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Question Number 175012 by lapache last updated on 16/Aug/22
prove: Use the residus form ∫_0 ^(+∞) ((xsinx)/((x^2 +1)^2 ))dx=(π/(4e))
$${prove}:\:{Use}\:{the}\:{residus}\:{form} \\ $$$$\int_{\mathrm{0}} ^{+\infty} \frac{{xsinx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{dx}=\frac{\pi}{\mathrm{4}{e}} \\ $$
Answered by Mathspace last updated on 16/Aug/22
letI=∫_0 ^∞ ((xsinx)/((x^2 +1)^2 )) ⇒ 2I=∫_(−∞) ^(+∞) ((xsinx)/((x^2 +1)^2 ))dx =Im(∫_(−∞) ^(+∞) ((xe^(ix) )/((x^2 +1)^2 ))dx) Ψ(z)=((ze^(iz) )/((z^2 +1)^2 )) poles of Ψ? Ψ(z)=((ze^(iz) )/((z−i)^2 (z+i)^2 )) so the poles are +^− i(doubles) ∫_(−∞) ^(+∞) Ψ(z)dz=2iπRes(Ψ,i) Res(Ψ,i)=lim_(z→i) (1/((2−1)!)){(z−i)^2 Ψ(z)}^((1)) =lim_(z→i) {((ze^(iz) )/((z+i)^2 ))}^((1)) =lim_(z→i) (((e^(iz) +ize^(iz) )(z+i)^2 −2(z+i)ze^(iz) )/((z+i)^4 )) =lim_(z→i) (((1+iz)e^(iz) (z+i)−2ze^(iz) )/((z+i)^3 )) =((−2ie^(−1) )/((2i)^3 ))=((−2ie^(−1) )/(−8i))=(1/(4e)) ∫_(−∞) ^(+∞) Ψ(z)dz=2iπ.(1/(4e))=((iπ)/(2e)) so 2I=(π/(2e)) ⇒★I=(π/(4e))★
$${letI}=\int_{\mathrm{0}} ^{\infty} \:\frac{{xsinx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\mathrm{2}{I}=\int_{−\infty} ^{+\infty} \frac{{xsinx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$$={Im}\left(\int_{−\infty} ^{+\infty} \frac{{xe}^{{ix}} }{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{dx}\right) \\ $$$$\Psi\left({z}\right)=\frac{{ze}^{{iz}} }{\left({z}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\:\:{poles}\:{of}\:\Psi? \\ $$$$\Psi\left({z}\right)=\frac{{ze}^{{iz}} }{\left({z}−{i}\right)^{\mathrm{2}} \left({z}+{i}\right)^{\mathrm{2}} } \\ $$$${so}\:{the}\:{poles}\:{are}\:\overset{−} {+}{i}\left({doubles}\right) \\ $$$$\int_{−\infty} ^{+\infty} \Psi\left({z}\right){dz}=\mathrm{2}{i}\pi{Res}\left(\Psi,{i}\right) \\ $$$${Res}\left(\Psi,{i}\right)={lim}_{{z}\rightarrow{i}} \frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\left({z}−{i}\right)^{\mathrm{2}} \Psi\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \left\{\frac{{ze}^{{iz}} }{\left({z}+{i}\right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\:\frac{\left({e}^{{iz}} +{ize}^{{iz}} \right)\left({z}+{i}\right)^{\mathrm{2}} −\mathrm{2}\left({z}+{i}\right){ze}^{{iz}} }{\left({z}+{i}\right)^{\mathrm{4}} } \\ $$$$={lim}_{{z}\rightarrow{i}} \frac{\left(\mathrm{1}+{iz}\right){e}^{{iz}} \left({z}+{i}\right)−\mathrm{2}{ze}^{{iz}} }{\left({z}+{i}\right)^{\mathrm{3}} } \\ $$$$=\frac{−\mathrm{2}{ie}^{−\mathrm{1}} }{\left(\mathrm{2}{i}\right)^{\mathrm{3}} }=\frac{−\mathrm{2}{ie}^{−\mathrm{1}} }{−\mathrm{8}{i}}=\frac{\mathrm{1}}{\mathrm{4}{e}} \\ $$$$\int_{−\infty} ^{+\infty} \Psi\left({z}\right){dz}=\mathrm{2}{i}\pi.\frac{\mathrm{1}}{\mathrm{4}{e}}=\frac{{i}\pi}{\mathrm{2}{e}} \\ $$$${so}\:\mathrm{2}{I}=\frac{\pi}{\mathrm{2}{e}}\:\Rightarrow\bigstar{I}=\frac{\pi}{\mathrm{4}{e}}\bigstar \\ $$
Commented by Mathspace last updated on 16/Aug/22
thank you
$${thank}\:{you}\: \\ $$
Commented by Tawa11 last updated on 16/Aug/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$

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