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Question Number 66620 by Mohamed Amine Bouguezzoul last updated on 18/Aug/19

$$find\underset{n\to \mathrm{\infty }}{\lim }{I}_n$$$$I_n={\int }_0^{\mathrm{\infty }}\frac{dx}{{(1+\coth \left(nx\right))}^n},n⩾1$$$$$$

Commented by mathmax by abdo last updated on 19/Aug/19

$$coth\left(nx\right)=\frac{ch\left(nx\right)}{sh\left(nx\right)}=\frac{e^{nx}+e^{-nx}}{e^{nx}-e^{-nx}}\Rightarrow 1+coth\left(nx\right)=1+\frac{e^{nx}+e^{-nx}}{e^{nx}-e^{-nx}}$$$$=\frac{e^{nx}-e^{-nx}+e^{nx}+e^{-nx}}{e^{nx}-e^{-nx}}=\frac{2e^{nx}}{e^{nx}(1-e^{-2nx})}=\frac{2}{1-e^{-2nx}}\Rightarrow$$$$I_n={\int }_0^{\mathrm{\infty }}\frac{dx}{{\left(\frac{2}{1-e^{-2nx}}\right)}^n}={\int }_0^{\mathrm{\infty }}\frac{1}{2^n}{(1-e^{-nx})}^{n}dx$$$$={\int }_0^{\mathrm{\infty }}\frac{1}{2^n}{e}^{nln(1-e^{-nx})}dx={\int }_0^{\mathrm{\infty }}{f}_n\left(x\right)dxwith{f}_n\left(x\right)=\frac{1}{2^n}{e}^{nln(1-e^{-nx})}$$$$wehaveln(1-e^{-nx})\sim -e^{-nx}andnln(1-e^{-nx})\sim -{ne}^{-nx}\Rightarrow$$$$f_n\left(x\right)\sim \frac{1}{2^n}{e}^{-{ne}^{-nx}}\to 0(n\to +\mathrm{\infty })thesequences\left(f_n\right)arecontinues$$$${lim}_{n\to +\mathrm{\infty }}{I}_n={\int }_{R^{+}}{lim}_{n\to +\mathrm{\infty }}{f}_n\left(x\right)=0$$

Commented by Mohamed Amine Bouguezzoul last updated on 20/Aug/19

$$excellentworkabdo.$$

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

$$youarewelcome.$$

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