Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 77752 by abdomathmax last updated on 09/Jan/20

let f(λ) =∫_(−∞) ^(+∞) ((sin( λe^x +e^(−x) ))/(x^2 +λ^2 ))dx with λ≥0 1) detdrmine a explicit form of f(λ) 2) calculate f^′ (λ) at form ofintergral and find its value.

$${let}\:{f}\left(\lambda\right)\:=\int_{−\infty} ^{+\infty} \:\frac{{sin}\left(\:\lambda{e}^{{x}} \:+{e}^{−{x}} \right)}{{x}^{\mathrm{2}} \:+\lambda^{\mathrm{2}} }{dx}\:{with}\:\lambda\geqslant\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{detdrmine}\:{a}\:{explicit}\:{form}\:{of}\:{f}\left(\lambda\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{f}^{'} \left(\lambda\right)\:{at}\:{form}\:{ofintergral}\:{and}\:{find} \\ $$$${its}\:{value}. \\ $$

Commented by mathmax by abdo last updated on 10/Jan/20

f(λ)=∫_(−∞) ^(+∞) ((sin(λe^x +e^(−x) ))/(x^2 +λ^2 )) ⇒f(λ)=_(x=λt) ∫_(−∞) ^(+∞) ((sin(λ e^(λt) +e^(−λt) ))/(λ^2 (t^2 +1)))λ(dt) =(1/λ) ∫_(−∞) ^(+∞) ((sin(λ e^(λt) +e^(−λt) ))/(t^2 +1))dt ( we take λ>0) ⇒ λf(λ) =Im(∫_(−∞) ^(+∞) (e^(i( λe^(λt) +e^(−λt) )) /(t^2 +1))dt)let W(z)=(e^(i(λ e^(λz) +e^(−λz) )) /(z^2 +1)) ⇒W(z) =(e^(i(λ e^(λz) +e^(−λz) )) /((z−i)(z+i))) and ∫_(−∞) ^(+∞) W(z)dz =2iπ Res(W,i) =2iπ ×(e^(i(λ e^(iλ) +e^(−iλ) )) /(2i)) we have i(λ e^(iλ) +e^(−iλ) ) =i(λ cos(λ)+λi sin(λ) +cos(λ)−isin(λ)) =i λ cos(λ)−λsin(λ) +icos(λ) +sinλ =(1−λ)sinλ +i(1+λ)cos(λ) ⇒ e^(i(λ e^(iλ) +e^(−iλ)) ) =e^((1−λ)sin(λ)) { cos(1+λ)cosλ +isin(1+λ)cosλ} ⇒ ∫_(−∞) ^(+∞) W(z)dz =π e^((1−λ)sinλ) cos((1+λ)cosλ) +i π e^((1−λ)sinλ) sin((1+λ)cosλ) ⇒λ f(λ) =π e^((1−λ)sinλ) sin{(1+λ)cosλ} ⇒ f(λ) =(π/λ)e^((1−λ)sinλ) sin{(1+λ)cosλ} (λ>0)

$${f}\left(\lambda\right)=\int_{−\infty} ^{+\infty} \:\frac{{sin}\left(\lambda{e}^{{x}} \:+{e}^{−{x}} \right)}{{x}^{\mathrm{2}} \:+\lambda^{\mathrm{2}} }\:\Rightarrow{f}\left(\lambda\right)=_{{x}=\lambda{t}} \:\:\int_{−\infty} ^{+\infty} \:\frac{{sin}\left(\lambda\:{e}^{\lambda{t}} \:+{e}^{−\lambda{t}} \right)}{\lambda^{\mathrm{2}} \left({t}^{\mathrm{2}} \:+\mathrm{1}\right)}\lambda\left({dt}\right) \\ $$$$=\frac{\mathrm{1}}{\lambda}\:\int_{−\infty} ^{+\infty} \:\frac{{sin}\left(\lambda\:{e}^{\lambda{t}} \:+{e}^{−\lambda{t}} \right)}{{t}^{\mathrm{2}} \:+\mathrm{1}}{dt}\:\:\:\:\left(\:\:{we}\:{take}\:\lambda>\mathrm{0}\right)\:\Rightarrow \\ $$$$\lambda{f}\left(\lambda\right)\:={Im}\left(\int_{−\infty} ^{+\infty} \:\frac{{e}^{{i}\left(\:\lambda{e}^{\lambda{t}} +{e}^{−\lambda{t}} \right)} }{{t}^{\mathrm{2}} \:+\mathrm{1}}{dt}\right){let}\:{W}\left({z}\right)=\frac{{e}^{{i}\left(\lambda\:{e}^{\lambda{z}} \:+{e}^{−\lambda{z}} \right)} }{{z}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$\Rightarrow{W}\left({z}\right)\:=\frac{{e}^{{i}\left(\lambda\:{e}^{\lambda{z}} \:+{e}^{−\lambda{z}} \right)} }{\left({z}−{i}\right)\left({z}+{i}\right)}\:{and}\:\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left({W},{i}\right) \\ $$$$=\mathrm{2}{i}\pi\:×\frac{{e}^{{i}\left(\lambda\:{e}^{{i}\lambda} \:+{e}^{−{i}\lambda} \right)} }{\mathrm{2}{i}}\:\:{we}\:{have}\: \\ $$$${i}\left(\lambda\:{e}^{{i}\lambda} \:+{e}^{−{i}\lambda} \right)\:={i}\left(\lambda\:{cos}\left(\lambda\right)+\lambda{i}\:{sin}\left(\lambda\right)\:+{cos}\left(\lambda\right)−{isin}\left(\lambda\right)\right) \\ $$$$={i}\:\lambda\:{cos}\left(\lambda\right)−\lambda{sin}\left(\lambda\right)\:+{icos}\left(\lambda\right)\:+{sin}\lambda \\ $$$$=\left(\mathrm{1}−\lambda\right){sin}\lambda\:+{i}\left(\mathrm{1}+\lambda\right){cos}\left(\lambda\right)\:\Rightarrow \\ $$$${e}^{{i}\left(\lambda\:{e}^{{i}\lambda} \:+{e}^{\left.−{i}\lambda\right)} \right.} \:={e}^{\left(\mathrm{1}−\lambda\right){sin}\left(\lambda\right)} \left\{\:{cos}\left(\mathrm{1}+\lambda\right){cos}\lambda\:+{isin}\left(\mathrm{1}+\lambda\right){cos}\lambda\right\}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\pi\:{e}^{\left(\mathrm{1}−\lambda\right){sin}\lambda} \:{cos}\left(\left(\mathrm{1}+\lambda\right){cos}\lambda\right)\:+{i}\:\pi\:{e}^{\left(\mathrm{1}−\lambda\right){sin}\lambda} {sin}\left(\left(\mathrm{1}+\lambda\right){cos}\lambda\right) \\ $$$$\Rightarrow\lambda\:{f}\left(\lambda\right)\:=\pi\:{e}^{\left(\mathrm{1}−\lambda\right){sin}\lambda} \:{sin}\left\{\left(\mathrm{1}+\lambda\right){cos}\lambda\right\}\:\Rightarrow \\ $$$${f}\left(\lambda\right)\:=\frac{\pi}{\lambda}{e}^{\left(\mathrm{1}−\lambda\right){sin}\lambda} \:{sin}\left\{\left(\mathrm{1}+\lambda\right){cos}\lambda\right\}\:\:\:\:\:\left(\lambda>\mathrm{0}\right) \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com