calculate-h-a-cos-at-ch-t-2-dt-

Question Number 54374 by maxmathsup by imad last updated on 02/Feb/19

c a l c u l a t e h (a) = \int_{- \infty}^{+ \infty} \frac{c o s (a t)}{c h (\frac{t}{2})} d t .

Commented by Abdo msup. last updated on 08/Feb/19

h (a) =_{\frac{t}{2} = x} 2 \int_{- \infty}^{+ \infty} \frac{c o s (2 a x)}{c h (x)} d x

= 4 \int_{0}^{\infty} \frac{c o s (2 a x)}{\frac{e^{x} + e^{- x}}{2}} d x = 8 \int_{0}^{\infty} \frac{e^{- x} c o s (2 a x)}{1 + e^{- 2 x}} d x

= 8 \int_{0}^{\infty} e^{- x} c o s (2 a x) (\sum_{n = 0}^{\infty} {(- 1)}^{n} e^{- 2 n x}) d x

= 8 \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{0}^{\infty} e^{- (2 n + 1) x} c o s (2 a x) d x

l e t A_{n} = \int_{0}^{\infty} e^{- (2 n + 1) x} c o s (2 a x) d x \Rightarrow

A_{n} = R e (\int_{0}^{\infty} e^{- (2 n + 1) x} e^{2 i a x} d x) b u t

\int_{0}^{\infty} e^{(- (2 n + 1) + 2 i a) x} d x = {[\frac{1}{- (2 n + 1) + 2 i a} e^{(- (2 n + 1) + 2 i a) x}]}_{0}^{+ \infty}

= \frac{1}{- (2 n + 1) + 2 i a} = - \frac{1}{(2 n + 1) - 2 i a}

= - \frac{2 n + 1 + 2 i a}{{(2 n + 1)}^{2} + 4 a^{2}} \Rightarrow A_{n} = - \frac{2 n + 1}{{(2 n + 1)}^{2} + 4 a^{2}} \Rightarrow

h (a) = 8 \sum_{n = 0}^{\infty} {(- 1)}^{n + 1} \frac{2 n + 1}{{(2 n + 1)}^{2} + 4 a^{2}}

a n d t h i s s e r i e c a n b e c a l c u l a t e d b y f o u r i e r s e r i e s

\dots b e c o n t i n u e d \dots