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Question Number 136999 by mnjuly1970 last updated on 28/Mar/21
......advanced......calculus....ϕ=∫0∞ln(x).sin(x).e−xdx=?
Answered by mnjuly1970 last updated on 28/Mar/21
ϕ=−Im∫0∞ln(x)e−ix−xdx=−Im∫0∞ln(x)e−x(1+i)dx=x(1+i)=y−12Im(1−i)∫0∞(ln(y)−ln(2eiπ4))e−ydy=−22Im(22−i22){∫0∞ln(y)e−ydy+(ln(2)+iπ4)[e−y]0∞}=−22Im{(22−i22)(−γ−ln(2)−iπ4)=−22(22.(−π4)+(γ+ln(2)2)2)=π8−γ2−14ln(2).....✓......ϕ=π8−14ln(2)−γ2
Answered by Dwaipayan Shikari last updated on 28/Mar/21
χ(α)=∫0∞xαe−xsin(x)dxχ′(0)=∫0∞log(x)e−xsin(x)dx⇒χ(α)=12i∫0∞xαe−(1−i)x−xαe−(1+i)xdx⇒χ(α)=12i.Γ(α+1)(1−i)α+1−Γ(α+1)2i(1+i)α+1⇒χ(α)=12α+32i(eπ4(α+1)i−e−π4(α+1)i)⇒χ(α)=sin(π4(α+1))Γ(α+1)2α+12⇒χ′(α)=(2)α+1π4cos(π4(α+1))Γ(α+1)+Γ′(α+1)(2)α+1sin(π4(α+1))−2a−12log(2)sin(π4(a+1))Γ(α+1)2α+1χ′(0)=π4−γ+log(2)22=π8−γ2−log(2)4
Commented by mnjuly1970 last updated on 28/Mar/21
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