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Question Number 104871 by I want to learn more last updated on 24/Jul/20

Answered by mathmax by abdo last updated on 24/Jul/20

$${\mathrm{A}}_{\mathrm{p}}={\int }_0^1\frac{{\tan }^2\mathrm{x}}{{\mathrm{x}}^{\mathrm{p}}}\mathrm{dx}\Rightarrow {\mathrm{A}}_{\mathrm{p}}={\int }_0^1{\left(\frac{\mathrm{tanx}}{\mathrm{x}}\right)}^2\times \frac{\mathrm{dx}}{{\mathrm{x}}^{\mathrm{p}-2}}$$$$\mathrm{t}\mathrm{he}\mathrm{function}\mathrm{x}\to {\left(\frac{\mathrm{tanx}}{\mathrm{x}}\right)}^2\times \frac{1}{{\mathrm{x}}^{\mathrm{p}-2}}\mathrm{is}\mathrm{continue}\mathrm{on}]0,1]\mathrm{so}\mathrm{integrable}$$$$\mathrm{at}\mathrm{V}\left(\mathrm{o}\right)\mathrm{f}\left(\mathrm{x}\right)\sim \frac{1}{{\mathrm{x}}^{\mathrm{p}-2}}\mathrm{and}\int \frac{\mathrm{dx}}{{\mathrm{x}}^{\mathrm{p}-2}}\mathrm{converge}\iff \mathrm{p}-2<1\iff \mathrm{p}<3$$

Commented by I want to learn more last updated on 24/Jul/20

$$\mathrm{Thanks}\mathrm{sir}.\mathrm{I}\mathrm{appreciate}.$$

Commented by mathmax by abdo last updated on 24/Jul/20

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

Answered by mathmax by abdo last updated on 24/Jul/20

$$4)\mathrm{let}\mathrm{S}=\sum _{\mathrm{n}=2}^{\mathrm{\infty }}\mathrm{n}{\mathrm{x}}^{\mathrm{n}-2}\Rightarrow \mathrm{S}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}(\mathrm{n}+2){\mathrm{x}}^{\mathrm{n}}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{nx}}^{\mathrm{n}}+2\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{x}}^{\mathrm{n}}$$$$\mathrm{for}\mid \mathrm{x}\mid <1\mathrm{we}\mathrm{get}\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{x}}^{\mathrm{n}}=\frac{1}{1-\mathrm{x}}\Rightarrow \sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{nx}}^{\mathrm{n}-1}=\frac{1}{{(1-\mathrm{x})}^2}\Rightarrow$$$$\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{nx}}^{\mathrm{n}}=\frac{\mathrm{x}}{{(1-\mathrm{x})}^2}\Rightarrow \mathrm{S}=\frac{\mathrm{x}}{{(1-\mathrm{x})}^2}+\frac{2}{1-\mathrm{x}}=\frac{\mathrm{x}+2(1-\mathrm{x})}{{(1-\mathrm{x})}^2}=\frac{\mathrm{x}+2-2\mathrm{x}}{{(1-\mathrm{x})}^2}\Rightarrow$$$$\mathrm{S}=\frac{2-\mathrm{x}}{{(1-\mathrm{x})}^2}.$$

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