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calculate-0-x-e-x-2-a-2-sin-bx-dx-with-a-gt-0-and-b-gt-0-




Question Number 63214 by mathmax by abdo last updated on 30/Jun/19
calculate  ∫_0 ^∞  x e^(−(x^2 /a^2 ))   sin(bx)dx  with  a>0 and b>0
calculate0xex2a2sin(bx)dxwitha>0andb>0
Commented by mathmax by abdo last updated on 01/Jul/19
let I =∫_0 ^∞   x e^(−(x^2 /a^2 ))  s in(bx)dx ⇒ 2I =∫_(−∞) ^(+∞)   x e^(−(x^2 /a^2 ))  sin(bx)dx  by parts  u^′  =x e^(−(x^2 /a^2 ))    and v =sin(bx) ⇒  2I =[−(a^2 /2) e^(−(x^2 /a^2 ))  sin(bx)]_(−∞) ^(+∞)  −∫_(−∞) ^(+∞)   −(a^2 /2) e^(−(x^2 /a^2 ))  b cos(bx) dx  =((a^2 b)/2) ∫_(−∞) ^(+∞)   e^(−(x^2 /a^2 ))  cos(bx)dx   let determine A_λ =∫_(−∞) ^(+∞)  e^(−λx^2 ) cos(bx)dx with λ>0  A_λ =Re (∫_(−∞) ^(+∞)  e^(−λx^2 +ibx) dx)  ∫_(−∞) ^(+∞)   e^(−λx^2  +ibx) dx =∫_(−∞) ^(+∞)   e^(−{ ((√λ)x)^2 −2(√λ)x((ib)/(2(√λ)))  −(b^2 /(4λ)) +(b^2 /(4λ))}) dx  =∫_(−∞) ^(+∞)   e^(−( (√λ)x +((ib)/(2(√λ))))^2 )  e^(−(b^2 /(4λ)))  dx =_((√λ)x +((ib)/(2(√υ)))=u)      e^(−(b^2 /(4λ))) ∫_(−∞) ^(+∞)  e^(−u^2 ) (du/( (√λ))) =((√π)/( (√λ))) ⇒ A_λ =((√π)/( (√λ))) e^(−(b^2 /(4λ)))  ⇒  2I =((a^2 b)/2) ((√π)/( (√λ))) e^(−(b^2 /(4λ)))         and λ =(1/a^2 ) ⇒  I =((a^2 b)/4) ((√π)/(1/a)) e^(−(b^2 /(4(1/a^2 )))) =((√π)/4) a^3 b e^(−((a^2 b^2 )/4))
letI=0xex2a2sin(bx)dx2I=+xex2a2sin(bx)dxbypartsu=xex2a2andv=sin(bx)2I=[a22ex2a2sin(bx)]++a22ex2a2bcos(bx)dx=a2b2+ex2a2cos(bx)dxletdetermineAλ=+eλx2cos(bx)dxwithλ>0Aλ=Re(+eλx2+ibxdx)+eλx2+ibxdx=+e{(λx)22λxib2λb24λ+b24λ}dx=+e(λx+ib2λ)2eb24λdx=λx+ib2υ=ueb24λ+eu2duλ=πλAλ=πλeb24λ2I=a2b2πλeb24λandλ=1a2I=a2b4π1aeb241a2=π4a3bea2b24

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