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Question Number 116744 by mnjuly1970 last updated on 06/Oct/20
              ... (( nice)/(calculus)) ...    prove  that ::                  ∫_(−1) ^( ∞) (e^(−4x) /( (√(x+1)))) dx =((√π)/2) e^4                  ... m.n .1970...
nicecalculusprovethat::1e4xx+1dx=π2e4m.n.1970
Answered by mindispower last updated on 06/Oct/20
x+1=t⇒  ∫_0 ^∞ (e^(−4t+4) /( (√t)))dt  4t=w⇒e^4 ∫_0 ^∞ (e^(−w) /( (√w))).(dw/2)=(e^4 /2).∫_0 ^∞ w^((1/2)−1) e^(−w) dw  =(e^4 /2)Γ((1/2))=((e^4 (√π))/2)
x+1=t0e4t+4tdt4t=we40eww.dw2=e42.0w121ewdw=e42Γ(12)=e4π2
Commented by mnjuly1970 last updated on 06/Oct/20
thank you so much ...
thankyousomuch
Answered by Dwaipayan Shikari last updated on 06/Oct/20
∫_(−1) ^∞ (e^(−4x) /( (√(1+x))))dx                      1+x=t^2 ⇒1=2t(dt/dx)  ∫_0 ^∞ ((e^(−4(t^2 −1)) 2t)/t)dt  2e^4 ∫_0 ^∞ e^(−4t^2 ) dt =2e^4 (1/4)∫_0 ^∞ e^(−u^2 ) du=((√π)/2)e^4
1e4x1+xdx1+x=t21=2tdtdx0e4(t21)2ttdt2e40e4t2dt=2e4140eu2du=π2e4
Commented by mnjuly1970 last updated on 06/Oct/20
grateful..
grateful..
Answered by 1549442205PVT last updated on 06/Oct/20
I=    ∫_(−1) ^( ∞) (e^(−4x) /( (√(x+1)))) dx =    ^(I.B.P) 2e^(−4x) (√(x+1))∣_(−1) ^(+∞)   +8∫_(−1) ^∞ e^(−4x) (√(x+1))dx=8∫_(−1) ^∞ e^(−4x) (√(x+1))dx  Put (√(x+1))=u⇒x+1=u^2 ⇒dx=2udu  J=∫_(−1) ^∞ e^(−4x) (√(x+1))dx=2∫_0 ^∞ e^(−4(u^2 −1)) u^2 du  =(1/4)e^4 ∫_0 ^∞ e^(−(2u)^2 ) (2u)^2 d(2u)=_(2u=v) (e^4 /4)∫_0 ^∞ e^(−v^2 ) v^2 dv  Put v^2 =t⇒2vdv=dt⇒dv=(dt/(2(√t)))  J=(e^4 /8)∫_0 ^∞ e^(−t) .t^(1/2) dt=(e^4 /8)∫_0 ^∞ e^(−t) .t^((3/2)−1) dt  =(e^4 /8)Γ((3/2))=(e^4 /8).((√π)/2).Therefore,  I=8J=((e^4 (√π))/2) (Q.E.D)
I=1e4xx+1dx=I.B.P2e4xx+11++81e4xx+1dx=81e4xx+1dxPutx+1=ux+1=u2dx=2uduJ=1e4xx+1dx=20e4(u21)u2duMissing \left or extra \rightPutv2=t2vdv=dtdv=dt2tJ=e480et.t12dt=e480et.t321dt=e48Γ(32)=e48.π2.Therefore,I=8J=e4π2(Q.E.D)
Commented by mnjuly1970 last updated on 06/Oct/20
thank you...
thankyou
Commented by 1549442205PVT last updated on 07/Oct/20
Thank Sir.You are welcome
ThankSir.Youarewelcome
Answered by mathmax by abdo last updated on 06/Oct/20
∫_(−1) ^∞  (e^(−4x) /( (√(x+1))))dx =_((√(x+1))=t)  ∫_0 ^∞   (e^(−4(t^2 −1)) /t)(2t)dt =2 ∫_0 ^∞  e^(−4t^2 +4)  dt  =2 e^4  ∫_0 ^∞   e^(−(2t)^2 ) dt =_(2t=z)    2e^4  ∫_0 ^∞  e^(−z^2 ) (dz/2) =e^4  ∫_0 ^∞ e^(−z^2 ) dz  =e^4 .((√π)/2) =((√π)/2)e^4
1e4xx+1dx=x+1=t0e4(t21)t(2t)dt=20e4t2+4dt=2e40e(2t)2dt=2t=z2e40ez2dz2=e40ez2dz=e4.π2=π2e4
Commented by mnjuly1970 last updated on 06/Oct/20
grateful  sir ..   very  nice  as  always..
gratefulsir..veryniceasalways..
Commented by Bird last updated on 07/Oct/20
you are welcome sir.
youarewelcomesir.

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