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Question Number 93860 by Ar Brandon last updated on 15/May/20

$$\mathrm{Consider}\mathrm{the}\mathrm{progression}{\left({\mathrm{I}}_{\mathrm{n}}\right)}_{\mathrm{n}\in \mathbb{N}}\mathrm{where}{\mathrm{I}}_{\mathrm{n}}={\int }_0^1\frac{\sin \left(\pi \mathrm{t}\right)}{\mathrm{t}+\mathrm{n}}\mathrm{dt}$$$$1\mathrm{\setminus }\mathrm{Show}\mathrm{that}:\mathrm{\forall }\mathrm{n}⩾0,0⩽{\mathrm{I}}_{\mathrm{n}+1}⩽{\mathrm{I}}_{\mathrm{n}}\mathrm{and}\mathrm{deduce}\mathrm{that}\mathrm{the}$$$$\mathrm{series}\mathrm{is}\mathrm{convergent}.$$$$2\mathrm{\setminus }\mathrm{Show}\mathrm{that}0⩽{\mathrm{I}}_{\mathrm{n}}⩽\ln \left(\frac{\mathrm{n}+1}{\mathrm{n}}\right)\mathrm{and}\mathrm{deduce}\mathrm{the}\mathrm{limit}$$$$\mathrm{the}\mathrm{series}{\left({\mathrm{I}}_{\mathrm{n}}\right)}_{\mathrm{n}\in \mathbb{N}}$$$$3\mathrm{\setminus }\mathrm{Calculate}\underset{\mathrm{n}\to +\mathrm{\infty }}{\lim }\left({\mathrm{nI}}_{\mathrm{n}}\right)$$

Commented by mathmax by abdo last updated on 15/May/20

$$1)I_n={\int }_0^1\frac{sin\left(\pi t\right)}{t+n}dtitsclearthat{I}_n⩾0$$$$wehave{I}_n-I_{n+1}={\int }_0^1\frac{sin\left(\pi t\right)}{t+n}dt-{\int }_0^1\frac{sin\left(\pi t\right)}{t+n+1}dt$$$$={\int }_0^{1}sin\left(\pi t\right)(\frac{1}{t+n}-\frac{1}{t+n+1})dt={\int }_0^1\frac{sin\left(\pi t\right)}{(t+n)(t+n+1)}dt⩾0\Rightarrow$$$$(sin\left(\pi t\right)⩾0on[0,1])I_{n}isdecrasingandminoredby0\Rightarrow$$$$I_{n}isconvergent.$$

Commented by mathmax by abdo last updated on 15/May/20

$$2)wehave{I}_n={\int }_0^1\frac{sin\left(\pi t\right)}{t+n}dt⩽{\int }_0^1\frac{dt}{t+n}={\left[ln(t+n)\right]}_0^1=ln\left(\frac{n+1}{n}\right)\Rightarrow$$$$0⩽I_n⩽ln\left(\frac{n+1}{n}\right)wehaveln\left(\frac{n+1}{n}\right)=ln(1+\frac{1}{n})\to 0(n\to +\mathrm{\infty })\Rightarrow$$$${lim}_{n\to +\mathrm{\infty }}{I}_n=0$$

Commented by mathmax by abdo last updated on 15/May/20

$$3)wehave{I}_n={\int }_0^1\frac{sin\left(\pi t\right)}{t+n}dt{=}_{\pi t=x}{\int }_0^{\pi }\frac{sin\left(x\right)}{\frac{x}{\pi }+n}\frac{dx}{\pi }$$$$={\int }_0^{\pi }\frac{sinx}{x+n\pi }dx=\frac{1}{n}{\int }_0^{\pi }\frac{sin\left(x\right)}{\frac{x}{n}+\pi }dx\Rightarrow {nI}_n={\int }_0^{\pi }\frac{sinx}{\frac{x}{n}+\pi }dx\Rightarrow$$$${lim}_{n\to +\mathrm{\infty }}{nI}_n=\frac{1}{\pi }{\int }_0^{\pi }sinxdx=\frac{1}{\pi }{[-cosx]}_0^{\pi }=\frac{2}{\pi }$$

Commented by Ar Brandon last updated on 15/May/20

Great job, thanks ��

Commented by mathmax by abdo last updated on 16/May/20

$$youarewelcomesir.$$

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