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Question Number 72046 by aliesam last updated on 23/Oct/19

Commented by mathmax by abdo last updated on 26/Oct/19

$$let{U}_n={\int }_0^{1}nln(1+{\left(\frac{x}{n}\right)}^{\alpha })dx\Rightarrow U_n{=}_{\frac{x}{n}=t}{\int }_0^{\frac{1}{n}}nln(1+t^{\alpha })ndt$$$$=n^2{\int }_0^{\frac{1}{n}}ln(1+t^{\alpha })dtbut0<t<\frac{1}{n}andn\to +\mathrm{\infty }ln(1+t^{\alpha })\sim t^{\alpha }\Rightarrow$$$$U_n\sim n^2{\int }_0^{\frac{1}{n}}{t}^{\alpha }dt=n^2{\left[\frac{1}{\alpha +1}{t}^{\alpha +1}\right]}_0^{\frac{1}{n}}=\frac{n^2}{\alpha +1}{\left(\frac{1}{n}\right)}^{\alpha +1}=\frac{1}{(\alpha +1)n^{\alpha +1-2}}$$$$=\frac{1}{(\alpha +1)n^{\alpha -1}}=\frac{n^{1-\alpha }}{(\alpha +1)}\to +\mathrm{\infty }because1-\alpha >0$$$$anotherway{U}_n={\int }_{R}nln(1+{\left(\frac{x}{n}\right)}^{\alpha }){\chi }_{[0,1]}\left(x\right)dx={\int }_R{f}_n\left(x\right)dx$$$$wehave{f}_n\left(x\right)\sim n{\left(\frac{x}{n}\right)}^{\alpha }=\frac{{nx}^n}{n^n}=\frac{x^{\alpha }}{n^{\alpha -1}}=n^{1-\alpha }{x}^{\alpha }but1-\alpha >0\Rightarrow$$$$f_n\left(x\right)\to +\mathrm{\infty }(n\to +\mathrm{\infty })\Rightarrow {lim}_{n\to +\mathrm{\infty }}{U}_n=+\mathrm{\infty }$$$$$$

Commented by mathmax by abdo last updated on 26/Oct/19

$$forgive{f}_n\left(x\right)\sim \frac{{nx}^{\alpha }}{n^{\alpha }}$$

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