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Question Number 136999 by mnjuly1970 last updated on 28/Mar/21

                   ......advanced  ......   calculus....             𝛗=∫_0 ^( ∞) ln(x).sin(x).e^(−x) dx=?

......advanced......calculus....ϕ=0ln(x).sin(x).exdx=?

Answered by mnjuly1970 last updated on 28/Mar/21

      𝛗=−Im∫_0 ^( ∞) ln(x)e^(−ix−x) dx          =−Im∫_0 ^( ∞) ln(x)e^(−x(1+i)) dx          =^(x(1+i)=y)  −(1/2)Im(1−i)∫_0 ^( ∞) (ln(y)−ln((√2) e^((iπ)/4) ))e^(−y) dy  =((−(√2))/2)Im(((√2)/2)−((i(√2))/2)){∫_0 ^( ∞) ln(y)e^(−y) dy+(ln((√2) )+((iπ)/4))[e^(−y) ]_0 ^∞ }  =((−(√2))/2) Im{(((√2)/2)−((i(√2))/2))(−γ−ln((√2) )−((iπ)/4))  =((−(√2))/2)(((√2)/2).(((−π)/4))+(((γ+ln((√2) ))/2))(√2) )  =(π/8)−(γ/2)−(1/4)ln(2) .....✓             ......𝛗=(π/8)−(1/4)ln(2)−(γ/2)

ϕ=Im0ln(x)eixxdx=Im0ln(x)ex(1+i)dx=x(1+i)=y12Im(1i)0(ln(y)ln(2eiπ4))eydy=22Im(22i22){0ln(y)eydy+(ln(2)+iπ4)[ey]0}=22Im{(22i22)(γln(2)iπ4)=22(22.(π4)+(γ+ln(2)2)2)=π8γ214ln(2)...........ϕ=π814ln(2)γ2

Answered by Dwaipayan Shikari last updated on 28/Mar/21

χ(α)=∫_0 ^∞ x^α e^(−x) sin(x)dx  χ′(0)=∫_0 ^∞ log(x)e^(−x) sin(x)dx  ⇒χ(α)=(1/(2i))∫_0 ^∞ x^α e^(−(1−i)x) −x^α e^(−(1+i)x) dx  ⇒χ(α)=(1/(2i)).((Γ(α+1))/((1−i)^(α+1) ))−((Γ(α+1))/(2i(1+i)^(α+1) ))⇒χ(α)=(1/(2^((α+3)/2) i))(e^((π/4)(α+1)i) −e^(−(π/4)(α+1)i) )  ⇒χ(α)=((sin((π/4)(α+1))Γ(α+1))/2^((α+1)/2) )  ⇒χ′(α)=((((√2))^(α+1) (π/4)cos((π/4)(α+1))Γ(α+1)+Γ′(α+1)((√2))^(α+1) sin((π/4)(α+1))−2^((a−1)/2) log(2)sin((π/4)(a+1))Γ(α+1))/2^(α+1) )  χ′(0)=(((π/4)−γ+((log(2))/2))/2)=(π/8)−(γ/2)−((log(2))/4)

χ(α)=0xαexsin(x)dxχ(0)=0log(x)exsin(x)dxχ(α)=12i0xαe(1i)xxαe(1+i)xdxχ(α)=12i.Γ(α+1)(1i)α+1Γ(α+1)2i(1+i)α+1χ(α)=12α+32i(eπ4(α+1)ieπ4(α+1)i)χ(α)=sin(π4(α+1))Γ(α+1)2α+12χ(α)=(2)α+1π4cos(π4(α+1))Γ(α+1)+Γ(α+1)(2)α+1sin(π4(α+1))2a12log(2)sin(π4(a+1))Γ(α+1)2α+1χ(0)=π4γ+log(2)22=π8γ2log(2)4

Commented by mnjuly1970 last updated on 28/Mar/21

   thank you so much..

thankyousomuch..

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