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Question Number 147344 by mathdanisur last updated on 20/Jul/21

$$\underset{n\to \mathrm{\infty }}{lim}\underset{0}{\stackrel{1}{\int }}log\left(\frac{1+{sin}^{\mathit{n}}\mathit{x}}{1+x+x^{\mathit{n}}}\right)dx=?$$

Answered by mathmax by abdo last updated on 20/Jul/21

$${\mathrm{f}}_{\mathrm{n}}\left(\mathrm{x}\right)=\log \left(\frac{1+{\sin }^{\mathrm{n}}\mathrm{x}}{1+\mathrm{x}+{\mathrm{x}}^{\mathrm{n}}}\right)\to \mathrm{cs}\mathrm{to}\mathrm{f}\left(\mathrm{x}\right)=\log \left(\frac{1}{1+\mathrm{x}}\right)=-\log (1+\mathrm{x})\Rightarrow$$$${\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\int }_0^1{\mathrm{f}}_{\mathrm{n}}\left(\mathrm{x}\right)\mathrm{dx}={\int }_0^1{\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{f}}_{\mathrm{n}}\left(\mathrm{x}\right)\mathrm{dx}=-{\int }_0^1\log (1+\mathrm{x})\mathrm{dx}$$$${=}_{1+\mathrm{x}=\mathrm{t}}-{\int }_1^2\log \left(\mathrm{t}\right)\mathrm{dt}=-{[\mathrm{tlot}-\mathrm{t}]}_1^2=-(2\log 2-2+1)$$$$=1-2\log 2$$

Commented by mathdanisur last updated on 20/Jul/21

$$thankyouSer$$

Commented by mathmax by abdo last updated on 20/Jul/21

$$\mathrm{you}\mathrm{are}\mathrm{welcome}\mathrm{sir}$$

Commented by mathdanisur last updated on 20/Jul/21

$$ThanksSer,buthowdidyouchange$$$$\left(interchangeof\right)thelimittothe$$$$integral,pleasenoteSer...$$

Commented by mathmax by abdo last updated on 21/Jul/21

$$\mathrm{theoreme}\mathrm{de}\mathrm{convergence}\mathrm{dominee}...$$

Commented by mathdanisur last updated on 21/Jul/21

$$Ser,canyoustatethattheorem,please$$

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