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 31107 by abdo imad last updated on 02/Mar/18

calculate ∫_(−∞) ^(+∞)       (dx/((x^2  +x+1)^n ))  with n>1.

$${calculate}\:\int_{−\infty} ^{+\infty} \:\:\:\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+{x}+\mathrm{1}\right)^{{n}} }\:\:{with}\:{n}>\mathrm{1}. \\ $$

Commented byabdo imad last updated on 06/Mar/18

let put A_n = ∫_(−∞) ^(+∞)     (dx/((x^2  +x+1)^n ))  with n≥2 we have  x^2  +x+1=x^2  +2x(1/2) +(1/4) +(3/4)=(x+(1/2))^2 +(3/4) the ch.  x+(1/2)=((√3)/2)t ⇒  I_n = ∫_(−∞) ^(+∞)      (1/(((3/4)(1+t^2 ))^n )) ((√3)/2)dt  =((4/3))^n ((√3)/2) ∫_(−∞) ^(+∞)     (dt/((1+t^2 )^n ))=(√3)((4/3))^n  ∫_0 ^∞    (dt/((1+t^2 )^n )) but  we have proved that ∫_0 ^∞   (dt/((1+t^2 )^n ))=π (((2n−2)!)/(2^(2n−1) ((n−1)!)^2 )) ⇒

$${let}\:{put}\:{A}_{{n}} =\:\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+{x}+\mathrm{1}\right)^{{n}} }\:\:{with}\:{n}\geqslant\mathrm{2}\:{we}\:{have} \\ $$ $${x}^{\mathrm{2}} \:+{x}+\mathrm{1}={x}^{\mathrm{2}} \:+\mathrm{2}{x}\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{4}}\:+\frac{\mathrm{3}}{\mathrm{4}}=\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}\:{the}\:{ch}. \\ $$ $${x}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{t}\:\Rightarrow\:\:{I}_{{n}} =\:\int_{−\infty} ^{+\infty} \:\:\:\:\:\frac{\mathrm{1}}{\left(\frac{\mathrm{3}}{\mathrm{4}}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\right)^{{n}} }\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{dt} \\ $$ $$=\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{{n}} \frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{{n}} }=\sqrt{\mathrm{3}}\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{{n}} \:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{{n}} }\:{but} \\ $$ $${we}\:{have}\:{proved}\:{that}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{{n}} }=\pi\:\frac{\left(\mathrm{2}{n}−\mathrm{2}\right)!}{\mathrm{2}^{\mathrm{2}{n}−\mathrm{1}} \left(\left({n}−\mathrm{1}\right)!\right)^{\mathrm{2}} }\:\Rightarrow \\ $$

Commented byabdo imad last updated on 06/Mar/18

⇒  A_n =π(√3) ((4/3))^n  (((2n−2)!)/(2^(2n−1) ((n−1)!)^2 )) .

$$\Rightarrow\:\:{A}_{{n}} =\pi\sqrt{\mathrm{3}}\:\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{{n}} \:\frac{\left(\mathrm{2}{n}−\mathrm{2}\right)!}{\mathrm{2}^{\mathrm{2}{n}−\mathrm{1}} \left(\left({n}−\mathrm{1}\right)!\right)^{\mathrm{2}} }\:. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com