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Question Number 60621 by fjdjdcjv last updated on 22/May/19

if π is rational then there exists a I_n =(v^(2n) /(n!))∫_0 ^π x^n (x−π)^n sin(x)dx can someone give a easier way to expaned this

$${if}\:\pi\:{is}\:{rational}\:{then}\:{there} \\ $$$${exists}\:{a}\:{I}_{{n}} =\frac{{v}^{\mathrm{2}{n}} }{{n}!}\underset{\mathrm{0}} {\overset{\pi} {\int}}{x}^{{n}} \left({x}−\pi\right)^{{n}} {sin}\left({x}\right){dx} \\ $$$${can}\:{someone}\:{give}\:{a}\:{easier}\:{way}\:{to}\:{expaned}\:{this} \\ $$

Commented by MJS last updated on 23/May/19

I don′t understand. we can solve the integral for any given n∈N even if π∉Q. and what is v?

$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{understand}.\:\mathrm{we}\:\mathrm{can}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{integral} \\ $$$$\mathrm{for}\:\mathrm{any}\:\mathrm{given}\:{n}\in\mathbb{N}\:\mathrm{even}\:\mathrm{if}\:\pi\notin\mathbb{Q}.\:\mathrm{and}\:\mathrm{what}\:\mathrm{is}\:{v}? \\ $$

Commented by fjdjdcjv last updated on 23/May/19

π=(u/v)

$$\pi=\frac{{u}}{{v}} \\ $$

Commented by MJS last updated on 23/May/19

ok. still the integral exists, quite apart from the nature of π another question: sin x with which period, 2π or 2(u/v)? is it (v^(2n) /(n!))∫_0 ^(u/v) x^n (x−(u/v))sin ((vπx)/u) dx? or shall we calculate I_n using the usual π and then show that π∉Q?

$$\mathrm{ok}.\:\mathrm{still}\:\mathrm{the}\:\mathrm{integral}\:\mathrm{exists},\:\mathrm{quite}\:\mathrm{apart}\:\mathrm{from} \\ $$$$\mathrm{the}\:\mathrm{nature}\:\mathrm{of}\:\pi \\ $$$$\mathrm{another}\:\mathrm{question}:\:\mathrm{sin}\:{x}\:\mathrm{with}\:\mathrm{which}\:\mathrm{period}, \\ $$$$\mathrm{2}\pi\:\mathrm{or}\:\mathrm{2}\frac{{u}}{{v}}? \\ $$$$\mathrm{is}\:\mathrm{it}\:\frac{{v}^{\mathrm{2}{n}} }{{n}!}\underset{\mathrm{0}} {\overset{\frac{{u}}{{v}}} {\int}}{x}^{{n}} \left({x}−\frac{{u}}{{v}}\right)\mathrm{sin}\:\frac{{v}\pi{x}}{{u}}\:{dx}? \\ $$$$\mathrm{or}\:\mathrm{shall}\:\mathrm{we}\:\mathrm{calculate}\:{I}_{{n}} \:\mathrm{using}\:\mathrm{the}\:\mathrm{usual}\:\pi \\ $$$$\mathrm{and}\:\mathrm{then}\:\mathrm{show}\:\mathrm{that}\:\pi\notin\mathbb{Q}? \\ $$

Commented by MJS last updated on 23/May/19

sorry this is quite confusing me

$$\mathrm{sorry}\:\mathrm{this}\:\mathrm{is}\:\mathrm{quite}\:\mathrm{confusing}\:\mathrm{me} \\ $$

Commented by MJS last updated on 23/May/19

I_1 =−4v^2 I_2 =2v^4 (−π^2 +12) I_3 =24v^6 (π^2 −10) I_4 =2v^8 (π^4 −180π^2 +1680) I_5 =60v^(10) (−π^4 +112π^2 −1008) ... all calculated with the usual π if π=(u/v) I_1 =−4v^2 I_2 =−2(u^2 −12v^2 )v^2 I_3 =24(u^2 −10v^2 )v^4 I_4 =2(u^4 −180u^2 v^2 +1680v^4 )v^4 I_5 =−60(u^4 −112u^2 v^2 +1008v^4 )v^6 ... but what do we know now?

$${I}_{\mathrm{1}} =−\mathrm{4}{v}^{\mathrm{2}} \\ $$$${I}_{\mathrm{2}} =\mathrm{2}{v}^{\mathrm{4}} \left(−\pi^{\mathrm{2}} +\mathrm{12}\right) \\ $$$${I}_{\mathrm{3}} =\mathrm{24}{v}^{\mathrm{6}} \left(\pi^{\mathrm{2}} −\mathrm{10}\right) \\ $$$${I}_{\mathrm{4}} =\mathrm{2}{v}^{\mathrm{8}} \left(\pi^{\mathrm{4}} −\mathrm{180}\pi^{\mathrm{2}} +\mathrm{1680}\right) \\ $$$${I}_{\mathrm{5}} =\mathrm{60}{v}^{\mathrm{10}} \left(−\pi^{\mathrm{4}} +\mathrm{112}\pi^{\mathrm{2}} −\mathrm{1008}\right) \\ $$$$... \\ $$$$\mathrm{all}\:\mathrm{calculated}\:\mathrm{with}\:\mathrm{the}\:\mathrm{usual}\:\pi \\ $$$$\mathrm{if}\:\pi=\frac{{u}}{{v}} \\ $$$${I}_{\mathrm{1}} =−\mathrm{4}{v}^{\mathrm{2}} \\ $$$${I}_{\mathrm{2}} =−\mathrm{2}\left({u}^{\mathrm{2}} −\mathrm{12}{v}^{\mathrm{2}} \right){v}^{\mathrm{2}} \\ $$$${I}_{\mathrm{3}} =\mathrm{24}\left({u}^{\mathrm{2}} −\mathrm{10}{v}^{\mathrm{2}} \right){v}^{\mathrm{4}} \\ $$$${I}_{\mathrm{4}} =\mathrm{2}\left({u}^{\mathrm{4}} −\mathrm{180}{u}^{\mathrm{2}} {v}^{\mathrm{2}} +\mathrm{1680}{v}^{\mathrm{4}} \right){v}^{\mathrm{4}} \\ $$$${I}_{\mathrm{5}} =−\mathrm{60}\left({u}^{\mathrm{4}} −\mathrm{112}{u}^{\mathrm{2}} {v}^{\mathrm{2}} +\mathrm{1008}{v}^{\mathrm{4}} \right){v}^{\mathrm{6}} \\ $$$$... \\ $$$$\mathrm{but}\:\mathrm{what}\:\mathrm{do}\:\mathrm{we}\:\mathrm{know}\:\mathrm{now}? \\ $$

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