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Question Number 34315 by prof Abdo imad last updated on 03/May/18
1) find F(x)= ∫_0 ^(+∞) ((e^(−at) −e^(−bt) )/t)sin(xt)dt with a>0 ,b>0 .
1)findF(x)=∫0+∞e−at−e−bttsin(xt)dtwitha>0,b>0.
Commented by math khazana by abdo last updated on 05/May/18
we have F^′ (x) = ∫_0 ^∞ (e^(−at) −e^(−bt) ) cos(xt)dt =Re( ∫_0 ^∞ (e^(−at) −e^(−bt) )e^(ixt) dt) =Re( ∫_0 ^∞ (e^((−a +ix)t) −e^((−b +ix)t) )dt but ∫_0 ^∞ e^((−a+ix)t) dt = [ (1/(−a +ix)) e^((−1+ix)t) ]_0 ^(+∞) = (1/(−a +ix)) =((−1)/(a−ix)) =−((a+ix)/(a^2 +x^2 )) by the same manner ∫_0 ^∞ e^((−b+ix)t) dt = −((b +ix)/(b^2 +x^2 )) ⇒ F^′ (x) = Re( ((b+ix)/(b^2 +x^2 )) −((a+ix)/(a^2 +x^2 ))) =(b/(b^2 +x^2 )) − (a/(a^2 +x^2 )) ⇒ F(x)= ∫_0 ^x (b/(b^2 +t^2 ))dt −∫_0 ^x (a/(a^2 +t^2 ))dt +λ but ∫_0 ^x (b/(b^2 +t^2 ))dt =_(t=bu) ∫_0 ^(x/b) (b/(b^2 (1+u^2 ))) bdu = arctan((x/b)) ⇒F(x)= arctan((x/b)) −arctan((x/a)) +λ λ =F(0) ⇒F(x)=arctan((x/b)) −arctan((x/a)) .
wehaveF′(x)=∫0∞(e−at−e−bt)cos(xt)dt=Re(∫0∞(e−at−e−bt)eixtdt)=Re(∫0∞(e(−a+ix)t−e(−b+ix)t)dtbut∫0∞e(−a+ix)tdt=[1−a+ixe(−1+ix)t]0+∞=1−a+ix=−1a−ix=−a+ixa2+x2bythesamemanner∫0∞e(−b+ix)tdt=−b+ixb2+x2⇒F′(x)=Re(b+ixb2+x2−a+ixa2+x2)=bb2+x2−aa2+x2⇒F(x)=∫0xbb2+t2dt−∫0xaa2+t2dt+λbut∫0xbb2+t2dt=t=bu∫0xbbb2(1+u2)bdu=arctan(xb)⇒F(x)=arctan(xb)−arctan(xa)+λλ=F(0)⇒F(x)=arctan(xb)−arctan(xa).
Commented by math khazana by abdo last updated on 08/May/18
we know that arctan α −arctanβ =arctan(((α−β)/(1+αβ))) ⇒F(x) = arctan((((x/b) −(x/a))/(1+(x^2 /(ab))))) = arctan(((ax−bx)/(x^2 +ab)) ) .
weknowthatarctanα−arctanβ=arctan(α−β1+αβ)⇒F(x)=arctan(xb−xa1+x2ab)=arctan(ax−bxx2+ab).

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