developp f(x)=e^(−αx) 2π periodic at Fourier serie with α>0.


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Question Number 28683 by abdo imad last updated on 28/Jan/18

$d e v e l o p p f (x) = e^{- α x} 2 π p e r i o d i c a t F o u r i e r s e r i e w i t h$ $α > 0 .$
Commented byabdo imad last updated on 31/Jan/18

$f (x) = \sum_{n = - \infty}^{+ \infty} c_{n} e^{i n x} a n d c_{n} = \frac{1}{T} \int_{[T]} f (x) e^{- i n x} d x$ $= \frac{1}{2 π} \int_{- π}^{π} e^{- α x} e^{- i n x} d x \Rightarrow 2 π c_{n} = \int_{- π}^{π} e^{- (α + i n) x} d x$ $= \frac{- 1}{α + i n} {[e^{- (α + i n) x}]}_{- π}^{π} = \frac{- 1}{α + i n} (e^{- (α + i n) π} - e^{(α + i n) π})$ $= \frac{- 1}{α + i n} ({(- 1)}^{n} e^{- α π} - {(- 1)}^{n} e^{α π}) = \frac{{(- 1)}^{n}}{α + i n} (e^{α π} - e^{- α π})$ $= \frac{2 {(- 1)}^{n}}{α + i n} s h (α π) \Rightarrow c_{n} = \frac{s h (α π)}{π} \frac{{(- 1)}^{n}}{α + i n} s o$ $f (x) = \sum_{n = - \infty}^{+ \infty} \frac{s h (α π)}{π} \frac{{(- 1)}^{n}}{α + i n} e^{i n x} .$
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