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Question Number 148376 by ArielVyny last updated on 27/Jul/21

$${\int }_1^{\mathrm{\infty }}{x}^{i}lnxdx{i}^2=-1$$

Answered by mathmax by abdo last updated on 27/Jul/21

$$\mathrm{I}={\int }_1^{\mathrm{\infty }}{\mathrm{x}}^{\mathrm{i}}\mathrm{logx}\mathrm{dx}\mathrm{cha}7\mathrm{gement}\mathrm{logx}=\mathrm{t}\mathrm{give}\mathrm{x}={\mathrm{e}}^{\mathrm{t}}\Rightarrow$$$$\mathrm{I}={\int }_0^{\mathrm{\infty }}{\left({\mathrm{e}}^{\mathrm{t}}\right)}^{\mathrm{i}}\mathrm{t}{\mathrm{e}}^{\mathrm{t}}\mathrm{dt}={\int }_0^{\mathrm{\infty }}\mathrm{t}{\mathrm{e}}^{\mathrm{it}+\mathrm{t}}\mathrm{dt}={\int }_0^{\mathrm{\infty }}\mathrm{t}{\mathrm{e}}^{(1+\mathrm{i})\mathrm{t}}\mathrm{dt}$$$${=}_{(1+\mathrm{i})\mathrm{t}=\mathrm{z}}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{z}}{1+\mathrm{i}}{\mathrm{e}}^{\mathrm{z}}\frac{\mathrm{dz}}{1+\mathrm{i}}=\frac{1}{{(1+\mathrm{i})}^2}{\int }_0^{\mathrm{\infty }}{\mathrm{ze}}^{\mathrm{z}}\mathrm{dz}\mathrm{by}\mathrm{parts}$$$$\mathrm{but}{\int }_0^{\mathrm{\infty }}{\mathrm{ze}}^{\mathrm{z}}\mathrm{dz}\mathrm{is}\mathrm{divergent}...!$$

Commented by ArielVyny last updated on 28/Jul/21

$$thanksir$$

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