find α and β from R /∫_0 ^π (αt^2 +βt)cos(nt)dt= (1/n^2 ) for all number n from N^(∗ ) then find Σ


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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
$let put I= ∫_0 ^π (αt^2 +βt)cos(nt)dt I= Re ( ∫_0 ^π (αt^2 +βt)^ e^(int) t) by parts u= αt^2 +βt and v^′ = e^(int) I= [(αt^2 +βt)(e^(int) /(in)) ]_0 ^π − (1/(in))∫_0 ^π (2αt+β)e^(int) dt =(1/(in))((απ^2 +βπ)(−1)^n ) −(1/(in)){[ (1/(in))(2αt+β) e^(int) ]_0 ^π −(1/(in))∫_0 ^π (2α)e^(int) dt} we find I= (1/n^2 ) ((2απ +β)(−1)^n −β) I = (1/n^2 ) ∀n∈N^∗ ⇔ (2απ+β)(−1)^n −β= 1 for alln ⇔ β =−1 and 2απ−1=0 ⇒ α=(1/(2π)) and β=−1 so (1/n^2 )= ∫_0 ^π ((1/(2π)) t^2 −t)cos(nt)dt Σ_(n=1) ^∝ (1/n^2 ) =∫_0 ^π ((1/(2π))t^2 −t)( Σ_(n=1) ^∝ cos(nt))dt but Σ_(n=1) ^∝ cos(nt)=Re( Σ_(n=0) ^∝ (^ e^(it) )^n )−1=Re( (1/(1−e^(it) )))−1 =Re( (1/(2sin^2 ((t/2))−2isin((t/2))cos((t/2))))) =Re ( (1/(−2isin((t/2)) e^(i(t/2)) ))) =Re( (1/2)i((cos((t/2))−i sin((t/2)))/(sin((t/2))))) = (1/2)−1=−(1/2) Σ_(n=1) ^∝ (1/n^2 ) =−(1/2) ∫_0 ^π ((1/(2π))t^2 −t)dt =−(1/(4π)) [ (t^3 /3)]_0 ^π +(1/2) [ (t^2 /2)]_0 ^π = (π^2 /4)−(1/(4π)) (π^3 /3)=(π^2 /4) −(π^2 /(12)) = ((3π^2 −π^2 )/(12)) = (π^2 /6).$
$l e t p u t I = \int_{0}^{π} (α t^{2} + β t) c o s (n t) d t$ $I = R e (\int_{0}^{π} {(α t^{2} + β t)}^{} e^{i n t} t) b y p a r t s$ $u = α t^{2} + β t a n d v^{^{'}} = e^{i n t}$ $I = {[(α t^{2} + β t) \frac{e^{i n t}}{i n}]}_{0}^{π} - \frac{1}{i n} \int_{0}^{π} (2 α t + β) e^{i n t} d t$ $= \frac{1}{i n} ((α π^{2} + β π) {(- 1)}^{n}) - \frac{1}{i n} {{[\frac{1}{i n} (2 α t + β) e^{i n t}]}_{0}^{π} - \frac{1}{i n} \int_{0}^{π} (2 α) e^{i n t} d t}$ $w e f i n d I = \frac{1}{n^{2}} ((2 α π + β) {(- 1)}^{n} - β)$ $I = \frac{1}{n^{2}} \forall n \in N^{*} \Leftrightarrow (2 α π + β) {(- 1)}^{n} - β = 1 f o r a l l n$ $\Leftrightarrow β = - 1 a n d 2 α π - 1 = 0 \Rightarrow α = \frac{1}{2 π} a n d β = - 1 s o$ $\frac{1}{n^{2}} = \int_{0}^{π} (\frac{1}{2 π} t^{2} - t) c o s (n t) d t$ $\sum_{n = 1}^{\propto} \frac{1}{n^{2}} = \int_{0}^{π} (\frac{1}{2 π} t^{2} - t) (\sum_{n = 1}^{\propto} c o s (n t)) d t b u t$ $\sum_{n = 1}^{\propto} c o s (n t) = R e (\sum_{n = 0}^{\propto} {(^{} e^{i t})}^{n}) - 1 = R e (\frac{1}{1 - e^{i t}}) - 1$ $= R e (\frac{1}{2 {s i n}^{2} (\frac{t}{2}) - 2 i s i n (\frac{t}{2}) c o s (\frac{t}{2})})$ $= R e (\frac{1}{- 2 i s i n (\frac{t}{2}) e^{i \frac{t}{2}}}) = R e (\frac{1}{2} i \frac{c o s (\frac{t}{2}) - i s i n (\frac{t}{2})}{s i n (\frac{t}{2})})$ $= \frac{1}{2} - 1 = - \frac{1}{2}$ $\sum_{n = 1}^{\propto} \frac{1}{n^{2}} = - \frac{1}{2} \int_{0}^{π} (\frac{1}{2 π} t^{2} - t) d t$ $= - \frac{1}{4 π} {[\frac{t^{3}}{3}]}_{0}^{π} + \frac{1}{2} {[\frac{t^{2}}{2}]}_{0}^{π} = \frac{π^{2}}{4} - \frac{1}{4 π} \frac{π^{3}}{3} = \frac{π^{2}}{4} - \frac{π^{2}}{12}$ $= \frac{3 π^{2} - π^{2}}{12} = \frac{π^{2}}{6} .$
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