Question Number 136999 by mnjuly1970 last updated on 28/Mar/21
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:……{advanced}\:\:……\:\:\:{calculus}…. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\infty} {ln}\left({x}\right).{sin}\left({x}\right).{e}^{−{x}} {dx}=? \\ $$$$\:\:\:\:\:\:\:\:\:\:\: \\ $$
Answered by mnjuly1970 last updated on 28/Mar/21
$$\:\:\:\:\:\:\boldsymbol{\phi}=−{Im}\int_{\mathrm{0}} ^{\:\infty} {ln}\left({x}\right){e}^{−{ix}−{x}} {dx} \\ $$$$\:\:\:\:\:\:\:\:=−{Im}\int_{\mathrm{0}} ^{\:\infty} {ln}\left({x}\right){e}^{−{x}\left(\mathrm{1}+{i}\right)} {dx} \\ $$$$\:\:\:\:\:\:\:\:\overset{{x}\left(\mathrm{1}+{i}\right)={y}} {=}\:−\frac{\mathrm{1}}{\mathrm{2}}{Im}\left(\mathrm{1}−{i}\right)\int_{\mathrm{0}} ^{\:\infty} \left({ln}\left({y}\right)−{ln}\left(\sqrt{\mathrm{2}}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\right){e}^{−{y}} {dy} \\ $$$$=\frac{−\sqrt{\mathrm{2}}}{\mathrm{2}}{Im}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}−\frac{{i}\sqrt{\mathrm{2}}}{\mathrm{2}}\right)\left\{\int_{\mathrm{0}} ^{\:\infty} {ln}\left({y}\right){e}^{−{y}} {dy}+\left({ln}\left(\sqrt{\mathrm{2}}\:\right)+\frac{{i}\pi}{\mathrm{4}}\right)\left[{e}^{−{y}} \right]_{\mathrm{0}} ^{\infty} \right\} \\ $$$$=\frac{−\sqrt{\mathrm{2}}}{\mathrm{2}}\:{Im}\left\{\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}−\frac{{i}\sqrt{\mathrm{2}}}{\mathrm{2}}\right)\left(−\gamma−{ln}\left(\sqrt{\mathrm{2}}\:\right)−\frac{{i}\pi}{\mathrm{4}}\right)\right. \\ $$$$=\frac{−\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}.\left(\frac{−\pi}{\mathrm{4}}\right)+\left(\frac{\gamma+{ln}\left(\sqrt{\mathrm{2}}\:\right)}{\mathrm{2}}\right)\sqrt{\mathrm{2}}\:\right) \\ $$$$=\frac{\pi}{\mathrm{8}}−\frac{\gamma}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}{ln}\left(\mathrm{2}\right)\:…..\checkmark \\ $$$$\:\:\:\:\:\:\:\:\:\:\:……\boldsymbol{\phi}=\frac{\pi}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{4}}{ln}\left(\mathrm{2}\right)−\frac{\gamma}{\mathrm{2}} \\ $$
Answered by Dwaipayan Shikari last updated on 28/Mar/21
$$\chi\left(\alpha\right)=\int_{\mathrm{0}} ^{\infty} {x}^{\alpha} {e}^{−{x}} {sin}\left({x}\right){dx} \\ $$$$\chi'\left(\mathrm{0}\right)=\int_{\mathrm{0}} ^{\infty} {log}\left({x}\right){e}^{−{x}} {sin}\left({x}\right){dx} \\ $$$$\Rightarrow\chi\left(\alpha\right)=\frac{\mathrm{1}}{\mathrm{2}{i}}\int_{\mathrm{0}} ^{\infty} {x}^{\alpha} {e}^{−\left(\mathrm{1}−{i}\right){x}} −{x}^{\alpha} {e}^{−\left(\mathrm{1}+{i}\right){x}} {dx} \\ $$$$\Rightarrow\chi\left(\alpha\right)=\frac{\mathrm{1}}{\mathrm{2}{i}}.\frac{\Gamma\left(\alpha+\mathrm{1}\right)}{\left(\mathrm{1}−{i}\right)^{\alpha+\mathrm{1}} }−\frac{\Gamma\left(\alpha+\mathrm{1}\right)}{\mathrm{2}{i}\left(\mathrm{1}+{i}\right)^{\alpha+\mathrm{1}} }\Rightarrow\chi\left(\alpha\right)=\frac{\mathrm{1}}{\mathrm{2}^{\frac{\alpha+\mathrm{3}}{\mathrm{2}}} {i}}\left({e}^{\frac{\pi}{\mathrm{4}}\left(\alpha+\mathrm{1}\right){i}} −{e}^{−\frac{\pi}{\mathrm{4}}\left(\alpha+\mathrm{1}\right){i}} \right) \\ $$$$\Rightarrow\chi\left(\alpha\right)=\frac{{sin}\left(\frac{\pi}{\mathrm{4}}\left(\alpha+\mathrm{1}\right)\right)\Gamma\left(\alpha+\mathrm{1}\right)}{\mathrm{2}^{\frac{\alpha+\mathrm{1}}{\mathrm{2}}} } \\ $$$$\Rightarrow\chi'\left(\alpha\right)=\frac{\left(\sqrt{\mathrm{2}}\right)^{\alpha+\mathrm{1}} \frac{\pi}{\mathrm{4}}{cos}\left(\frac{\pi}{\mathrm{4}}\left(\alpha+\mathrm{1}\right)\right)\Gamma\left(\alpha+\mathrm{1}\right)+\Gamma'\left(\alpha+\mathrm{1}\right)\left(\sqrt{\mathrm{2}}\right)^{\alpha+\mathrm{1}} {sin}\left(\frac{\pi}{\mathrm{4}}\left(\alpha+\mathrm{1}\right)\right)−\mathrm{2}^{\frac{{a}−\mathrm{1}}{\mathrm{2}}} {log}\left(\mathrm{2}\right){sin}\left(\frac{\pi}{\mathrm{4}}\left({a}+\mathrm{1}\right)\right)\Gamma\left(\alpha+\mathrm{1}\right)}{\mathrm{2}^{\alpha+\mathrm{1}} } \\ $$$$\chi'\left(\mathrm{0}\right)=\frac{\frac{\pi}{\mathrm{4}}−\gamma+\frac{{log}\left(\mathrm{2}\right)}{\mathrm{2}}}{\mathrm{2}}=\frac{\pi}{\mathrm{8}}−\frac{\gamma}{\mathrm{2}}−\frac{{log}\left(\mathrm{2}\right)}{\mathrm{4}} \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 28/Mar/21
$$\:\:\:{thank}\:{you}\:{so}\:{much}.. \\ $$