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Question Number 17514 by sma3l2996 last updated on 07/Jul/17

S(n)=∫_0 ^1 x^(2n) sin(2nπx)dx

$${S}\left({n}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}{n}} {sin}\left(\mathrm{2}{n}\pi{x}\right){dx} \\ $$

Commented by alex041103 last updated on 07/Jul/17

Is n a whole number?

$$\mathrm{Is}\:{n}\:\mathrm{a}\:\mathrm{whole}\:\mathrm{number}? \\ $$

Commented by sma3l2996 last updated on 07/Jul/17

No n∈N

$${No}\:{n}\in{N} \\ $$

Answered by alex041103 last updated on 08/Jul/17

For n∈N S(n) = Σ_(i=0) ^(n−1) ((1/(2πn)))^(2i+1) (((2n)!)/((2n−2i)!)) (−1)^(i+1) For n∈R S(n) = Σ_(i=0) ^∞ (((−1)^i )/(2(n+i+1)))(((2πn)^(2i+1) )/((2i+1)!)) Do you want the hole solution?

$$\mathrm{For}\:\mathrm{n}\in\mathbb{N} \\ $$$${S}\left({n}\right)\:=\:\underset{{i}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2}\pi{n}}\right)^{\mathrm{2}{i}+\mathrm{1}} \frac{\left(\mathrm{2}{n}\right)!}{\left(\mathrm{2}{n}−\mathrm{2}{i}\right)!}\:\:\left(−\mathrm{1}\right)^{{i}+\mathrm{1}} \\ $$$$ \\ $$$$\mathrm{For}\:\mathrm{n}\in\mathbb{R} \\ $$$${S}\left({n}\right)\:=\:\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\:\:\frac{\left(−\mathrm{1}\right)^{{i}} }{\mathrm{2}\left({n}+{i}+\mathrm{1}\right)}\frac{\left(\mathrm{2}\pi{n}\right)^{\mathrm{2}{i}+\mathrm{1}} }{\left(\mathrm{2}{i}+\mathrm{1}\right)!} \\ $$$$\mathrm{Do}\:\mathrm{you}\:\mathrm{want}\:\mathrm{the}\:\mathrm{hole}\:\mathrm{solution}? \\ $$

Commented by alex041103 last updated on 09/Jul/17

This is corrected and its the right answer

$$\mathrm{This}\:\mathrm{is}\:\mathrm{corrected}\:\mathrm{and}\:\mathrm{its}\:\mathrm{the}\:\mathrm{right}\:\mathrm{answer} \\ $$

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