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


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

$i f π i s r a t i o n a l t h e n t h e r e$ $e x i s t s a I_{n} = \frac{v^{2 n}}{n!} \underset{0}{\int^{π}} x^{n} {(x - π)}^{n} s i n (x) d x$ $c a n s o m e o n e g i v e a e a s i e r w a y t o e x p a n e d t h i s$
Commented by MJS last updated on 23/May/19

$I {don}^{'} t understand . we can solve the integral$ $for any given n \in N even if π \notin Q . and what is v ?$
Commented by fjdjdcjv last updated on 23/May/19

$π = \frac{u}{v}$
Commented by MJS last updated on 23/May/19

$ok . still the integral exists, quite apart from$ $the nature of π$ $another question : \sin x with which period,$ $2 π or 2 \frac{u}{v} ?$ $is it \frac{v^{2 n}}{n!} \underset{0}{\int^{\frac{u}{v}}} x^{n} (x - \frac{u}{v}) \sin \frac{v π x}{u} d x ?$ $or shall we calculate I_{n} using the usual π$ $and then show that π \notin Q ?$
Commented by MJS last updated on 23/May/19

$sorry this is quite confusing me$
Commented by MJS last updated on 23/May/19

$I_{1} = - 4 v^{2}$ $I_{2} = 2 v^{4} (- π^{2} + 12)$ $I_{3} = 24 v^{6} (π^{2} - 10)$ $I_{4} = 2 v^{8} (π^{4} - 180 π^{2} + 1680)$ $I_{5} = 60 v^{10} (- π^{4} + 112 π^{2} - 1008)$ $. . .$ $all calculated with the usual π$ $if π = \frac{u}{v}$ $I_{1} = - 4 v^{2}$ $I_{2} = - 2 (u^{2} - 12 v^{2}) v^{2}$ $I_{3} = 24 (u^{2} - 10 v^{2}) v^{4}$ $I_{4} = 2 (u^{4} - 180 u^{2} v^{2} + 1680 v^{4}) v^{4}$ $I_{5} = - 60 (u^{4} - 112 u^{2} v^{2} + 1008 v^{4}) v^{6}$ $. . .$ $but what do we know now ?$
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