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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=?
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:……{advanced}\:\:……\:\:\:{calculus}…. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\infty} {ln}\left({x}\right).{sin}\left({x}\right).{e}^{−{x}} {dx}=? \\ $$$$\:\:\:\:\:\:\:\:\:\:\: \\ $$
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)
$$\:\:\:\:\:\:\boldsymbol{\phi}=−{Im}\int_{\mathrm{0}} ^{\:\infty} {ln}\left({x}\right){e}^{−{ix}−{x}} {dx} \\ $$$$\:\:\:\:\:\:\:\:=−{Im}\int_{\mathrm{0}} ^{\:\infty} {ln}\left({x}\right){e}^{−{x}\left(\mathrm{1}+{i}\right)} {dx} \\ $$$$\:\:\:\:\:\:\:\:\overset{{x}\left(\mathrm{1}+{i}\right)={y}} {=}\:−\frac{\mathrm{1}}{\mathrm{2}}{Im}\left(\mathrm{1}−{i}\right)\int_{\mathrm{0}} ^{\:\infty} \left({ln}\left({y}\right)−{ln}\left(\sqrt{\mathrm{2}}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\right){e}^{−{y}} {dy} \\ $$$$=\frac{−\sqrt{\mathrm{2}}}{\mathrm{2}}{Im}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}−\frac{{i}\sqrt{\mathrm{2}}}{\mathrm{2}}\right)\left\{\int_{\mathrm{0}} ^{\:\infty} {ln}\left({y}\right){e}^{−{y}} {dy}+\left({ln}\left(\sqrt{\mathrm{2}}\:\right)+\frac{{i}\pi}{\mathrm{4}}\right)\left[{e}^{−{y}} \right]_{\mathrm{0}} ^{\infty} \right\} \\ $$$$=\frac{−\sqrt{\mathrm{2}}}{\mathrm{2}}\:{Im}\left\{\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}−\frac{{i}\sqrt{\mathrm{2}}}{\mathrm{2}}\right)\left(−\gamma−{ln}\left(\sqrt{\mathrm{2}}\:\right)−\frac{{i}\pi}{\mathrm{4}}\right)\right. \\ $$$$=\frac{−\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}.\left(\frac{−\pi}{\mathrm{4}}\right)+\left(\frac{\gamma+{ln}\left(\sqrt{\mathrm{2}}\:\right)}{\mathrm{2}}\right)\sqrt{\mathrm{2}}\:\right) \\ $$$$=\frac{\pi}{\mathrm{8}}−\frac{\gamma}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}{ln}\left(\mathrm{2}\right)\:…..\checkmark \\ $$$$\:\:\:\:\:\:\:\:\:\:\:……\boldsymbol{\phi}=\frac{\pi}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{4}}{ln}\left(\mathrm{2}\right)−\frac{\gamma}{\mathrm{2}} \\ $$
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)
$$\chi\left(\alpha\right)=\int_{\mathrm{0}} ^{\infty} {x}^{\alpha} {e}^{−{x}} {sin}\left({x}\right){dx} \\ $$$$\chi'\left(\mathrm{0}\right)=\int_{\mathrm{0}} ^{\infty} {log}\left({x}\right){e}^{−{x}} {sin}\left({x}\right){dx} \\ $$$$\Rightarrow\chi\left(\alpha\right)=\frac{\mathrm{1}}{\mathrm{2}{i}}\int_{\mathrm{0}} ^{\infty} {x}^{\alpha} {e}^{−\left(\mathrm{1}−{i}\right){x}} −{x}^{\alpha} {e}^{−\left(\mathrm{1}+{i}\right){x}} {dx} \\ $$$$\Rightarrow\chi\left(\alpha\right)=\frac{\mathrm{1}}{\mathrm{2}{i}}.\frac{\Gamma\left(\alpha+\mathrm{1}\right)}{\left(\mathrm{1}−{i}\right)^{\alpha+\mathrm{1}} }−\frac{\Gamma\left(\alpha+\mathrm{1}\right)}{\mathrm{2}{i}\left(\mathrm{1}+{i}\right)^{\alpha+\mathrm{1}} }\Rightarrow\chi\left(\alpha\right)=\frac{\mathrm{1}}{\mathrm{2}^{\frac{\alpha+\mathrm{3}}{\mathrm{2}}} {i}}\left({e}^{\frac{\pi}{\mathrm{4}}\left(\alpha+\mathrm{1}\right){i}} −{e}^{−\frac{\pi}{\mathrm{4}}\left(\alpha+\mathrm{1}\right){i}} \right) \\ $$$$\Rightarrow\chi\left(\alpha\right)=\frac{{sin}\left(\frac{\pi}{\mathrm{4}}\left(\alpha+\mathrm{1}\right)\right)\Gamma\left(\alpha+\mathrm{1}\right)}{\mathrm{2}^{\frac{\alpha+\mathrm{1}}{\mathrm{2}}} } \\ $$$$\Rightarrow\chi'\left(\alpha\right)=\frac{\left(\sqrt{\mathrm{2}}\right)^{\alpha+\mathrm{1}} \frac{\pi}{\mathrm{4}}{cos}\left(\frac{\pi}{\mathrm{4}}\left(\alpha+\mathrm{1}\right)\right)\Gamma\left(\alpha+\mathrm{1}\right)+\Gamma'\left(\alpha+\mathrm{1}\right)\left(\sqrt{\mathrm{2}}\right)^{\alpha+\mathrm{1}} {sin}\left(\frac{\pi}{\mathrm{4}}\left(\alpha+\mathrm{1}\right)\right)−\mathrm{2}^{\frac{{a}−\mathrm{1}}{\mathrm{2}}} {log}\left(\mathrm{2}\right){sin}\left(\frac{\pi}{\mathrm{4}}\left({a}+\mathrm{1}\right)\right)\Gamma\left(\alpha+\mathrm{1}\right)}{\mathrm{2}^{\alpha+\mathrm{1}} } \\ $$$$\chi'\left(\mathrm{0}\right)=\frac{\frac{\pi}{\mathrm{4}}−\gamma+\frac{{log}\left(\mathrm{2}\right)}{\mathrm{2}}}{\mathrm{2}}=\frac{\pi}{\mathrm{8}}−\frac{\gamma}{\mathrm{2}}−\frac{{log}\left(\mathrm{2}\right)}{\mathrm{4}} \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 28/Mar/21
thank you so much..
$$\:\:\:{thank}\:{you}\:{so}\:{much}.. \\ $$

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