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Question Number 48255 by Abdo msup. last updated on 21/Nov/18
calculate A_λ   =∫_0 ^∞   ((cos(λsinx)−sin(λcosx))/(x^2  +λ^2 ))dx  λ from R.
$${calculate}\:{A}_{\lambda} \:\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left(\lambda{sinx}\right)−{sin}\left(\lambda{cosx}\right)}{{x}^{\mathrm{2}} \:+\lambda^{\mathrm{2}} }{dx} \\ $$$$\lambda\:{from}\:{R}. \\ $$
Commented by Abdo msup. last updated on 23/Nov/18
we have A_λ =∫_0 ^∞  ((cos(λsinx))/(x^2  +λ^2 ))dx−∫_0 ^∞  ((sin(λcosx))/(x^2  +λ^2 ))dx  =H−K  case 1  λ>0  2H =∫_(−∞) ^(+∞)  ((cos(λsinx))/(x^2  +λ^2 ))dx=Re(∫_(−∞) ^(+∞)  (e^(iλsinx) /(x^2  +λ^2 ))dx)let  ϕ(z)=(e^(iλsinz) /(z^2  +λ^2 )) ⇒ϕ(z)=(e^(iλsinz) /((z−iλ)(z+iλ)))  ∫_(−∞) ^(+∞) ϕ(z)dz =2iπRes(ϕ,iλ)  =2iπ (e^(iλsin(iλ)) /(2iλ)) =(π/λ) e^(iλsin(iλ))   but sin(iλ)=((e^(i(iλ)) −e^(−i(iλ)) )/(2i))  =((e^λ −e^(−λ) )/2) ⇒∫_(−∞) ^(+∞) ϕ(z)dz =(π/λ) e^(iλ((e^λ −e^(−λ) )/2))   =(π/λ){cos((λ/2)(e^λ −e^(−λ) ))+i sin((λ/2)(e^λ −e^(−λ) )) ⇒  H =(π/(2λ))cos((λ/2)(e^λ −e^(−λ) )) also  2K =∫_(−∞) ^(+∞)    ((sin(λcosx))/(x^2  +λ^2 ))dx =Im( ∫_(−∞) ^(+∞)  (e^(iλcosx) /(x^2  +λ^2 ))dx) but  ∫_(−∞) ^(+∞)    (e^(iλ cosx) /(x^2  +λ^2 ))dx?=2iπ (e^(iλcos(iλ)) /(2iλ)) =(π/λ) e^(iλcos(iλ))  but  cos(iλ) =((e^(i(iλ))  +e^(−i(iλ)) )/2) =((e^λ  +e^(−λ) )/2) ⇒  ∫_(−∞) ^(+∞)   (e^(iλcosx) /(x^2  +λ^2 ))dx =(π/λ) e^(iλ((e^λ  +e^(−λ) )/2))   =(π/λ){cos((λ/2)(e^λ  +e^(−λ) ))+isin((λ/2)(e^λ  +e^(−λ) ))} ⇒  K =(π/(2λ)) sin((λ/2)(e^λ  +e^(−λ) )) ⇒  A_λ =(π/(2λ)){ cos((λ/2)(e^λ  −e^(−λ) )−sin((λ/2)(e^λ  +e^(−λ) ))}  if λ<0 we follow the same method.
$${we}\:{have}\:{A}_{\lambda} =\int_{\mathrm{0}} ^{\infty} \:\frac{{cos}\left(\lambda{sinx}\right)}{{x}^{\mathrm{2}} \:+\lambda^{\mathrm{2}} }{dx}−\int_{\mathrm{0}} ^{\infty} \:\frac{{sin}\left(\lambda{cosx}\right)}{{x}^{\mathrm{2}} \:+\lambda^{\mathrm{2}} }{dx} \\ $$$$={H}−{K}\:\:{case}\:\mathrm{1}\:\:\lambda>\mathrm{0} \\ $$$$\mathrm{2}{H}\:=\int_{−\infty} ^{+\infty} \:\frac{{cos}\left(\lambda{sinx}\right)}{{x}^{\mathrm{2}} \:+\lambda^{\mathrm{2}} }{dx}={Re}\left(\int_{−\infty} ^{+\infty} \:\frac{{e}^{{i}\lambda{sinx}} }{{x}^{\mathrm{2}} \:+\lambda^{\mathrm{2}} }{dx}\right){let} \\ $$$$\varphi\left({z}\right)=\frac{{e}^{{i}\lambda{sinz}} }{{z}^{\mathrm{2}} \:+\lambda^{\mathrm{2}} }\:\Rightarrow\varphi\left({z}\right)=\frac{{e}^{{i}\lambda{sinz}} }{\left({z}−{i}\lambda\right)\left({z}+{i}\lambda\right)} \\ $$$$\int_{−\infty} ^{+\infty} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi{Res}\left(\varphi,{i}\lambda\right) \\ $$$$=\mathrm{2}{i}\pi\:\frac{{e}^{{i}\lambda{sin}\left({i}\lambda\right)} }{\mathrm{2}{i}\lambda}\:=\frac{\pi}{\lambda}\:{e}^{{i}\lambda{sin}\left({i}\lambda\right)} \:\:{but}\:{sin}\left({i}\lambda\right)=\frac{{e}^{{i}\left({i}\lambda\right)} −{e}^{−{i}\left({i}\lambda\right)} }{\mathrm{2}{i}} \\ $$$$=\frac{{e}^{\lambda} −{e}^{−\lambda} }{\mathrm{2}}\:\Rightarrow\int_{−\infty} ^{+\infty} \varphi\left({z}\right){dz}\:=\frac{\pi}{\lambda}\:{e}^{{i}\lambda\frac{{e}^{\lambda} −{e}^{−\lambda} }{\mathrm{2}}} \\ $$$$=\frac{\pi}{\lambda}\left\{{cos}\left(\frac{\lambda}{\mathrm{2}}\left({e}^{\lambda} −{e}^{−\lambda} \right)\right)+{i}\:{sin}\left(\frac{\lambda}{\mathrm{2}}\left({e}^{\lambda} −{e}^{−\lambda} \right)\right)\:\Rightarrow\right. \\ $$$${H}\:=\frac{\pi}{\mathrm{2}\lambda}{cos}\left(\frac{\lambda}{\mathrm{2}}\left({e}^{\lambda} −{e}^{−\lambda} \right)\right)\:{also} \\ $$$$\mathrm{2}{K}\:=\int_{−\infty} ^{+\infty} \:\:\:\frac{{sin}\left(\lambda{cosx}\right)}{{x}^{\mathrm{2}} \:+\lambda^{\mathrm{2}} }{dx}\:={Im}\left(\:\int_{−\infty} ^{+\infty} \:\frac{{e}^{{i}\lambda{cosx}} }{{x}^{\mathrm{2}} \:+\lambda^{\mathrm{2}} }{dx}\right)\:{but} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\:\frac{{e}^{{i}\lambda\:{cosx}} }{{x}^{\mathrm{2}} \:+\lambda^{\mathrm{2}} }{dx}?=\mathrm{2}{i}\pi\:\frac{{e}^{{i}\lambda{cos}\left({i}\lambda\right)} }{\mathrm{2}{i}\lambda}\:=\frac{\pi}{\lambda}\:{e}^{{i}\lambda{cos}\left({i}\lambda\right)} \:{but} \\ $$$${cos}\left({i}\lambda\right)\:=\frac{{e}^{{i}\left({i}\lambda\right)} \:+{e}^{−{i}\left({i}\lambda\right)} }{\mathrm{2}}\:=\frac{{e}^{\lambda} \:+{e}^{−\lambda} }{\mathrm{2}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\:\frac{{e}^{{i}\lambda{cosx}} }{{x}^{\mathrm{2}} \:+\lambda^{\mathrm{2}} }{dx}\:=\frac{\pi}{\lambda}\:{e}^{{i}\lambda\frac{{e}^{\lambda} \:+{e}^{−\lambda} }{\mathrm{2}}} \\ $$$$=\frac{\pi}{\lambda}\left\{{cos}\left(\frac{\lambda}{\mathrm{2}}\left({e}^{\lambda} \:+{e}^{−\lambda} \right)\right)+{isin}\left(\frac{\lambda}{\mathrm{2}}\left({e}^{\lambda} \:+{e}^{−\lambda} \right)\right)\right\}\:\Rightarrow \\ $$$${K}\:=\frac{\pi}{\mathrm{2}\lambda}\:{sin}\left(\frac{\lambda}{\mathrm{2}}\left({e}^{\lambda} \:+{e}^{−\lambda} \right)\right)\:\Rightarrow \\ $$$${A}_{\lambda} =\frac{\pi}{\mathrm{2}\lambda}\left\{\:{cos}\left(\frac{\lambda}{\mathrm{2}}\left({e}^{\lambda} \:−{e}^{−\lambda} \right)−{sin}\left(\frac{\lambda}{\mathrm{2}}\left({e}^{\lambda} \:+{e}^{−\lambda} \right)\right)\right\}\right. \\ $$$${if}\:\lambda<\mathrm{0}\:{we}\:{follow}\:{the}\:{same}\:{method}. \\ $$

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