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Question Number 70870 by Abdo msup. last updated on 09/Oct/19
let f(x)=∫_(−∞) ^(+∞) (dt/((t^2 −2t +x^2 )^4 )) with ∣x∣>1 and n integr natural 1)find a explicit form for f(x) 2) determine also g(x)=∫_(−∞) ^(+∞) (dt/((t^2 −2t +x^2 )^5 )) 3)find the values of integrals ∫_(−∞) ^(+∞) (dt/((t^2 −2t +3)^4 )) and ∫_(−∞) ^(+∞) (dt/((t^2 −2t +3)^5 ))
$${let}\:{f}\left({x}\right)=\int_{−\infty} ^{+\infty} \:\:\frac{{dt}}{\left({t}^{\mathrm{2}} −\mathrm{2}{t}\:+{x}^{\mathrm{2}} \right)^{\mathrm{4}} }\:\:{with}\:\:\mid{x}\mid>\mathrm{1} \\ $$$${and}\:{n}\:{integr}\:{natural} \\ $$$$\left.\mathrm{1}\right){find}\:{a}\:{explicit}\:{form}\:{for}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{determine}\:{also}\:{g}\left({x}\right)=\int_{−\infty} ^{+\infty} \:\:\frac{{dt}}{\left({t}^{\mathrm{2}} −\mathrm{2}{t}\:+{x}^{\mathrm{2}} \right)^{\mathrm{5}} } \\ $$$$\left.\mathrm{3}\right){find}\:{the}\:{values}\:{of}\:{integrals}\: \\ $$$$\int_{−\infty} ^{+\infty} \:\frac{{dt}}{\left({t}^{\mathrm{2}} −\mathrm{2}{t}\:+\mathrm{3}\right)^{\mathrm{4}} }\:\:{and}\:\int_{−\infty} ^{+\infty} \:\frac{{dt}}{\left({t}^{\mathrm{2}} −\mathrm{2}{t}\:+\mathrm{3}\right)^{\mathrm{5}} } \\ $$
Commented by mathmax by abdo last updated on 13/Oct/19
1) f(x)=∫_(−∞) ^(+∞) (dt/((t^2 −2t +x^2 )^4 )) let ϕ(z)=(1/((z^2 −2z +x^2 )^4 )) z^2 −2z+x^2 =0 →Δ^′ =1−x^2 <0 ⇒Δ^′ =(i(√(x^2 −1)))^2 z_1 =1+i(√(x^2 −1)) and z_2 =1−i(√(x^2 −1)) ∣z_1 ∣=(√(1+x^2 −1))=∣x∣ ⇒z_1 =∣x∣ e^(iarctan((√(x^2 −1)))) z_2 =∣x∣ e^(−iarctan((√(x^2 −1)))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,∣x∣e^(iarctan((√(x^2 −1)))) ) Res(ϕ,∣x∣ e^(iarctan((√(x^2 −1)))) =lim_(z→∣x∣ e^(iarctan((√(x^2 −1)))) ) (1/((4−1)!)){(z−∣x∣e^(iarctan((√(x^2 −1)))) )^4 ϕ(z)}^((3)) =lim_(z→(...)) (1/(3!)){ (1/((z−z_2 )^4 ))}^((3)) ....be continued...
$$\left.\mathrm{1}\right)\:{f}\left({x}\right)=\int_{−\infty} ^{+\infty} \:\frac{{dt}}{\left({t}^{\mathrm{2}} −\mathrm{2}{t}\:+{x}^{\mathrm{2}} \right)^{\mathrm{4}} }\:{let}\:\varphi\left({z}\right)=\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} −\mathrm{2}{z}\:+{x}^{\mathrm{2}} \right)^{\mathrm{4}} } \\ $$$${z}^{\mathrm{2}} −\mathrm{2}{z}+{x}^{\mathrm{2}} =\mathrm{0}\:\rightarrow\Delta^{'} =\mathrm{1}−{x}^{\mathrm{2}} <\mathrm{0}\:\Rightarrow\Delta^{'} =\left({i}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)^{\mathrm{2}} \\ $$$${z}_{\mathrm{1}} =\mathrm{1}+{i}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:\:{and}\:{z}_{\mathrm{2}} =\mathrm{1}−{i}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\mid{z}_{\mathrm{1}} \mid=\sqrt{\mathrm{1}+{x}^{\mathrm{2}} −\mathrm{1}}=\mid{x}\mid\:\Rightarrow{z}_{\mathrm{1}} =\mid{x}\mid\:{e}^{{iarctan}\left(\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)} \\ $$$${z}_{\mathrm{2}} =\mid{x}\mid\:{e}^{−{iarctan}\left(\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)} \\ $$$$\int_{−\infty} ^{+\infty} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\:{Res}\left(\varphi,\mid{x}\mid{e}^{{iarctan}\left(\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)} \right) \\ $$$${Res}\left(\varphi,\mid{x}\mid\:{e}^{{iarctan}\left(\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)} \right. \\ $$$$={lim}_{{z}\rightarrow\mid{x}\mid\:{e}^{{iarctan}\left(\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)} } \:\:\:\:\frac{\mathrm{1}}{\left(\mathrm{4}−\mathrm{1}\right)!}\left\{\left({z}−\mid{x}\mid{e}^{{iarctan}\left(\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)} \right)^{\mathrm{4}} \varphi\left({z}\right)\right\}^{\left(\mathrm{3}\right)} \\ $$$$={lim}_{{z}\rightarrow\left(…\right)} \:\:\:\frac{\mathrm{1}}{\mathrm{3}!}\left\{\:\:\frac{\mathrm{1}}{\left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{4}} }\right\}^{\left(\mathrm{3}\right)} \:….{be}\:{continued}… \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 13/Oct/19
{(z−z_2 )^(−4) }^((3)) ={−4(z−z_2 )^(−5) }^((2)) ={20(z−z_2 )^(−6) }^((1)) ={−120(z−z_2 )^(−7) } ⇒ Res(ϕ,∣x∣ e^(iarctan((√(x^2 −1))) )=(1/6)(−120)(z_1 −z_2 )^(−7) =−20 (2iIm(z_1 ))^(−7) =−20(2isin(∣x∣arctan((√(x^2 −1))))^(−7) =((−20)/((2i)^7 sin^7 (∣x∣ arctan((√(x^2 −1)))))) =((−20)/(2^7 (−i)sin^7 (∣x∣arctan((√(x^2 −1))))) i^7 =(i^2 )^3 i =−i ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ ((2^2 ×5)/(2^3 ×2^4 (i)sin^7 (∣x∣ arctan((√(x^2 −1)))))) =((5π)/(16 sin^7 (∣x∣ arctan((√(x^2 −1))))) ⇒ f(x)=((5π)/(16 sin^7 (∣x∣arctan((√(x^2 −1)))))
$$\left\{\left({z}−{z}_{\mathrm{2}} \right)^{−\mathrm{4}} \right\}^{\left(\mathrm{3}\right)} =\left\{−\mathrm{4}\left({z}−{z}_{\mathrm{2}} \right)^{−\mathrm{5}} \right\}^{\left(\mathrm{2}\right)} =\left\{\mathrm{20}\left({z}−{z}_{\mathrm{2}} \right)^{−\mathrm{6}} \right\}^{\left(\mathrm{1}\right)} \\ $$$$=\left\{−\mathrm{120}\left({z}−{z}_{\mathrm{2}} \right)^{−\mathrm{7}} \right\}\:\Rightarrow \\ $$$${Res}\left(\varphi,\mid{x}\mid\:{e}^{{iarctan}\left(\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right.} \right)=\frac{\mathrm{1}}{\mathrm{6}}\left(−\mathrm{120}\right)\left({z}_{\mathrm{1}} −{z}_{\mathrm{2}} \right)^{−\mathrm{7}} \\ $$$$=−\mathrm{20}\:\left(\mathrm{2}{iIm}\left({z}_{\mathrm{1}} \right)\right)^{−\mathrm{7}} \:=−\mathrm{20}\left(\mathrm{2}{isin}\left(\mid{x}\mid{arctan}\left(\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)\right)^{−\mathrm{7}} \right. \\ $$$$=\frac{−\mathrm{20}}{\left(\mathrm{2}{i}\right)^{\mathrm{7}} \:{sin}^{\mathrm{7}} \left(\mid{x}\mid\:{arctan}\left(\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)\right)}\:=\frac{−\mathrm{20}}{\mathrm{2}^{\mathrm{7}} \left(−{i}\right){sin}^{\mathrm{7}} \left(\mid{x}\mid{arctan}\left(\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)\right.} \\ $$$${i}^{\mathrm{7}} =\left({i}^{\mathrm{2}} \right)^{\mathrm{3}} {i}\:=−{i} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\:\frac{\mathrm{2}^{\mathrm{2}} ×\mathrm{5}}{\mathrm{2}^{\mathrm{3}} ×\mathrm{2}^{\mathrm{4}} \left({i}\right){sin}^{\mathrm{7}} \left(\mid{x}\mid\:{arctan}\left(\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)\right)} \\ $$$$=\frac{\mathrm{5}\pi}{\mathrm{16}\:{sin}^{\mathrm{7}} \left(\mid{x}\mid\:{arctan}\left(\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)\right.}\:\Rightarrow \\ $$$${f}\left({x}\right)=\frac{\mathrm{5}\pi}{\mathrm{16}\:{sin}^{\mathrm{7}} \left(\mid{x}\mid{arctan}\left(\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)\right.} \\ $$
Commented by mathmax by abdo last updated on 13/Oct/19
2) we have f^′ (x)=−∫_(−∞) ^(+∞) ((4(2x)(t^2 −2t+x^2 )^3 )/((t^2 −2t+x^2 )^8 ))dt =−8x ∫_(−∞) ^(+∞) (dx/((t^2 −2t +x^2 )^5 )) =−8x g(x) ⇒g(x)=−(1/(8x))f^′ (x) rest the calculus of f^′ (x) ....
$$\left.\mathrm{2}\right)\:{we}\:{have}\:{f}^{'} \left({x}\right)=−\int_{−\infty} ^{+\infty} \:\frac{\mathrm{4}\left(\mathrm{2}{x}\right)\left({t}^{\mathrm{2}} −\mathrm{2}{t}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }{\left({t}^{\mathrm{2}} −\mathrm{2}{t}+{x}^{\mathrm{2}} \right)^{\mathrm{8}} }{dt} \\ $$$$=−\mathrm{8}{x}\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{dx}}{\left({t}^{\mathrm{2}} −\mathrm{2}{t}\:+{x}^{\mathrm{2}} \right)^{\mathrm{5}} }\:=−\mathrm{8}{x}\:{g}\left({x}\right)\:\Rightarrow{g}\left({x}\right)=−\frac{\mathrm{1}}{\mathrm{8}{x}}{f}^{'} \left({x}\right) \\ $$$${rest}\:{the}\:{calculus}\:{of}\:{f}^{'} \left({x}\right)\:…. \\ $$
Commented by mathmax by abdo last updated on 13/Oct/19
∫_(−∞) ^(+∞) (dt/((t^2 −2t +3)^4 )) =f((√3)) =((5π)/(16sin^7 ((√3)arctan((√(3−1))))) =((5π)/(16 sin^7 ((√3)arctan(2))))
$$\int_{−\infty} ^{+\infty} \:\:\frac{{dt}}{\left({t}^{\mathrm{2}} −\mathrm{2}{t}\:+\mathrm{3}\right)^{\mathrm{4}} }\:={f}\left(\sqrt{\mathrm{3}}\right)\:=\frac{\mathrm{5}\pi}{\mathrm{16}{sin}^{\mathrm{7}} \left(\sqrt{\mathrm{3}}{arctan}\left(\sqrt{\mathrm{3}−\mathrm{1}}\right)\right.} \\ $$$$=\frac{\mathrm{5}\pi}{\mathrm{16}\:{sin}^{\mathrm{7}} \left(\sqrt{\mathrm{3}}{arctan}\left(\mathrm{2}\right)\right)} \\ $$

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