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find-the-value-of-I-0-e-t-sint-t-dt-and-J-0-e-t-cos-t-t-dt-study-first-the-convergence-




Question Number 62262 by maxmathsup by imad last updated on 18/Jun/19
find the value of   I =∫_0 ^∞   ((e^(−t) sint)/( (√t)))dt  and J =∫_0 ^∞   ((e^(−t) cos(t))/( (√t)))dt  ,study first the convergence.
findthevalueofI=0etsinttdtandJ=0etcos(t)tdt,studyfirsttheconvergence.
Commented by maxmathsup by imad last updated on 19/Jun/19
the convergence o I and J is proved by sir smail let passe to the values  we have J−iI =∫_0 ^∞   (e^(−t) /( (√t)))(cost −isint)dt  =∫_0 ^∞   ((e^(−t ) e^(−it) )/( (√t))) dt =∫_0 ^∞   (e^(−(i+1)t) /( (√t))) dt  changement (√t)=x give  J−iI =∫_0 ^∞  (e^(−(1+i)x^2 ) /x) (2x)dx =2 ∫_0 ^∞   e^(−(1+i)x^2 ) dx =∫_(−∞) ^(+∞)   e^(−(1+i)x^2 ) dx  =_((√(1+i))x =u)     ∫_(−∞) ^(+∞)  e^(−u^2 ) (du/( (√(1+i)))) =(1/( (√(1+i)))) ∫_(−∞) ^(+∞)  e^(−u^2 ) du =((√π)/( (√(1+i))))  1+i =(√2)e^((iπ)/4)  ⇒(√(1+i))=^4 (√2)e^((iπ)/8)  ⇒J−iI =((√π)/((^4 (√2)))) e^(−((iπ)/8))   =((√π)/((^4 (√2)))){cos((π/8))−isin((π/8))} ⇒ I =((√π)/((^4 (√2)))) sin((π/8)) and J =((√π)/((^4 (√2)))) cos((π/8)) ⇒  I =((√π)/((^4 (√2))))((√(2−(√2)))/2)    and  J =((√π)/((^4 (√2)))) ((√(2+(√2)))/2)  .
theconvergenceoIandJisprovedbysirsmailletpassetothevalueswehaveJiI=0ett(costisint)dt=0eteittdt=0e(i+1)ttdtchangementt=xgiveJiI=0e(1+i)x2x(2x)dx=20e(1+i)x2dx=+e(1+i)x2dx=1+ix=u+eu2du1+i=11+i+eu2du=π1+i1+i=2eiπ41+i=42eiπ8JiI=π(42)eiπ8=π(42){cos(π8)isin(π8)}I=π(42)sin(π8)andJ=π(42)cos(π8)I=π(42)222andJ=π(42)2+22.
Answered by Smail last updated on 18/Jun/19
sint≤1⇔e^(−t) sint≤e^(−t)   ((e^(−t) sint)/( (√t)))≤(e^(−t) /( (√t)))⇔∫_0 ^∞ ((e^(−t) sint)/( (√t)))dt≤∫_0 ^∞ (e^(−t) /( (√t)))dt  Knowing that ∫_0 ^∞ (e^(−t) /( (√t)))dt=2∫_0 ^∞ e^(−x^2 ) dx=(√π)  which means ∫_0 ^∞ (e^(−t) /( (√t)))dt converges.  That means I=∫_0 ^∞ ((e^(−t) sint)/( (√t)))dt converges   because I≤(√π)   Also  J≤∫_0 ^∞ (e^(−t) /( (√t)))dt meaning  J also converges  J−iI=∫_0 ^∞ ((e^(−t) e^(−it) )/( (√t)))dt  =∫_0 ^∞ (e^(−t(1+i)) /( (√t)))dt  Let  x=(√t)((√(1+i)))⇒(dt/( (√t)))=((2dx)/( (√(1+i))))  J−iI=(2/( (2)^(1/4) e^(i(π/8)) ))∫_0 ^∞ e^(−x^2 ) dx=((√π)/( (2)^(1/4) e^(i(π/8)) ))  =((√π)/( (2)^(1/4) ))(cos((π/8))−isin((π/8)))  J=((√π)/( (2)^(1/4) ))cos((π/8))=((√π)/( (2)^(1/4) ))(√(((√2)+1)/(2(√2))))=  J=((√(π((√2)+1)))/2)  I=((√(π((√2)−1)))/2)
sint1etsintetetsinttett0etsinttdt0ettdtKnowingthat0ettdt=20ex2dx=πwhichmeans0ettdtconverges.ThatmeansI=0etsinttdtconvergesbecauseIπAlsoJ0ettdtmeaningJalsoconvergesJiI=0eteittdt=0et(1+i)tdtLetx=t(1+i)dtt=2dx1+iJiI=224eiπ80ex2dx=π24eiπ8=π24(cos(π8)isin(π8))J=π24cos(π8)=π242+122=J=π(2+1)2I=π(21)2
Commented by maxmathsup by imad last updated on 18/Jun/19
thanks sir
thankssir

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