Question Number 213138 by Spillover last updated on 31/Oct/24

Commented by Frix last updated on 31/Oct/24

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)))](https://www.tinkutara.com/question/Q213145.png)
Commented by Frix last updated on 31/Oct/24

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](https://www.tinkutara.com/question/Q213157.png)
Commented by Frix last updated on 31/Oct/24

Commented by Berbere last updated on 01/Nov/24
