calculate-0-sin-x-sh-2x-dx-

Question Number 43906 by abdo.msup.com last updated on 17/Sep/18

c a l c u l a t e \int_{0}^{\infty} \frac{s i n (x)}{s h (2 x)} d x

Commented by maxmathsup by imad last updated on 20/Sep/18

l e t A = \int_{0}^{\infty} \frac{s i n x}{s h (2 x)} d x \Rightarrow A = \int_{0}^{\infty} \frac{2 s i n x}{e^{2 x} - e^{- 2 x}} d x

= \int_{0}^{\infty} \frac{e^{- 2 x} s i n x}{1 - e^{- 4 x}} d x = \int_{0}^{\infty} e^{- 2 x} (\sum_{n = 0}^{\infty} e^{- 4 n x}) s i n x d x

= \sum_{n = 0}^{\infty} \int_{0}^{\infty} e^{- (4 n + 2) x} s i n x d x = \sum_{n = 0}^{\infty} A_{n} a n d A_{n} = I m (\int_{0}^{\infty} e^{- (4 n + 2 - i) x} d x) b u t

\int_{0}^{\infty} e^{- (4 n + 2 + i) x} d x = {[- \frac{1}{4 n + 2 - i} e^{- (4 n + 2 + i) x}]}_{0}^{+ \infty} = \frac{1}{4 n + 2 - i}

\frac{4 n + 2 + i}{{(4 n + 2)}^{2} + 1} \Rightarrow A_{n} = \frac{1}{{(4 n + 2)}^{2} + 1} \Rightarrow A = \sum_{n = 0}^{\infty} \frac{1}{4 {(2 n + 1)}^{2} + 1}

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