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0-sinxdx-x-




Question Number 27410 by math1967 last updated on 06/Jan/18
∫_0 ^∞ ((sinxdx)/x)
0sinxdxx
Commented by math1967 last updated on 07/Jan/18
Thank you sir
Thankyousir
Commented by abdo imad last updated on 06/Jan/18
let introduce the fonction f(t)= ∫_0 ^∞ ((sinx)/x) e^(−tx) dx with t≥0  after verifying that f is derivable we have  f^′ (t)= −∫_0 ^∞ sinx e^(−tx) dx=−Im(∫_0 ^∞ e^((i−t)x) dx)  =−[ (1/(i−t)) e^((i−t)x)  ]_(x=0) ^(x−>∝) = (1/(i−t)) =((i+t)/(−1−t^2 ))=−(t/(1+t^2 )) −(i/(1+t^2 ))  ⇒f^′ (t)= −(1/(1+t^2 ))  ⇒f(t)=λ−arctan(t)and due to f continue  ∃ M>0//f(t)≤M∫_0 ^∞  e^(−tx) dx=_(t−>∝so) (M/t)−>0  so λ=(π/2) and f(t)=(π/2) −arctan(t)  ∫_0 ^∞ ((sinx)/x)=f(0)=(π/2)  .
letintroducethefonctionf(t)=0sinxxetxdxwitht0afterverifyingthatfisderivablewehavef(t)=0sinxetxdx=Im(0e(it)xdx)=[1ite(it)x]x=0x>∝=1it=i+t1t2=t1+t2i1+t2f(t)=11+t2f(t)=λarctan(t)andduetofcontinueM>0//f(t)M0etxdx=t>∝soMt>0soλ=π2andf(t)=π2arctan(t)0sinxx=f(0)=π2.

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