n-1-sin-2n-1-2n-1-2-2-log-2-sin-log-sin-4-2-F-1-1-2-1-2-3-2-sin-2-sin-16-2-F-1-1-2-1-2-1-2-3-2-3-2-sin-2-Prove-or-disprove-0-lt-lt-pi-

Question Number 136420 by Dwaipayan Shikari last updated on 21/Mar/21

\underset{n = 1}{\sum^{\infty}} \frac{s i n (2 n + 1) θ}{{(2 n + 1)}^{2}} = \frac{θ}{2} l o g (2) + s i n θ \frac{l o g (s i n θ)}{4}_{2} F_{1} (\frac{1}{2}, \frac{1}{2}; \frac{3}{2}; {s i n}^{2} θ) + \frac{s i n θ}{16}_{2} F_{1} (\frac{1}{2}, \frac{1}{2}, \frac{1}{2}; \frac{3}{2}; \frac{3}{2}; {s i n}^{2} θ)

P r o v e o r d i s p r o v e 0 < θ < π

Answered by mindispower last updated on 21/Mar/21

θ = π, s i n (2 n + 1) θ = 0

l e f t = 0

r i g h t = \frac{π l n (2)}{2} + \frac{1}{4} + \frac{1}{16}

Commented by Dwaipayan Shikari last updated on 21/Mar/21

S o r r y i h a v e e d i t e d t h e q u e s t i o n

Commented by Dwaipayan Shikari last updated on 21/Mar/21

V a l i d f o r 0 < θ < π

Commented by mindispower last updated on 22/Mar/21

o k i w i l l t r y b u t t c h e k i t a g a i n

Answered by mindispower last updated on 22/Mar/21

\sum_{n ⩾ 0} \frac{s i n (2 n + 1) θ}{{(2 n + 1)}^{2}} \int = S = \sum_{n ⩾ 1} \frac{s i n (n θ)}{n^{2}} - \sum_{n ⩾ 1} \frac{s i n (2 θ)}{4 n^{2}}

r e c a l \sum_{n ⩾ 1} \frac{s i n (n θ)}{n^{2}} = {C l}_{2} (θ), c l a u s e n f u n c t i o n

w e g e t S = {C l}_{2} (θ) - \frac{{C l}_{2} (2 θ)}{4}

c l a u s e n i n t e g r a t i o n r e p r e s e n t a t i o n

{C l}_{2} (θ) = - 2 \int_{0}^{\frac{θ}{2}} l n (2 s i n (t)) d t

= u = s i n (t) = - 2 \int_{0}^{s i n (\frac{θ}{2})} \frac{l n (2 u)}{\sqrt{1 - u^{2}}} d u = {c l}_{2} (θ)

{(1 + x)}^{a} = 1 + \sum_{n ⩾ 1} \frac{\prod_{k ⩽ n - 1} (a - k)}{n!} . x^{n}

{c l}_{2} (θ) = - 2 \int_{0}^{s i n (\frac{θ}{2})} (1 + \sum_{n ⩾ 1} {(- 1)}^{n} \frac{\prod_{k ⩽ n - 1} (- \frac{1}{2} - k) u^{2 n}}{n!}) l n (2 u) d u

\int u^{2 n} l n (2 u) d u = \frac{u^{2 n + 1}}{2 n + 1} l n (2 u) - \frac{u^{2 n + 1}}{{(2 n + 1)}^{2}}

i t s e e m s g o o d s t a r t e w e c a n f i n_{p} F_{q}

s i n l o g (s i n . .) a l s o i t h i n k v e r r y l o n g w o r c k

= - 2 (1 + \sum_{n ⩾ 1} \frac{{(\frac{1}{2})}_{n}}{n!}) (\frac{{s i n}^{2 n + 1} (\frac{θ}{2}) l n (2 s i n (\frac{θ}{2}))}{(2 n + 1)} - \frac{{s i n}^{2 n + 1} (\frac{θ}{2})}{{(2 n + 1)}^{2}})

= - 2 (1 + \sum_{n ⩾ 1} \frac{{(\frac{1}{2})}_{n} {(\frac{1}{2})}_{n}}{2 . {(\frac{3}{2})}_{n}}

Commented by Dwaipayan Shikari last updated on 22/Mar/21

T h a n k s s i r! I w i l l p o s t m y w o r k l a t e r

Commented by mindispower last updated on 22/Mar/21

y o u a r e w e l c o m