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Question Number 88547 by ajfour last updated on 11/Apr/20
prove for (0<a<2) ∫_0 ^( ∞) ((x^(a−1) dx)/(1+x+x^2 )) = ((2π)/( (√3)))cos (((2πa+π)/6))cosec πa .
provefor(0<a<2)∫0∞xa−1dx1+x+x2=2π3cos(2πa+π6)cosecπa.
Answered by mind is power last updated on 12/Apr/20
∫_(−∞) ^(+∞) (x^(a−1) /(1+x+x^2 ))dx=∫_0 ^∞ ((x^(a−1) dx)/(1+x+x^2 ))+∫_0 ^(+∞) (((−1)^(a−1) x^(a−1) )/(1−x+x^2 ))dx =∫_C_R (x^(a−1) /(1+x+x^2 ))dx≤∫_0 ^π ((R^a e^(i(a−1)π) )/(R^2 −R+1))→0 ⇒∫_0 ^(+∞) ((e^(iπ(a−1)) x^(a−1) )/(x^2 −x+1))dx+∫_0 ^∞ ((x^(a−1) dx)/(x^2 +x+1))=2iπRes((z^(a−1) /(z^2 +z+1)),e^(i((2π)/3)) ) =((2iπe^(i(a−1)((2π)/3)) )/((3e^((2iπ)/3) +1)))=−isin(πa)∫_0 ^∞ ((x^(a−1) dx)/(x^2 −x+1))+−cos(πa)∫_0 ^(+∞) ((x^(a−1) dx)/(x^2 −x+1)) +∫_0 ^∞ ((x^(a−1) dx)/(x^2 +x+1)) ((2iπe^(i(a−1)((2π)/3)) )/(2(−(1/2)+((i(√3))/2))+1))=((2iπe^(i(a−1)((2π)/3)) )/(i(√3)))=((2π)/( (√3)))(cos(((2π(a−1))/3))+isin(((2π(a−1))/3))) =−isin(πa)∫_0 ^∞ ((x^(a−1) dx)/(x^2 −x+1))−cos(πa)∫_0 ^∞ ((x^(a−1) dx)/(x^2 −x+1))+∫_0 ^(+∞) ((x^(a−1) dx)/(x^2 +x+1)) ⇒((2iπ)/( (√3)))sin(((2π(a−1))/3))=−isin(πa)∫_0 ^(+∞) ((x^(a−1) dx)/(x^2 −x+1)) ⇒∫_0 ^(+∞) ((x^(a−1) dx)/(x^2 −x+1))=((−2πsin(((2π(a−1))/3)))/(sin(πa)(√3))) ∫_0 ^(+∞) ((x^(a−1) dx)/(x^2 +x+1))=((2π)/( (√3)))cos(((2π(a−1))/3))+((−2πsin(((2π(a−1))/3))cos(πa))/(sin(πa)(√3))) =((2π)/( (√3))).(1/(sin(πa))){cos(((2π(a−1))/3))sin(πa)−sin(((2π(a−1))/3))cos(πa)) =((2π)/(sin(πa)(√3))){sin(πa−((2π(a−1))/3))} =((2π)/( (√3)sin(πa))){sin(((πa)/3)+((2π)/3))} =((2π)/(sin(πa)(√3)))cos((π/2)−((πa)/3)−((2π)/3))=((2π)/(sin(πa)(√3)))(cos(((−πa)/3)−(π/6))) =((2π)/(sin(πa)(√3)))cos(((2πa+π)/6))=((2π)/( (√3)))cos(((2πa+π)/6))cosec(πa)
∫−∞+∞xa−11+x+x2dx=∫0∞xa−1dx1+x+x2+∫0+∞(−1)a−1xa−11−x+x2dx=∫CRxa−11+x+x2dx⩽∫0πRaei(a−1)πR2−R+1→0⇒∫0+∞eiπ(a−1)xa−1x2−x+1dx+∫0∞xa−1dxx2+x+1=2iπRes(za−1z2+z+1,ei2π3)=2iπei(a−1)2π3(3e2iπ3+1)=−isin(πa)∫0∞xa−1dxx2−x+1+−cos(πa)∫0+∞xa−1dxx2−x+1+∫0∞xa−1dxx2+x+12iπei(a−1)2π32(−12+i32)+1=2iπei(a−1)2π3i3=2π3(cos(2π(a−1)3)+isin(2π(a−1)3))=−isin(πa)∫0∞xa−1dxx2−x+1−cos(πa)∫0∞xa−1dxx2−x+1+∫0+∞xa−1dxx2+x+1⇒2iπ3sin(2π(a−1)3)=−isin(πa)∫0+∞xa−1dxx2−x+1⇒∫0+∞xa−1dxx2−x+1=−2πsin(2π(a−1)3)sin(πa)3∫0+∞xa−1dxx2+x+1=2π3cos(2π(a−1)3)+−2πsin(2π(a−1)3)cos(πa)sin(πa)3=2π3.1sin(πa){cos(2π(a−1)3)sin(πa)−sin(2π(a−1)3)cos(πa))=2πsin(πa)3{sin(πa−2π(a−1)3)}=2π3sin(πa){sin(πa3+2π3)}=2πsin(πa)3cos(π2−πa3−2π3)=2πsin(πa)3(cos(−πa3−π6))=2πsin(πa)3cos(2πa+π6)=2π3cos(2πa+π6)cosec(πa)
Commented by ajfour last updated on 12/Apr/20
Thank you sir, hope you liked solving it.
Thankyousir,hopeyoulikedsolvingit.
Commented by mind is power last updated on 12/Apr/20
nice one Sir since 4 or 5 month ago i lost my motivation lost pleasur of solving problemes i dont know why!
niceoneSirsince4or5monthagoilostmymotivationlostpleasurofsolvingproblemesidontknowwhy!

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