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Question Number 34314 by prof Abdo imad last updated on 03/May/18
let f(x)= ∫_0 ^(+∞) ((1−cos(xt))/t^2 ) e^(−t) dt calculate f(x) .
letf(x)=∫0+∞1−cos(xt)t2e−tdtcalculatef(x).
Commented by prof Abdo imad last updated on 06/May/18
we have f^′ (x) = ∫_0 ^∞ ((t sin(xt))/t^2 ) e^(−t) dt = ∫_0 ^∞ ((sin(xt))/t) e^(−t) dt ⇒ f^(′′) (x)= ∫_0 ^∞ ((t cos(xt))/t) e^(−t) dt =∫_0 ^∞ e^(−t) cos(xt)dt =Re( ∫_0 ^∞ e^(−t) e^(ixt) dt) =Re( ∫_0 ^∞ e^((−1+ix)t) dt) but ∫_0 ^∞ e^((−1+ix)t) dt =[ (1/(−1 +ix))e^((−1+ix)t) ]_0 ^(+∞) =((−1)/(−1+ix)) = (1/(1−ix)) = ((1+ix)/(1+x^2 )) ⇒ f^(′′) (x)= (1/(1+x^2 )) ⇒ f^′ (x)= arctanx +λ but λ =f^′ (0)=0 ⇒ f^′ (x)= arctanx ⇒ f(x) = ∫ arctanxdx +c ∫ arctanxdx = x arctanx −∫ (x/(1+x^2 ))dx = x arctanx −(1/2)ln(1+x^2 ) ⇒ f(x)= x arctanx −(1/2)ln(1+x^2 ) +c c =f(0) =0 ⇒ f(x)= x arctanx −(1/2) ln(1+x^2 ) .
wehavef′(x)=∫0∞tsin(xt)t2e−tdt=∫0∞sin(xt)te−tdt⇒f″(x)=∫0∞tcos(xt)te−tdt=∫0∞e−tcos(xt)dt=Re(∫0∞e−teixtdt)=Re(∫0∞e(−1+ix)tdt)but∫0∞e(−1+ix)tdt=[1−1+ixe(−1+ix)t]0+∞=−1−1+ix=11−ix=1+ix1+x2⇒f″(x)=11+x2⇒f′(x)=arctanx+λbutλ=f′(0)=0⇒f′(x)=arctanx⇒f(x)=∫arctanxdx+c∫arctanxdx=xarctanx−∫x1+x2dx=xarctanx−12ln(1+x2)⇒f(x)=xarctanx−12ln(1+x2)+cc=f(0)=0⇒f(x)=xarctanx−12ln(1+x2).

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