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Question-213138




Question Number 213138 by Spillover last updated on 31/Oct/24
Commented by Frix last updated on 31/Oct/24
We must first solve ∫sin t^3  dt which might  be possible using  sin t^3  =((e^(it^3 ) −e^(−it^3 ) )/(2i))=((1−e^(2t^3 ) )/(2e^t^3  ))i  But how?  Then ∫sin t^2  dt (Fresnel−Integral)  I don′t think many here among us can solve  this.  Approximating I get  ∫_0 ^1  ∫_0 ^(1/2)  ∫_0 ^(1/3) sin (x+y^2 +z^3 ) dz dy dx≈.088997
Wemustfirstsolvesint3dtwhichmightbepossibleusingsint3=eit3eit32i=1e2t32et3iButhow?Thensint2dt(FresnelIntegral)Idontthinkmanyhereamonguscansolvethis.ApproximatingIget10120130sin(x+y2+z3)dzdydx.088997
Answered by MrGaster last updated on 31/Oct/24
=∫_0 ^1 ∫_0 ^(1/2) [−((cos(x+y^2 +z^3 ))/(3z^2 ))]_0 ^(1/3) dydx  =∫_0 ^1 ∫_0 ^(1/2) (−((cos(x+y^2 +(1/(27)))/(3∙(1/9)))+((cos(x+y^2 ))/(3∙0^2 )))dydx  =∫_0 ^1 ∫_0 ^(1/2) (−3cos(x+y^2 +(1/(27)))+cos2(x+y^2 ))dydx  =∫_0 ^1 [sin(x+y^2 )−3sin(x+y^2 +(1/(27)))]_0 ^(1/2) dx  =∫_0 ^1 (sin(x+(1/4))−3sin(x+(1/4)+(1/(27)))−sin(x)+3sin(x+(1/(27))))dx  =[−cos(x+(1/4))−3sin(x+(1/4)+(1/(27)))+cos(x)−3cos(x+(1/(27)))]_0 ^1   =−cos(1+(1/4))+3cos(1+(1/4)+(1/(27)))+cos(1)−3cos(1+(1/(27)))+cos((1/4))−3cos((1/4)+(1/(27)))−cos(0)+cos((1/(27)))
=01012[cos(x+y2+z3)3z2]013dydx=01012(cos(x+y2+127319+cos(x+y2)302)dydx=01012(3cos(x+y2+127)+cos2(x+y2))dydx=01[sin(x+y2)3sin(x+y2+127)]012dx=01(sin(x+14)3sin(x+14+127)sin(x)+3sin(x+127))dx=[cos(x+14)3sin(x+14+127)+cos(x)3cos(x+127)]01=cos(1+14)+3cos(1+14+127)+cos(1)3cos(1+127)+cos(14)3cos(14+127)cos(0)+cos(127)
Commented by Frix last updated on 31/Oct/24
No.
No.
Answered by Berbere last updated on 31/Oct/24
Im ∫_0 ^1 ∫_0 ^(1/2) ∫_0 ^(1/3) e^(i(x+y^2 +z^3 )) dxdydz  =Im∫_0 ^1 e^(ix) ∫_0 ^(1/2) e^(iy^2 ) dy∫_0 ^(1/3) e^(iz^3 ) dz  =Im[−ie^i +i].A.B  −y=x^n ⇒x=y^(1/n) e^(i(π/n)) =(ye^(iπ) )^(1/n)   ∫e^x^n  dx=∫e^(−y) .(1/n)e^(i(π/n)) y^((1/n)−1) dy=−(1/n)e^(i(π/n)) Γ((1/n),y)  Γ(a,z)=∫_z ^∞ t^(a−1) e^(−t) dt [incomplete Gamma function]  −(1/n)e^(i(π/n)) Γ((1/n),−x^n )+c  ⇒A=∫_0 ^((1/2) )  e^(iy^2 ) dy=(1/2)∫_0 e^(−z) i^(1/2) z^((1/2)−1) dz  =−(1/2)i^(1/2) Γ((1/2),z)  A=−(1/2)i^(1/2) Γ((1/2),−iy^2 )]_0 ^(1/2) =−(1/2)e^((iπ)/4) Γ((1/2),−(i/4)]+(1/2)e^(i(π/4)) Γ((1/2),0]  B=−(1/3)i^(1/3) Γ((1/3),−(i/(27)))+(i^(1/3) /3)Γ((1/3),0)  ∫_0 ^1 ∫_0 ^(1/2) ∫_0 ^(1/3) sin(x+y^2 +z^3 )dxdydz  =TM(i(1−e^i )(−(1/2)e^(i(π/4)) (Γ((1/2),−(i/4))+Γ((1/2),0)).(−(1/3)e^(i(π/6)) Γ((1/3),−(i/(27)),)+(e^(i(π/6)) /3)Γ((1/3),0))  IM imaginary Part
Im01012013ei(x+y2+z3)dxdydz=Im01eix012eiy2dy013eiz3dz=Im[iei+i].A.By=xnx=y1neiπn=(yeiπ)1nexndx=ey.1neiπny1n1dy=1neiπnΓ(1n,y)Γ(a,z)=zta1etdt[incompleteGammafunction]1neiπnΓ(1n,xn)+cA=012eiy2dy=120ezi12z121dz=12i12Γ(12,z)A=12i12Γ(12,iy2)]012=12eiπ4Γ(12,i4]+12eiπ4Γ(12,0]B=13i13Γ(13,i27)+i133Γ(13,0)01012013sin(x+y2+z3)dxdydz=TM(i(1ei)(12eiπ4(Γ(12,i4)+Γ(12,0)).(13eiπ6Γ(13,i27,)+eiπ63Γ(13,0))IMimaginaryPart
Commented by Frix last updated on 31/Oct/24
I was hoping you show us the solution,  thank you!
Iwashopingyoushowusthesolution,thankyou!
Commented by Berbere last updated on 01/Nov/24
withe pleasur God blessYou
withepleasurGodblessYou

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