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Question Number 37288 by math khazana by abdo last updated on 11/Jun/18
calculate f(α) = ∫_(−∞) ^(+∞) ((cos(2x))/(1+ax^2 )) dx with a>0 2) find the value of ∫_(−∞) ^(+∞) ((cos(2x))/(1+3x^2 )) dx .
$${calculate}\:\:{f}\left(\alpha\right)\:=\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{cos}\left(\mathrm{2}{x}\right)}{\mathrm{1}+{ax}^{\mathrm{2}} }\:{dx}\:{with}\:{a}>\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\:\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{cos}\left(\mathrm{2}{x}\right)}{\mathrm{1}+\mathrm{3}{x}^{\mathrm{2}} }\:{dx}\:. \\ $$
Commented by prof Abdo imad last updated on 16/Jun/18
1) f(a) =Re(∫_(−∞) ^(+∞) (e^(i2x) /(1+ax^2 ))dx) let ϕ(z) = (e^(2iz) /(1+az^2 )) ϕ(z)= (e^(2iz) /(((√a)z)^2 −i^2 )) = (e^(2iz) /(((√a)z−i)((√a)z+i))) =(e^(2iz) /(a(z−(i/( (√a))))(z+(i/( (√a)))))) so the poles of ϕ are (i/( (√a))) and ((−i)/( (√a))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,(i/( (√a)))) but we have Res(ϕ,(i/( (√a)))) = (e^(2i(i/( (√a)))) /(a((2i)/( (√a))))) = (e^((−2)/( (√a))) /(2i(√a))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz = 2iπ (e^(−(2/( (√a)))) /(2i(√a))) = (π/( (√a))) e^(−(2/( (√a)))) ⇒ f(a) = (π/( (√a))) e^(−(2/( (√a)))) 2) ∫_(−∞) ^(+∞) ((cos(2x))/(1+3x^2 )) dx =f(3) =(π/( (√3))) e^(−(2/( (√3)))) .
$$\left.\mathrm{1}\right)\:{f}\left({a}\right)\:={Re}\left(\int_{−\infty} ^{+\infty} \:\:\:\frac{{e}^{{i}\mathrm{2}{x}} }{\mathrm{1}+{ax}^{\mathrm{2}} }{dx}\right)\:\:{let}\: \\ $$$$\varphi\left({z}\right)\:=\:\frac{{e}^{\mathrm{2}{iz}} }{\mathrm{1}+{az}^{\mathrm{2}} } \\ $$$$\varphi\left({z}\right)=\:\frac{{e}^{\mathrm{2}{iz}} }{\left(\sqrt{{a}}{z}\right)^{\mathrm{2}} −{i}^{\mathrm{2}} }\:=\:\frac{{e}^{\mathrm{2}{iz}} }{\left(\sqrt{{a}}{z}−{i}\right)\left(\sqrt{{a}}{z}+{i}\right)} \\ $$$$=\frac{{e}^{\mathrm{2}{iz}} }{{a}\left({z}−\frac{{i}}{\:\sqrt{{a}}}\right)\left({z}+\frac{{i}}{\:\sqrt{{a}}}\right)}\:\:{so}\:{the}\:{poles}\:{of}\:\varphi\:{are}\:\frac{{i}}{\:\sqrt{{a}}} \\ $$$${and}\:\frac{−{i}}{\:\sqrt{{a}}} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,\frac{{i}}{\:\sqrt{{a}}}\right)\:{but}\:{we}\:{have} \\ $$$${Res}\left(\varphi,\frac{{i}}{\:\sqrt{{a}}}\right)\:=\:\frac{{e}^{\mathrm{2}{i}\frac{{i}}{\:\sqrt{{a}}}} }{{a}\frac{\mathrm{2}{i}}{\:\sqrt{{a}}}}\:=\:\frac{{e}^{\frac{−\mathrm{2}}{\:\sqrt{{a}}}} }{\mathrm{2}{i}\sqrt{{a}}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\:\mathrm{2}{i}\pi\:\:\frac{{e}^{−\frac{\mathrm{2}}{\:\sqrt{{a}}}} }{\mathrm{2}{i}\sqrt{{a}}}\:=\:\frac{\pi}{\:\sqrt{{a}}}\:{e}^{−\frac{\mathrm{2}}{\:\sqrt{{a}}}} \:\:\:\Rightarrow \\ $$$${f}\left({a}\right)\:=\:\frac{\pi}{\:\sqrt{{a}}}\:{e}^{−\frac{\mathrm{2}}{\:\sqrt{{a}}}} \\ $$$$\left.\mathrm{2}\right)\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{cos}\left(\mathrm{2}{x}\right)}{\mathrm{1}+\mathrm{3}{x}^{\mathrm{2}} }\:{dx}\:={f}\left(\mathrm{3}\right) \\ $$$$=\frac{\pi}{\:\sqrt{\mathrm{3}}}\:{e}^{−\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}} \:\:. \\ $$$$ \\ $$$$ \\ $$

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