1-calculate-u-n-0-sin-nx-sh-2x-dx-with-n-integr-natural-2-calculate-n-0-u-n-

Question Number 47064 by maxmathsup by imad last updated on 04/Nov/18

1) c a l c u l a t e u_{n} = \int_{0}^{\infty} \frac{s i n (n x)}{s h (2 x)} d x w i t h n i n t e g r n a t u r a l

2) c a l c u l a t e \sum_{n = 0}^{\infty} u_{n} .

Commented by maxmathsup by imad last updated on 07/Nov/18

1) w e h a v e u_{n} = \int_{0}^{\infty} \frac{2 s i n (n x)}{e^{2 x} - e^{- 2 x}} d x = 2 \int_{0}^{\infty} \frac{e^{- 2 x} s i n (n x)}{1 - e^{- 4 x}} d x

= 2 \int_{0}^{\infty} e^{- 2 x} s i n (n x) (\sum_{p = 0}^{\infty} e^{- 4 p x}) e x

= 2 \sum_{p = 0}^{\infty} \int_{0}^{\infty} e^{- (2 + 4 p) x} s i n (n x) d x

=_{(2 + 4 p) x = u} 2 \sum_{p = 0}^{\infty} \int_{0}^{\infty} e^{- u} s i n (n \frac{u}{2 + 4 p}) \frac{d u}{2 + 4 p}

= \sum_{p = 0}^{\infty} \frac{1}{2 p + 1} \int_{0}^{\infty} e^{- u} s i n (\frac{n u}{4 p + 2}) d u l e t d e t e r m i n e I_{λ} = \int_{0}^{\infty} e^{- u} s i n (λ u) d u (λ > 0)

I_{λ} = I m (\int_{0}^{\infty} e^{- u + i λ u} d u) = I m (\int_{0}^{\infty} e^{(- 1 + i λ) u} d u) b u t

\int_{0}^{\infty} e^{(- 1 + i λ) u} d u = {[\frac{1}{- 1 + i λ} e^{(- 1 + i λ) u}]}_{0}^{+ \infty} = \frac{- 1}{- 1 + i λ} = \frac{1}{1 - i λ} = \frac{1 + i λ}{1 + λ^{2}} \Rightarrow

I_{λ} = \frac{λ}{1 + λ^{2}} \Rightarrow u_{n} = \sum_{p = 0}^{\infty} \frac{1}{2 p + 1} (\frac{n}{(4 p + 2) (1 + {(\frac{n}{4 p + 2})}^{2})})

= \sum_{p = 0}^{\infty} \frac{n}{(2 p + 1) (4 p + 2 + \frac{n^{2}}{4 p + 2})} = \sum_{p = 0}^{\infty} \frac{n (4 p + 2)}{(2 p + 1) ({(4 p + 2)}^{2} + n^{2})} = \sum_{p = 0}^{\infty} \frac{2 n}{{(4 p + 2)}^{2} + n^{2}}

u_{n} 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 \dots . b e c o n t i n u e d \dots .