find-n-1-sin-n-n-x-n-with-1-lt-x-lt-1-

Question Number 29973 by abdo imad last updated on 14/Feb/18

f i n d \sum_{n = 1}^{\infty} \frac{s i n (n α)}{n} x^{n} w i t h - 1 < x < 1 .

Commented by abdo imad last updated on 16/Feb/18

l e t p u t S (x) = \sum_{n = 1}^{\infty} \frac{s i n (n α)}{n} x^{n} d u e t o u n i f o r m c o n v e r g e n c e

w e h a v e S^{^{'}} (x) = \sum_{n = 1}^{\infty} s i n (n α) x^{n - 1} = \sum_{n = 0}^{\infty} s i n ((n + 1) α) x^{n}

= I m (\sum_{n = 0}^{\infty} e^{i (n + 1) α} x^{n}) = I m (e^{i α} \sum_{n = 0}^{\infty} {(x e^{i α})}^{n}) w e h a v e

∣ x e^{i α} ∣< 1 \Rightarrow e^{i α} \sum_{n = 0}^{\infty} {({x e}^{i α})}^{n} = e^{i α} \frac{1}{1 - {x e}^{i α}} = \frac{1}{e^{- i α} - x}

= \frac{1}{c o s α - i s i n α - x} = \frac{1}{c o s α - x - i s i n α}

= \frac{c o s α - x + i s i n α}{{(c o s α - x)}^{2} + {s i n}^{2} α} \Rightarrow I m (Σ (\dots)) = \frac{s i n α}{{(x - c o s α)}^{2} + {s i n}^{2} α} \Rightarrow

S (x) = \int_{0}^{x} \frac{s i n α}{{(t - c o s α)}^{2} + {s i n}^{2} α} d t + λ b u t λ = S (0) = 0

t h e c h . t - c o s α = s i n α u g i v e

S (x) = \int_{- c o t a n α}^{\frac{x - c o s α}{s i n α}} \frac{s i n α}{{s i n}^{2} u^{2} + {s i n}^{2} α} s i n α d u

= \int_{- c o t a n α}^{\frac{x - c o s α}{s i n α}} \frac{d u}{1 + u^{2}} = {[a r c t a n u]}_{- c o t a n α}^{\frac{x - c o s α}{s i n α}}

= a r t a n (\frac{x - c o s α}{s i n α}) + a r t a n (c o t a n α) .