Question Number 146057 by mnjuly1970 last updated on 10/Jul/21
Answered by mathmax by abdo last updated on 10/Jul/21
$$\mathrm{J}=\mathrm{Im}\left(\int_{\mathrm{0}} ^{\infty} \:\mathrm{x}^{−\frac{\mathrm{1}}{\mathrm{2}}} \:\mathrm{e}^{−\mathrm{x}+\mathrm{ix}} \mathrm{dx}\right)\:\:\mathrm{we}\:\mathrm{have} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\mathrm{x}^{−\frac{\mathrm{1}}{\mathrm{2}}} \:\mathrm{e}^{\left(−\mathrm{1}+\mathrm{i}\right)\mathrm{x}} \:\mathrm{dx}\:=_{\left(\mathrm{1}−\mathrm{i}\right)\mathrm{x}=\mathrm{t}} \:\:\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{t}}{\mathrm{1}−\mathrm{i}}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \:\mathrm{e}^{−\mathrm{t}} \frac{\mathrm{dt}}{\mathrm{1}−\mathrm{i}} \\ $$$$=\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{i}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} }\int_{\mathrm{0}} ^{\infty} \:\mathrm{t}^{−\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{e}^{−\mathrm{t}} \:\mathrm{dt}\:=\frac{\mathrm{1}}{\left(\sqrt{\mathrm{2}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} }\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} \:\mathrm{e}^{−\mathrm{t}} \:\mathrm{dt} \\ $$$$=\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right).\frac{\mathrm{1}}{\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right)}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{8}}} \:=\frac{\sqrt{\pi}}{\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right)}\left\{\mathrm{cos}\left(\frac{\pi}{\mathrm{8}}\right)+\mathrm{isin}\left(\frac{\pi}{\mathrm{8}}\right)\right\}\:\Rightarrow \\ $$$$\mathrm{J}=\frac{\sqrt{\pi}}{\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right)}\mathrm{sin}\left(\frac{\pi}{\mathrm{8}}\right)\:\Rightarrow\lambda=\frac{\sqrt{\pi}}{\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right)} \\ $$
Commented by SANOGO last updated on 30/Aug/21
