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we-give-for-t-gt-0-0-sinx-x-e-tx-dx-pi-2-arctant-use-this-result-to-find-the-value-of-0-1-e-x-sinx-x-2-dx-




Question Number 33979 by abdo imad last updated on 28/Apr/18
we give for t>0 ∫_0 ^∞   ((sinx)/x) e^(−tx) dx =(π/2) −arctant  use this result to find the value of ∫_0 ^∞   (((1−e^(−x) )sinx)/x^2 )dx .
wegivefort>00sinxxetxdx=π2arctantusethisresulttofindthevalueof0(1ex)sinxx2dx.
Commented by abdo mathsup 649 cc last updated on 03/May/18
we know that  ∫_0 ^∞    ((sinx)/x) e^(−tx)  dx=(π/2) −arctant ⇒  ∫_0 ^1  ((π/2) −arctant)dt = ∫_0 ^1  ( ∫_0 ^∞  ((sinx)/x) e^(−tx) dx)  =∫_0 ^∞   (∫_0 ^1   e^(−tx) dt) ((sinx)/x)dx ( by fubini)  =∫_0 ^∞  ([−(1/x) e^(−tx) ]_(t=0) ^(t=1) )((sinx)/x)dx  =∫_0 ^∞ (1/x^2 )(1−e^(−x) ) sinx dx but  ∫_0 ^1 ( (π/2) −arctant)dt =(π/2) −∫_0 ^1   arctan t dt  by parts  ∫_0 ^1   arctan t dt = [tarctant]_0 ^1  −∫_0 ^1  (t/(1+t^2 ))dt  =(π/4) −(1/2)[ln(1+t^2 )]_0 ^1   = (π/4) −(1/2)ln(2) ⇒  ∫_0 ^∞    (((1−e^(−x) )sinx)/x^2 ) dx = (π/4) +(1/2)ln(2) .
weknowthat0sinxxetxdx=π2arctant01(π2arctant)dt=01(0sinxxetxdx)=0(01etxdt)sinxxdx(byfubini)=0([1xetx]t=0t=1)sinxxdx=01x2(1ex)sinxdxbut01(π2arctant)dt=π201arctantdtbyparts01arctantdt=[tarctant]0110t1+t2dt=π412[ln(1+t2)]01=π412ln(2)0(1ex)sinxx2dx=π4+12ln(2).

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