))cos(x^3 +3x+(√2))}
$$\left.\mathrm{2}\right){f}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{{sin}\left({xt}^{\mathrm{2}} +\sqrt{\mathrm{2}}\right)}{{t}^{\mathrm{2}} +{x}^{\mathrm{2}} +\mathrm{3}}{dt}\:\Rightarrow \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{+\infty} \:\frac{{sin}\left({xt}^{\mathrm{2}} +\sqrt{\mathrm{2}}\right)}{{t}^{\mathrm{2}} +{x}^{\mathrm{2}} \:+\mathrm{3}}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{Im}\left(\int_{−\infty} ^{+\infty} \:\frac{{e}^{{i}\left({xt}^{\mathrm{2}} +\sqrt{\mathrm{2}}\right)} }{{t}^{\mathrm{2}} +{x}^{\mathrm{2}} +\mathrm{3}}{dt}\right) \\ $$$${g}\left({z}\right)=\frac{{e}^{{i}\left({xz}^{\mathrm{2}} +\sqrt{\mathrm{2}}\right)} }{{z}^{\mathrm{2}} +{x}^{\mathrm{2}} +\mathrm{3}}\:\Rightarrow \\ $$$${g}\left({z}\right)\:=\frac{{e}^{{i}\left({xz}^{\mathrm{2}} +\sqrt{\mathrm{2}}\right)} }{\left({z}−{i}\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}\right)\left({z}+{i}\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}\right)} \\ $$$$\int_{−\infty} ^{+\infty} {g}\left({x}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left({g},{i}\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}\right) \\ $$$$=\mathrm{2}{i}\pi×\frac{{e}^{{i}\left({x}\left(−{x}^{\mathrm{2}} −\mathrm{3}\right)+\sqrt{\mathrm{2}}\right)} }{\mathrm{2}{i}\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}} \\ $$$$=\frac{\pi}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}}.{e}^{{i}\left(−{x}^{\mathrm{3}} −\mathrm{3}{x}+\sqrt{\mathrm{2}}\right)} \\ $$$$=\frac{\pi}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}}\left\{{cos}\left(−{x}^{\mathrm{3}} −\mathrm{3}{x}+\sqrt{\mathrm{2}}\right)\right. \\ $$$$\left.+{isin}\left(−{x}^{\mathrm{3}} −\mathrm{3}{x}+\sqrt{\mathrm{2}}\right)\right\}\:\Rightarrow \\ $$$${f}\left({x}\right)=\frac{\pi}{\mathrm{2}\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}}{sin}\left(−{x}^{\mathrm{3}} −\mathrm{3}{x}+\sqrt{\mathrm{2}}\right) \\ $$$$=−\frac{\pi}{\mathrm{2}\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}}{sin}\left({x}^{\mathrm{3}} +\mathrm{3}{x}−\sqrt{\mathrm{2}}\right) \\ $$$${f}\left({x}\right)=−\frac{\pi}{\mathrm{2}}\left({x}^{\mathrm{2}} +\mathrm{3}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} {sin}\left({x}^{\mathrm{3}} +\mathrm{3}{x}−\sqrt{\mathrm{2}}\right)\:\Rightarrow \\ $$$${f}^{'} \left({x}\right)=−\frac{\pi}{\mathrm{2}}\left\{−{x}\left({x}^{\mathrm{2}} +\mathrm{3}\right)^{−\frac{\mathrm{3}}{\mathrm{2}}} {sin}\left({x}^{\mathrm{3}} +\mathrm{3}{x}−\sqrt{\mathrm{2}}\right)+\left({x}^{\mathrm{2}} +\mathrm{3}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}\right){cos}\left({x}^{\mathrm{3}} +\mathrm{3}{x}−\sqrt{\mathrm{2}}\right)\right\} \\ $$$${f}^{'} \left({x}\right)\:=−\frac{\pi}{\mathrm{2}}\left\{−\frac{{x}}{\left({x}^{\mathrm{2}} +\mathrm{3}\right)\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}}{sin}\left({x}^{\mathrm{3}} +\mathrm{3}{x}−\sqrt{\mathrm{2}}\right)+\frac{\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}}{cos}\left({x}^{\mathrm{3}} +\mathrm{3}{x}+\sqrt{\mathrm{2}}\right)\right\} \\ $$

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