sin-2-1-3-sin-2-2-2-3-sin-3-2-3-3-pi-b-1-a-b-pi-b-Find-a-b-

Question Number 136070 by Dwaipayan Shikari last updated on 18/Mar/21

\frac{s i n (\sqrt{2})}{1^{3}} + \frac{s i n (2 \sqrt{2})}{2^{3}} + \frac{s i n (3 \sqrt{2})}{3^{3}} + \dots = \frac{π^{b} + 1}{a \sqrt{b}} - \frac{π}{b}

F i n d a - b

Answered by mnjuly1970 last updated on 18/Mar/21

Ω (x) = \underset{n = 1}{\sum^{\infty}} \frac{s i n (n x)}{n^{3}}

- I m (l n (1 - e^{i x})) = \sum_{n = 1} = \infty \frac{s i n (n x)}{n}

- I m l n ((1 - c o s (x)) - i s i n (x)) = \underset{n = 1}{\sum^{\infty}} \frac{s i n (n x)}{n}

- I m {l n (2 s i n (\frac{x}{2}) - {i t a n}^{- 1} (\frac{s i n (x)}{1 - c o s (x)}) = \underset{n = 1}{\sum^{\infty}} \frac{s i n (n x)}{n = 1}

\frac{π}{2} - \frac{x}{2} = \underset{n = 1}{\sum^{\infty}} \frac{s i n (n x)}{n}

\frac{π}{2} x - \frac{x^{2}}{4} = - \underset{n = 1}{\sum^{\infty}} \frac{c o s (n x)}{n^{2}} + C

x = π

\frac{π^{2}}{4} = \frac{π^{2}}{12} + C \Rightarrow C = \frac{π^{2}}{6}

\frac{π^{2}}{6} + \frac{x^{2}}{4} - \frac{π}{2} x = \underset{n = 1}{\sum^{\infty}} \frac{c o s (n x)}{n^{2}}

\frac{π^{2}}{6} x + \frac{x^{3}}{12} - \frac{π}{4} x^{2} = \underset{n = 1}{\sum^{\infty}} \frac{s i n (n x)}{n^{3}}

x = \sqrt{2}

\underset{n = 1}{\sum^{\infty}} \frac{s i n (n \sqrt{2})}{n^{3}} = \frac{π^{2}}{6} \sqrt{2} + \frac{\sqrt{2}}{6} - \frac{π}{2}

Ω = \frac{π^{2} + 1}{3 \sqrt{2}} - \frac{π}{2}

h e n c e :: a - b = 3 - 2 = 1

p l e a s e c h e c k m y s o l u t i o n \dots .

Commented by Dwaipayan Shikari last updated on 18/Mar/21

G r e a t s i r!

Commented by mnjuly1970 last updated on 18/Mar/21

t h a n k y o u f o r y o u r n i c e q u e s t i o n s \dots