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Question Number 111104 by Lordose last updated on 02/Sep/20

Commented by mathdave last updated on 02/Sep/20

the question is not integradable

thequestionisnotintegradable

Commented by Dwaipayan Shikari last updated on 02/Sep/20

∫(−1)^x dx=∫e^(xlog(−1)) dx=∫e^(iπx) dx=(1/(iπ))e^(iπx) +C

(1)xdx=exlog(1)dx=eiπxdx=1iπeiπx+C

Answered by abdomsup last updated on 02/Sep/20

this integral can be solved...

thisintegralcanbesolved...

Commented by mathdave last updated on 02/Sep/20

then solve it since it is integradable

thensolveitsinceitisintegradable

Answered by mathmax by abdo last updated on 03/Sep/20

I =∫_(−∞) ^(+∞)  (((−1)^x^2  )/((x^2  +x+1)^2 )) dx ⇒ I =∫_(−∞) ^(+∞)  (e^(iπx^2 ) /((x^2  +x+1)^2 ))dx let  ϕ(z) =(e^(iπz^2 ) /((z^2  +z+1)^2 ))  poles of ϕ?  z^2  +z +1 =0→Δ =1−4 =−3 ⇒z_1 =((−1+i(√3))/2) =e^((i2π)/3)   z_2 =((−1−i(√3))/2) =e^(−((i2π)/3))     ⇒ϕ(z) =(e^(iπz^2 ) /((z−e^((i2π)/3) )^2 (z−e^(−((i2π)/3)) )^2 ))  residus theorem give ∫_(−∞) ^(+∞)  ϕ(z)dz =2iπ Res(ϕ,e^((i2π)/3) )  Res(ϕ ,e^((i2π)/3) ) =lim_(z→e^((i2π)/3) )   (1/((2−1)!)){ (z−e^((i2π)/3) )^2 ϕ(z)}^((1))   =lim_(z→e^((i2π)/3) )   {(e^(iπz^2 ) /((z−e^(−((i2π)/3)) )^2 ))}^((1))   =lim_(z→e^((i2π)/3) )     ((2iπze^(iπz^2 ) (z−e^(−((i2π)/3)) )^2 −2(z−e^(−((i2π)/3)) )e^(iπz^2 ) )/((z−e^(−((i2π)/3)) )^4 ))  =lim_(z→e^((i2π)/3) )     ((2iπz e^(iπz^2 ) (z−e^(−((i2π)/3)) )−2e^(iπz^2 ) )/((z−e^(−((i2π)/3)) )^3 ))  =lim_(z→e^((i2π)/3) )     ((2e^(iπz^2 ) {iπz (z−e^(−((i2π)/3)) )−1})/((z−e^(−((i2π)/3)) )^3 ))  =((2e^(iπe^(i((4π)/3)) ) {iπ e^((i2π)/3) (2isin(((2π)/3))−1})/((2isin(((2π)/3)))^3 )) =((2e^(iπ{−e^((iπ)/3) }) {−π(√3)e^((i2π)/3) −1})/(−8i(((√3)/2))^3 ))  =2×((e^(−iπ{(1/2)+((i(√3))/2)}) {π(√3)(−(1/2)+((i(√3))/2)+1})/(3i(√3)))  =(2/(3i(√3))) × (−i)e^((π(√3))/2) {((π(√3))/2)  +((3π)/2)i} =−(1/(3(√3))) e^((π(√3))/2) {π(√3)+3iπ} ⇒  ∫_(−∞) ^(+∞ ) ϕ(z)dz =2iπ(−(1/(3(√3))))e^((π(√3))/2) {π(√3)+3iπ} =I

I=+(1)x2(x2+x+1)2dxI=+eiπx2(x2+x+1)2dxletφ(z)=eiπz2(z2+z+1)2polesofφ?z2+z+1=0Δ=14=3z1=1+i32=ei2π3z2=1i32=ei2π3φ(z)=eiπz2(zei2π3)2(zei2π3)2residustheoremgive+φ(z)dz=2iπRes(φ,ei2π3)Res(φ,ei2π3)=limzei2π31(21)!{(zei2π3)2φ(z)}(1)=limzei2π3{eiπz2(zei2π3)2}(1)=limzei2π32iπzeiπz2(zei2π3)22(zei2π3)eiπz2(zei2π3)4=limzei2π32iπzeiπz2(zei2π3)2eiπz2(zei2π3)3=limzei2π32eiπz2{iπz(zei2π3)1}(zei2π3)3=2eiπei4π3{iπei2π3(2isin(2π3)1}(2isin(2π3))3=2eiπ{eiπ3}{π3ei2π31}8i(32)3=2×eiπ{12+i32}{π3(12+i32+1}3i3=23i3×(i)eπ32{π32+3π2i}=133eπ32{π3+3iπ}+φ(z)dz=2iπ(133)eπ32{π3+3iπ}=I

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