let-f-x-1-1-sinx-2pi-periodic-even-developp-f-at-fourier-serie-

Question Number 39039 by maxmathsup by imad last updated on 01/Jul/18

l e t f (x) = \frac{1}{1 + ∣ s i n x ∣} (2 π p e r i o d i c e v e n)

d e v e l o p p f a t f o u r i e r s e r i e .

Commented by abdo mathsup 649 cc last updated on 05/Jul/18

f (x) = \frac{a_{0}}{2} + \sum_{n = 1}^{\infty} a_{n} c o s (n x) w i t h

a_{n} = \frac{2}{T} \int_{[T]} f (x) c o s (n x) d x

= \frac{2}{2 π} \int_{- π}^{π} \frac{1}{1 + ∣ s i n x ∣} c o s (n x) d x

= \frac{2}{π} \int_{0}^{π} \frac{c o s (n x)}{1 + s i n x} d x \Rightarrow \frac{π}{2} a_{n} = \int_{0}^{π} \frac{c o s (n x)}{1 + s i n x} d x

b u t \int_{0}^{π} \frac{c o s (n x)}{1 + s i n x} d x = R e (\int_{0}^{π} \frac{e^{i n x}}{1 + s i n x} d x)

c h a n g e m e n t x = 2 t g i v e

\int_{0}^{π} \frac{e^{i n x}}{1 + s i n x} d x = \int_{0}^{2 π} \frac{2 e^{i 2 n t}}{1 + s i n (2 t)} d t a l s o

c h a n g . e^{i t} = z g i v e

\int_{0}^{2 π} \frac{2 e^{i 2 n t}}{1 + s i n (2 t)} d t = \int_{∣ z ∣= 1} \frac{2 z^{2 n}}{1 + \frac{z^{2} - z^{- 2}}{2 i}} \frac{d z}{i z}

= \int_{∣ z ∣= 1} \frac{4 i z^{2 n}}{i z (2 i + z^{2} - z^{- 2})} d z

= \int_{∣ z ∣= 1} \frac{4 z^{2 n - 1}}{2 i + z^{2} - \frac{1}{z^{2}}} d z

= \int_{∣ z ∣= 1} \frac{4 z^{2 n + 1}}{2 {i z}^{2} + z^{4} - 1} d z l e t

φ (z) = \frac{4 z^{2 n + 1}}{z^{4} + 2 {i z}^{2} - 1}

φ (z) = \frac{4 z^{2 n + 1}}{{(z^{2} + i)}^{2}} = \frac{4 z^{2 n + 1}}{{(z - \sqrt{- i})}^{2} {(2 + \sqrt{- i})}^{2}}

= \frac{4 z^{2 n + 1}}{{(z - e^{- \frac{i π}{4}})}^{2} {(z + e^{- \frac{i π}{4}})}^{2}} s o φ h a v e a d o u b l e

p o l e s \bar{+} e^{- \frac{i π}{4}}

\int_{∣ z ∣= 1} φ (2) d z = 2 i π {R e s (φ, e^{- \frac{i π}{4}}) + R e s (φ, - e^{- \frac{i π}{4}})}

Commented by abdo mathsup 649 cc last updated on 05/Jul/18

R e s (φ, e^{- \frac{i π}{4}}) = {l i m}_{z \to e^{- \frac{i π}{4}}} {{(z - e^{- \frac{i π}{4}})}^{2} φ (z)}^{(1)}

= {l i m}_{z \to e^{- \frac{i π}{4}}} {\frac{4 z^{2 n + 1}}{{(z + e^{- \frac{i π}{4}})}^{2}}}^{(1)}

= {l i m}_{z \to e^{- \frac{i π}{4}}} 4 \frac{(2 n + 1) z^{2 n} {(z + e^{- \frac{i π}{4}})}^{2} - 2 (z + e^{- \frac{i π}{4}}) z^{2 n + 1}}{{(z + e^{- \frac{i π}{4}})}^{4}}

= {l i m}_{z \to e^{- \frac{i π}{4}}} 4 \frac{(2 n + 1) z^{2 n} (z + e^{- \frac{i π}{4}}) - 2 z^{2 n + 1}}{{(z + e^{- \frac{i π}{4}})}^{3}}

= 4 \frac{(2 n + 1) e^{- \frac{i n π}{2}} (2 e^{- \frac{i π}{4}}) - 2 e^{- i \frac{(2 n + 1) π}{4}}}{{(2 e^{- \frac{i π}{4}})}^{3}}

= 8 \frac{(2 n + 1) e^{- i (\frac{n}{2} + \frac{1}{4}) π} - e^{- i \frac{(2 n + 1) π}{4}}}{8 e^{- i \frac{3 π}{4}}}

= \frac{2 n e^{- i \frac{(2 n + 1) π}{4}}}{e^{- i \frac{3 π}{4}}} = 2 n e^{- i {\frac{2 n + 1}{4} + \frac{3}{4}} π}

= 2 n e^{- i \frac{(n + 2) π}{2}} = - 2 n e^{- \frac{i n π}{2}}

= - 2 n {c o s (\frac{n π}{2}) - i s i n (\frac{n π}{2})}