find-and-from-R-0-pi-t-2-t-cos-nt-dt-1-n-2-for-all-number-n-from-N-then-find-n-1-1-n-2-

Question Number 27495 by abdo imad last updated on 07/Jan/18

f i n d α a n d β f r o m R / \int_{0}^{π} (α t^{2} + β t) c o s (n t) d t = \frac{1}{n^{2}}

f o r a l l n u m b e r n f r o m N^{*} t h e n f i n d

\sum_{n = 1}^{\propto} \frac{1}{n^{2}} .

Commented by abdo imad last updated on 19/Jan/18

l e t p u t I = \int_{0}^{π} (α t^{2} + β t) c o s (n t) d t

I = R e (\int_{0}^{π} {(α t^{2} + β t)}^{} e^{i n t} t) b y p a r t s

u = α t^{2} + β t a n d v^{^{'}} = e^{i n t}

I = {[(α t^{2} + β t) \frac{e^{i n t}}{i n}]}_{0}^{π} - \frac{1}{i n} \int_{0}^{π} (2 α t + β) e^{i n t} d t

= \frac{1}{i n} ((α π^{2} + β π) {(- 1)}^{n}) - \frac{1}{i n} {{[\frac{1}{i n} (2 α t + β) e^{i n t}]}_{0}^{π} - \frac{1}{i n} \int_{0}^{π} (2 α) e^{i n t} d t}

w e f i n d I = \frac{1}{n^{2}} ((2 α π + β) {(- 1)}^{n} - β)

I = \frac{1}{n^{2}} \forall n \in N^{*} \Leftrightarrow (2 α π + β) {(- 1)}^{n} - β = 1 f o r a l l n

\Leftrightarrow β = - 1 a n d 2 α π - 1 = 0 \Rightarrow α = \frac{1}{2 π} a n d β = - 1 s o

\frac{1}{n^{2}} = \int_{0}^{π} (\frac{1}{2 π} t^{2} - t) c o s (n t) d t

\sum_{n = 1}^{\propto} \frac{1}{n^{2}} = \int_{0}^{π} (\frac{1}{2 π} t^{2} - t) (\sum_{n = 1}^{\propto} c o s (n t)) d t b u t

\sum_{n = 1}^{\propto} c o s (n t) = R e (\sum_{n = 0}^{\propto} {(^{} e^{i t})}^{n}) - 1 = R e (\frac{1}{1 - e^{i t}}) - 1

= R e (\frac{1}{2 {s i n}^{2} (\frac{t}{2}) - 2 i s i n (\frac{t}{2}) c o s (\frac{t}{2})})

= R e (\frac{1}{- 2 i s i n (\frac{t}{2}) e^{i \frac{t}{2}}}) = R e (\frac{1}{2} i \frac{c o s (\frac{t}{2}) - i s i n (\frac{t}{2})}{s i n (\frac{t}{2})})

= \frac{1}{2} - 1 = - \frac{1}{2}

\sum_{n = 1}^{\propto} \frac{1}{n^{2}} = - \frac{1}{2} \int_{0}^{π} (\frac{1}{2 π} t^{2} - t) d t

= - \frac{1}{4 π} {[\frac{t^{3}}{3}]}_{0}^{π} + \frac{1}{2} {[\frac{t^{2}}{2}]}_{0}^{π} = \frac{π^{2}}{4} - \frac{1}{4 π} \frac{π^{3}}{3} = \frac{π^{2}}{4} - \frac{π^{2}}{12}

= \frac{3 π^{2} - π^{2}}{12} = \frac{π^{2}}{6} .