1-sin-x-x-

Question Number 122883 by kolos last updated on 20/Nov/20

\sum_{1}^{\infty} (\frac{\sin (x)}{x}) = ?

Commented by Dwaipayan Shikari last updated on 20/Nov/20

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

= \frac{1}{2 i} (- l o g (1 - e^{i}) + l o g (1 - e^{- i})) = \frac{1}{2 i} l o g (\frac{1 - e^{- i}}{1 - e^{i}})

= \frac{1}{2 i} l o g (- e^{- i}) = \frac{1}{2 i} l o g (e^{i π}) + \frac{1}{2 i} ({l o g e}^{- i}) = \frac{π}{2} - \frac{1}{2}

Commented by kolos last updated on 21/Nov/20

t h a n k s o m u c h, e v e r y s i n g l e o n e

Answered by TANMAY PANACEA last updated on 20/Nov/20

i t h i n k

q = \underset{x = 1}{\sum^{\infty}} \frac{s i n x}{x}

p = \underset{x = 1}{\sum^{\infty}} \frac{c o s x}{x}

p + i q = \underset{x = 1}{\sum^{\infty}} \frac{e^{i x}}{x} = \frac{e^{i}}{1} + \frac{e^{2 i}}{2} + \frac{e^{3 i}}{3} + \dots \infty

p + i q = - l n (1 - e^{i}) = - l n (1 - c o s 1 - i s i n 1)

p + i q = - l n (2 {s i n}^{2} \frac{1}{2} - i \times 2 s i n \frac{1}{2} x c o s \frac{1}{2})

p + i q = - l n {- 2 s i n \frac{1}{2} (s i n \frac{1}{2} + i c o s \frac{1}{2})}

p + i q = - l n {2 s i n (π + \frac{1}{2}) (c o s (\frac{π}{2} - \frac{1}{2}) + i s i n (\frac{π}{2} - \frac{1}{2})}

p + i a q = - l n (2 s i n (π + \frac{1}{2})) - l n (e^{i (\frac{π}{2} - \frac{1}{2})}

p + i q = - l n (2 s i n (π + \frac{1}{2}) - i (\frac{π}{2} - \frac{1}{2})

q = - (\frac{π - 1}{2})

l n (1 - t) = - t - \frac{t^{2}}{2} - \frac{t^{3}}{3} - \frac{t^{4}}{4} \dots \infty

- l n (1 - t) = t + \frac{t^{2}}{2} + \frac{t^{3}}{3} + . . \infty

Answered by Dwaipayan Shikari last updated on 20/Nov/20

I n t h e s a m e w a y

\underset{n = 1}{\sum^{\infty}} \frac{c o s x}{x} = \frac{1}{2} \underset{n = 1}{\sum^{\infty}} \frac{e^{i x}}{x} + \frac{e^{- i x}}{x} = \frac{1}{2} (l o g (\frac{1}{(1 - e^{i}) (1 - e^{- i})})

= \frac{1}{2} l o g (\frac{1}{1 - (e^{i} + e^{- i}) + 1}) = \frac{1}{2} l o g (\frac{1}{2 - 2 c o s (1)})

Answered by mathmax by abdo last updated on 21/Nov/20

i think the Q is \sum_{n = 1}^{\infty} \frac{\sin (n)}{n}

let S (x) = \sum_{n = 1}^{\infty} \frac{\sin (nx)}{n} \Rightarrow S (x) = Im (\sum_{n = 1}^{\infty} \frac{e^{inx}}{n})

w (x) = \sum_{n = 1}^{\infty} \frac{e^{inx}}{n} \Rightarrow w^{^{'}} (x) = \sum_{n = 1}^{\infty} \frac{in e^{inx}}{n} = i \sum_{n = 1}^{\infty} {(e^{ix})}^{n}

= i \times \frac{1}{1 - e^{ix}} \Rightarrow w (x) = - \ln (1 - e^{ix}) + c with w) 0) = 0 = c \Rightarrow

w (x) = - \ln (1 - e^{ix}) \Rightarrow S (x) = Im (- \ln (1 - e^{ix})) we have

\ln (1 - e^{ix}) = \ln (1 - cosx - isinx) = \ln ({2 \sin}^{2} (\frac{x}{2}) - 2 isin (\frac{x}{2}) \cos (\frac{x}{2}))

= \ln (2) + \ln (- isin (\frac{x}{2}) (\cos (\frac{x}{2}) + isin (\frac{x}{2})) = \ln (2) + \ln (- i) + \ln (\sin (\frac{x}{2})) + \ln (e^{\frac{ix}{2}})

= \ln (2) - \frac{i π}{2} + \ln (\sin (\frac{x}{2})) + \frac{ix}{2} = \ln (2 \sin (\frac{x}{2})) + \frac{x - π}{2} i \Rightarrow

S (x) = \frac{π - x}{2} \Rightarrow \sum_{n = 1}^{\infty} \frac{\sin (n)}{n} = S (1) = \frac{π - 1}{2}

Answered by mnjuly1970 last updated on 21/Nov/20

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

= \frac{1}{2 i} [- l n (1 - e^{i}) + l n (1 - e^{- i})]

= \frac{1}{2 i} l n (\frac{1 - e^{- i}}{1 - e^{i}}) = \frac{1}{2 i} l n (- e^{- i})

= \frac{1}{2 i} [i π - i] = \frac{π - 1}{2}