Menu Close

calculate-dx-x-2-x-1-3-

Tinku Tara 0 Comments
Pin




Question Number 65129 by turbo msup by abdo last updated on 25/Jul/19
calculate ∫_(−∞) ^(+∞) (dx/((x^2 +x+1)^3 ))
$${calculate}\:\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+{x}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$
Commented by mathmax by abdo last updated on 26/Jul/19
let A =∫_(−∞) ^(+∞) (dx/((x^2 +x +1)^3 )) let turn to magical number i we have x^2 +x+1 =(x+(1/2))^2 +(3/4) and chang. x+(1/2) =((√3)/2)t give A = ∫_(−∞) ^(+∞) (1/(((3/4))^3 (t^2 +1)^3 )) ((√3)/2) dt =(4^3 /3^3 ) ((√3)/2) ∫_(−∞) ^(+∞) (dt/((t^2 +1)^3 )) =((32(√3))/(27)) ∫_(−∞) ^(+∞) (dt/((t^2 +1)^3 )) let ϕ(z) =(1/((z^2 +1)^3 )) ⇒ϕ(z) =(1/((z−i)^3 (z+i)^3 )) the poles of ϕ are +^− i (triples) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) Res(ϕ,i) =lim_(z→i) (1/((3−1)!)){(z−i)^3 ϕ(z)}^((2)) =lim_(z→i) (1/2){(z+i)^(−3) }^((2)) =lim_(z→i) (1/2){−3(z+i)^(−4) }^((1)) =lim_(z→i) 6(z+i)^(−5) =6(2i)^(−5) =(6/((2i)^5 )) =(6/(2^5 i)) =(6/(32i)) =(3/(16i)) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (3/(16i)) =((3π)/8) ⇒ A =((32(√3))/(27)) ((3π)/8) =((3π(√3) .8.4)/(3.9 .8)) A =((4π(√3))/9) .
$${let}\:{A}\:=\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+{x}\:+\mathrm{1}\right)^{\mathrm{3}} }\:\:\:{let}\:{turn}\:{to}\:{magical}\:{number}\:{i} \\ $$$${we}\:{have}\:{x}^{\mathrm{2}} \:+{x}+\mathrm{1}\:=\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{4}}\:\:{and}\:{chang}.\:{x}+\frac{\mathrm{1}}{\mathrm{2}}\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{t}\:{give} \\ $$$${A}\:=\:\int_{−\infty} ^{+\infty} \:\:\:\frac{\mathrm{1}}{\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{3}} \left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{3}} }\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{dt}\:=\frac{\mathrm{4}^{\mathrm{3}} }{\mathrm{3}^{\mathrm{3}} }\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{dt}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{32}\sqrt{\mathrm{3}}}{\mathrm{27}}\:\int_{−\infty} ^{+\infty} \:\:\frac{{dt}}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{3}} }\:\:{let}\:\varphi\left({z}\right)\:=\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{3}} }\:\Rightarrow\varphi\left({z}\right)\:=\frac{\mathrm{1}}{\left({z}−{i}\right)^{\mathrm{3}} \left({z}+{i}\right)^{\mathrm{3}} } \\ $$$${the}\:{poles}\:{of}\:\varphi\:{are}\:\overset{−} {+}{i}\:\left({triples}\right)\:\:{residus}\:{theorem}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{i}\right) \\ $$$${Res}\left(\varphi,{i}\right)\:={lim}_{{z}\rightarrow{i}} \:\frac{\mathrm{1}}{\left(\mathrm{3}−\mathrm{1}\right)!}\left\{\left({z}−{i}\right)^{\mathrm{3}} \varphi\left({z}\right)\right\}^{\left(\mathrm{2}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\:\frac{\mathrm{1}}{\mathrm{2}}\left\{\left({z}+{i}\right)^{−\mathrm{3}} \right\}^{\left(\mathrm{2}\right)} \:={lim}_{{z}\rightarrow{i}} \:\frac{\mathrm{1}}{\mathrm{2}}\left\{−\mathrm{3}\left({z}+{i}\right)^{−\mathrm{4}} \right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\mathrm{6}\left({z}+{i}\right)^{−\mathrm{5}} \:\:=\mathrm{6}\left(\mathrm{2}{i}\right)^{−\mathrm{5}} \:=\frac{\mathrm{6}}{\left(\mathrm{2}{i}\right)^{\mathrm{5}} }\:=\frac{\mathrm{6}}{\mathrm{2}^{\mathrm{5}} {i}}\:=\frac{\mathrm{6}}{\mathrm{32}{i}}\:=\frac{\mathrm{3}}{\mathrm{16}{i}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\frac{\mathrm{3}}{\mathrm{16}{i}}\:=\frac{\mathrm{3}\pi}{\mathrm{8}}\:\Rightarrow\:{A}\:=\frac{\mathrm{32}\sqrt{\mathrm{3}}}{\mathrm{27}}\:\frac{\mathrm{3}\pi}{\mathrm{8}}\:\:=\frac{\mathrm{3}\pi\sqrt{\mathrm{3}}\:.\mathrm{8}.\mathrm{4}}{\mathrm{3}.\mathrm{9}\:.\mathrm{8}} \\ $$$${A}\:=\frac{\mathrm{4}\pi\sqrt{\mathrm{3}}}{\mathrm{9}}\:. \\ $$$$ \\ $$
Answered by MJS last updated on 29/Jul/19
∫(dx/((x^2 +x+1)^3 ))= Ostrogradski′s Method ∫((p(x))/(q(x)))=((p_1 (x))/(q_1 (x)))+∫((p_2 (x))/(q_2 (x))) p(x)=1 q(x)=(x^2 +x+1)^3 q_1 (x)=(x^2 +x+1)^2 q_2 (x)=x^2 +x+1 p_1 (x)=(((2x+1)(2x^2 +2x+3))/6) p_2 =(2/3) =(((2x+1)(2x^2 +2x+3))/(6(x^2 +x+1)^2 ))+(2/3)∫(dx/(x^2 +x+1))= =(((2x+1)(2x^2 +2x+3))/(6(x^2 +x+1)^2 ))+((4(√3))/9)arctan (((√3)/3)(2x+1)) +C ∫_(−∞) ^(+∞) (dx/((x^2 +x+1)^3 ))=((4(√3))/9)π
$$\int\frac{{dx}}{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{3}} }= \\ $$$$\:\:\:\:\:\mathrm{Ostrogradski}'\mathrm{s}\:\mathrm{Method} \\ $$$$\:\:\:\:\:\int\frac{{p}\left({x}\right)}{{q}\left({x}\right)}=\frac{{p}_{\mathrm{1}} \left({x}\right)}{{q}_{\mathrm{1}} \left({x}\right)}+\int\frac{{p}_{\mathrm{2}} \left({x}\right)}{{q}_{\mathrm{2}} \left({x}\right)} \\ $$$$\:\:\:\:\:{p}\left({x}\right)=\mathrm{1}\:\:\:\:\:{q}\left({x}\right)=\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{3}} \\ $$$$\:\:\:\:\:{q}_{\mathrm{1}} \left({x}\right)=\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} \:\:\:\:\:{q}_{\mathrm{2}} \left({x}\right)={x}^{\mathrm{2}} +{x}+\mathrm{1} \\ $$$$\:\:\:\:\:{p}_{\mathrm{1}} \left({x}\right)=\frac{\left(\mathrm{2}{x}+\mathrm{1}\right)\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}\right)}{\mathrm{6}}\:\:\:\:\:{p}_{\mathrm{2}} =\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$=\frac{\left(\mathrm{2}{x}+\mathrm{1}\right)\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}\right)}{\mathrm{6}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{2}}{\mathrm{3}}\int\frac{{dx}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}= \\ $$$$=\frac{\left(\mathrm{2}{x}+\mathrm{1}\right)\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}\right)}{\mathrm{6}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{9}}\mathrm{arctan}\:\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\left(\mathrm{2}{x}+\mathrm{1}\right)\right)\:+{C} \\ $$$$\underset{−\infty} {\overset{+\infty} {\int}}\frac{{dx}}{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{3}} }=\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{9}}\pi \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *