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Question Number 146057 by mnjuly1970 last updated on 10/Jul/21

Answered by mathmax by abdo last updated on 10/Jul/21

$$\mathrm{J}=\mathrm{Im}\left({\int }_0^{\mathrm{\infty }}{\mathrm{x}}^{-\frac{1}{2}}{\mathrm{e}}^{-\mathrm{x}+\mathrm{ix}}\mathrm{dx}\right)\mathrm{we}\mathrm{have}$$$${\int }_0^{\mathrm{\infty }}{\mathrm{x}}^{-\frac{1}{2}}{\mathrm{e}}^{(-1+\mathrm{i})\mathrm{x}}\mathrm{dx}{=}_{(1-\mathrm{i})\mathrm{x}=\mathrm{t}}{\int }_0^{\mathrm{\infty }}{\left(\frac{\mathrm{t}}{1-\mathrm{i}}\right)}^{-\frac{1}{2}}{\mathrm{e}}^{-\mathrm{t}}\frac{\mathrm{dt}}{1-\mathrm{i}}$$$$=\frac{1}{{(1-\mathrm{i})}^{\frac{1}{2}}}{\int }_0^{\mathrm{\infty }}{\mathrm{t}}^{-\frac{1}{2}}{\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}=\frac{1}{{\left(\sqrt{2}{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}}\right)}^{\frac{1}{2}}}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{\frac{1}{2}-1}{\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}$$$$=\mathrm{\Gamma }\left(\frac{1}{2}\right).\frac{1}{\left({}^4\sqrt{2}\right)}{\mathrm{e}}^{\frac{\mathrm{i}\pi }{8}}=\frac{\sqrt{\pi }}{\left({}^4\sqrt{2}\right)}\{\cos \left(\frac{\pi }{8}\right)+\mathrm{isin}\left(\frac{\pi }{8}\right)\}\Rightarrow$$$$\mathrm{J}=\frac{\sqrt{\pi }}{\left({}^4\sqrt{2}\right)}\sin \left(\frac{\pi }{8}\right)\Rightarrow \lambda =\frac{\sqrt{\pi }}{\left({}^4\sqrt{2}\right)}$$

Commented by SANOGO last updated on 30/Aug/21

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