Question Number 133537 by mathmax by abdo last updated on 22/Feb/21
$$\mathrm{calculate}\:\int_{\mathrm{1}} ^{\infty} \:\frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\right)^{\mathrm{9}} } \\ $$
Answered by Ñï= last updated on 23/Feb/21
![∫_1 ^∞ (dx/((1+x+x^2 )^9 )) =∫_1 ^∞ (dx/([(x+(1/2))^2 +(3/4)]^9 )) =(√(3/4))∫_(π/3) ^(π/2) ((sec^2 θdθ)/(((3/4)tan^2 θ+(3/4))^9 )) =((√(3/4))/( ((3/4))^9 ))∫_(π/3) ^(π/2) cos^(16) θdθ =((3/4))^(−17/2) Re(∫_(π/3) ^(π/2) e^(16iθ) dθ) =((3/4))^(−17/2) Re((1/(16i))(e^(8iπ) −e^(16iπ/3) ) =(2^(12) /3^8 )](https://www.tinkutara.com/question/Q133639.png)
$$\int_{\mathrm{1}} ^{\infty} \frac{{dx}}{\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} \right)^{\mathrm{9}} } \\ $$$$=\int_{\mathrm{1}} ^{\infty} \frac{{dx}}{\left[\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}\right]^{\mathrm{9}} } \\ $$$$=\sqrt{\frac{\mathrm{3}}{\mathrm{4}}}\int_{\frac{\pi}{\mathrm{3}}} ^{\frac{\pi}{\mathrm{2}}} \frac{{sec}^{\mathrm{2}} \theta{d}\theta}{\left(\frac{\mathrm{3}}{\mathrm{4}}{tan}^{\mathrm{2}} \theta+\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{9}} } \\ $$$$=\frac{\sqrt{\frac{\mathrm{3}}{\mathrm{4}}}}{\:\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{9}} }\int_{\pi/\mathrm{3}} ^{\pi/\mathrm{2}} {cos}^{\mathrm{16}} \theta{d}\theta \\ $$$$=\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{−\mathrm{17}/\mathrm{2}} {Re}\left(\int_{\pi/\mathrm{3}} ^{\pi/\mathrm{2}} {e}^{\mathrm{16}{i}\theta} {d}\theta\right) \\ $$$$=\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{−\mathrm{17}/\mathrm{2}} {Re}\left(\frac{\mathrm{1}}{\mathrm{16}{i}}\left({e}^{\mathrm{8}{i}\pi} −{e}^{\mathrm{16}{i}\pi/\mathrm{3}} \right)\right. \\ $$$$=\frac{\mathrm{2}^{\mathrm{12}} }{\mathrm{3}^{\mathrm{8}} } \\ $$
Answered by mathmax by abdo last updated on 24/Feb/21
![Φ=∫_1 ^∞ (dx/((x^2 +x+1)^9 )) ⇒Φ =∫_1 ^(∞ ) (dx/(((x+(1/2))^2 +(3/4))^9 )) =_(x+(1/2)=((√3)/2)t) ∫_(√3) ^∞ ((4/3))^9 (1/((t^2 +1)^9 ))((√3)/2)dt =((√3)/2)((4/3))^9 ∫_(√3) ^∞ (dt/((t^2 +1)^9 )) wehave ∫_(√3) ^∞ (dt/((t^2 +1)^9 )) =∫_(√3) ^∞ (dt/((t−i)^9 (t+i)^9 ))=∫_(√3) ^∞ (dt/((((t−i)/(t+i)))^9 (t+i)^(18) )) changement ((t−i)/(t+i))=z give t−i=zt+iz ⇒(1−z)t=i(1+z) ⇒ t=i((1+iz)/(1−iz)) ⇒(dt/dz)=i((i(1−iz)−(1+iz)(−i))/((1−iz)^2 ))=i((i+z+i−z)/((1−iz)^2 ))=((−2)/((1−iz)^2 )) and t+i =((i−z)/(1−iz))+i =((i−z+i+z)/((1−iz)))=((2i)/((1−iz))) ⇒ Φ =∫_(((√3)−i)/( (√3)+i)) ^1 ((−2)/((1−iz)^2 z^9 (((2i)/(1−iz)))^(18) ))dz =(2/((2i)^(18) ))∫_1 ^(((√3)−i)/( (√3)+i)) (((1−iz)^(16) )/z^9 )dz =(2/((2i)^(18) ))∫_1 ^(((√3)−i)/( (√3)+i)) ((Σ_(k=0) ^(16) C_(16) ^k (iz)^k (−1)^(16−k) )/z^9 )dz =−(2/(2^(18) )) ∫_1 ^(((√3)−i)/( (√3) +i)) Σ_(k=0) ^(16) (−1)^k C_(16) ^k i^k z^(k−9) dz =−(1/2^(17) )Σ_(k=0 ,k≠8) ^(16) (−i)^k C_(16) ^k [(1/(k−8))z^(k−8) ]_1 ^(((√3)−i)/( (√(3 ))+i)) −(1/2^(17) )C_(16) ^8 [lnz]_1 ^(((√3)−i)/( (√3)+i)) =−(1/2^(17) )Σ_(k=0 and k≠8) ^(16) (((−1)^k C_(16) ^k )/(k−8)){((((√3)−i)/( (√3)+i)))^(k−8) −1} −(1/2^(17) )C_(16) ^8 ln((((√3)−i)/( (√3)+i)))](https://www.tinkutara.com/question/Q133756.png)
$$\Phi=\int_{\mathrm{1}} ^{\infty} \:\frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{x}+\mathrm{1}\right)^{\mathrm{9}} }\:\Rightarrow\Phi\:=\int_{\mathrm{1}} ^{\infty\:} \:\frac{\mathrm{dx}}{\left(\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{9}} } \\ $$$$=_{\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{t}} \:\:\:\:\int_{\sqrt{\mathrm{3}}} ^{\infty} \:\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{\mathrm{9}} \:\frac{\mathrm{1}}{\left(\mathrm{t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{9}} }\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{dt}\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{\mathrm{9}} \:\int_{\sqrt{\mathrm{3}}} ^{\infty} \frac{\mathrm{dt}}{\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{9}} } \\ $$$$\mathrm{wehave}\:\:\int_{\sqrt{\mathrm{3}}} ^{\infty} \:\frac{\mathrm{dt}}{\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{9}} }\:=\int_{\sqrt{\mathrm{3}}} ^{\infty} \:\frac{\mathrm{dt}}{\left(\mathrm{t}−\mathrm{i}\right)^{\mathrm{9}} \left(\mathrm{t}+\mathrm{i}\right)^{\mathrm{9}} }=\int_{\sqrt{\mathrm{3}}} ^{\infty} \:\frac{\mathrm{dt}}{\left(\frac{\mathrm{t}−\mathrm{i}}{\mathrm{t}+\mathrm{i}}\right)^{\mathrm{9}} \left(\mathrm{t}+\mathrm{i}\right)^{\mathrm{18}} } \\ $$$$\mathrm{changement}\:\frac{\mathrm{t}−\mathrm{i}}{\mathrm{t}+\mathrm{i}}=\mathrm{z}\:\mathrm{give}\:\mathrm{t}−\mathrm{i}=\mathrm{zt}+\mathrm{iz}\:\Rightarrow\left(\mathrm{1}−\mathrm{z}\right)\mathrm{t}=\mathrm{i}\left(\mathrm{1}+\mathrm{z}\right)\:\Rightarrow \\ $$$$\mathrm{t}=\mathrm{i}\frac{\mathrm{1}+\mathrm{iz}}{\mathrm{1}−\mathrm{iz}}\:\Rightarrow\frac{\mathrm{dt}}{\mathrm{dz}}=\mathrm{i}\frac{\mathrm{i}\left(\mathrm{1}−\mathrm{iz}\right)−\left(\mathrm{1}+\mathrm{iz}\right)\left(−\mathrm{i}\right)}{\left(\mathrm{1}−\mathrm{iz}\right)^{\mathrm{2}} }=\mathrm{i}\frac{\mathrm{i}+\mathrm{z}+\mathrm{i}−\mathrm{z}}{\left(\mathrm{1}−\mathrm{iz}\right)^{\mathrm{2}} }=\frac{−\mathrm{2}}{\left(\mathrm{1}−\mathrm{iz}\right)^{\mathrm{2}} } \\ $$$$\mathrm{and}\:\mathrm{t}+\mathrm{i}\:=\frac{\mathrm{i}−\mathrm{z}}{\mathrm{1}−\mathrm{iz}}+\mathrm{i}\:=\frac{\mathrm{i}−\mathrm{z}+\mathrm{i}+\mathrm{z}}{\left(\mathrm{1}−\mathrm{iz}\right)}=\frac{\mathrm{2i}}{\left(\mathrm{1}−\mathrm{iz}\right)}\:\Rightarrow \\ $$$$\Phi\:=\int_{\frac{\sqrt{\mathrm{3}}−\mathrm{i}}{\:\sqrt{\mathrm{3}}+\mathrm{i}}} ^{\mathrm{1}} \:\:\frac{−\mathrm{2}}{\left(\mathrm{1}−\mathrm{iz}\right)^{\mathrm{2}} \mathrm{z}^{\mathrm{9}} \left(\frac{\mathrm{2i}}{\mathrm{1}−\mathrm{iz}}\right)^{\mathrm{18}} }\mathrm{dz} \\ $$$$=\frac{\mathrm{2}}{\left(\mathrm{2i}\right)^{\mathrm{18}} }\int_{\mathrm{1}} ^{\frac{\sqrt{\mathrm{3}}−\mathrm{i}}{\:\sqrt{\mathrm{3}}+\mathrm{i}}} \:\:\:\:\:\frac{\left(\mathrm{1}−\mathrm{iz}\right)^{\mathrm{16}} }{\mathrm{z}^{\mathrm{9}} }\mathrm{dz}\:=\frac{\mathrm{2}}{\left(\mathrm{2i}\right)^{\mathrm{18}} }\int_{\mathrm{1}} ^{\frac{\sqrt{\mathrm{3}}−\mathrm{i}}{\:\sqrt{\mathrm{3}}+\mathrm{i}}} \:\frac{\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{16}} \:\mathrm{C}_{\mathrm{16}} ^{\mathrm{k}} \left(\mathrm{iz}\right)^{\mathrm{k}} \left(−\mathrm{1}\right)^{\mathrm{16}−\mathrm{k}} }{\mathrm{z}^{\mathrm{9}} }\mathrm{dz} \\ $$$$=−\frac{\mathrm{2}}{\mathrm{2}^{\mathrm{18}} \:}\:\int_{\mathrm{1}} ^{\frac{\sqrt{\mathrm{3}}−\mathrm{i}}{\:\sqrt{\mathrm{3}}\:+\mathrm{i}}} \:\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{16}} \:\left(−\mathrm{1}\right)^{\mathrm{k}} \:\mathrm{C}_{\mathrm{16}} ^{\mathrm{k}} \:\mathrm{i}^{\mathrm{k}} \:\mathrm{z}^{\mathrm{k}−\mathrm{9}} \:\mathrm{dz} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{17}} }\sum_{\mathrm{k}=\mathrm{0}\:,\mathrm{k}\neq\mathrm{8}} ^{\mathrm{16}} \:\left(−\mathrm{i}\right)^{\mathrm{k}} \:\mathrm{C}_{\mathrm{16}} ^{\mathrm{k}} \:\left[\frac{\mathrm{1}}{\mathrm{k}−\mathrm{8}}\mathrm{z}^{\mathrm{k}−\mathrm{8}} \right]_{\mathrm{1}} ^{\frac{\sqrt{\mathrm{3}}−\mathrm{i}}{\:\sqrt{\mathrm{3}\:}+\mathrm{i}}} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{17}} }\mathrm{C}_{\mathrm{16}} ^{\mathrm{8}} \:\left[\mathrm{lnz}\right]_{\mathrm{1}} ^{\frac{\sqrt{\mathrm{3}}−\mathrm{i}}{\:\sqrt{\mathrm{3}}+\mathrm{i}}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{17}} }\sum_{\mathrm{k}=\mathrm{0}\:\mathrm{and}\:\mathrm{k}\neq\mathrm{8}} ^{\mathrm{16}} \:\:\:\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} \:\mathrm{C}_{\mathrm{16}} ^{\mathrm{k}} }{\mathrm{k}−\mathrm{8}}\left\{\left(\frac{\sqrt{\mathrm{3}}−\mathrm{i}}{\:\sqrt{\mathrm{3}}+\mathrm{i}}\right)^{\mathrm{k}−\mathrm{8}} −\mathrm{1}\right\} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{17}} }\mathrm{C}_{\mathrm{16}} ^{\mathrm{8}} \:\mathrm{ln}\left(\frac{\sqrt{\mathrm{3}}−\mathrm{i}}{\:\sqrt{\mathrm{3}}+\mathrm{i}}\right) \\ $$