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Question Number 29975 by abdo imad last updated on 14/Feb/18

l e t g i v e 0 < α < 1

1) p r o v e t h a t π c o t h (π α) - \frac{1}{α} = \sum_{n = 1}^{\infty} \frac{2 α}{α^{2} + n^{2}} .

2) b y i n t e g r a t i o n o n [0, 1] f i n d \prod_{n = 1}^{\infty} (1 + \frac{1}{n^{2}}) .

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

1) l e t d e v e l o p p t h e 2 π p e r i o d i c f u n c t i o n f (x) = c h (α x)

f (x) = \frac{a_{0}}{2} + \sum_{n = 1}^{\infty} a_{n} c o s (n x)

a_{n} = \frac{2}{T} \int_{[T]} f (x) c o s (n x) d x = \frac{2}{2 π} \int_{- π}^{π} c h (α x) c o s (n x) d x

= \frac{2}{π} \int_{0}^{π} c h (α x) c o s (n x) d x \Rightarrow \frac{π}{2} a_{n} = \int_{0}^{π} \frac{e^{α x} + e^{- α x}}{2} c o s (n x) d x

π a_{n} = \int_{0}^{π} e^{α x} c o s (n x) d x + \int_{0}^{π} e^{- α x} c o s (n x) d x = I (α) + I (- α)

I (α) = R e (\int_{0}^{π} e^{α x + i n x} d x) = R e (\int_{0}^{π} e^{(α + i n) x} d x) b u t

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

= \frac{α - i n}{α^{2} + n^{2}} ({(- 1)}^{n} e^{α π} - 1) \Rightarrow I (α) = \frac{α ({(- 1)}^{n} e^{α π} - 1)}{α^{2} + n^{2}}

I (- α) = \frac{- α ({(- 1)}^{n} e^{- α π} - 1)}{α^{2} + n^{2}}

π a_{n} = \frac{α {(- 1)}^{n} (e^{α π} - e^{- α π})}{α^{2} + n^{2}} = \frac{2 α {(- 1)}^{n} s h (α π)}{α^{2} + n^{2}} \Rightarrow

a_{n} = \frac{2}{π} α s h (α π) \frac{{(- 1)}^{n}}{α^{2} + n^{2}} a n d a_{0} = \frac{2}{π} \frac{s h (α π)}{α} \Rightarrow

c h (α x) = \frac{s h (α π)}{π α} + \frac{2 α s h (α π)}{π} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{α^{2} + n^{2}} c o s (n x)

x = π \Rightarrow c h (π α) = \frac{s h (α π)}{π α} + \frac{2 α s h (α π)}{π} \sum_{n = 1}^{\infty} \frac{1}{α^{2} + n^{2}} \Rightarrow

c o t h (π α) = \frac{1}{π α} + \frac{2 α}{π} \sum_{n = 1}^{\infty} \frac{1}{α^{2} + n^{2}} \Rightarrow

π c o t h (π α) - \frac{1}{α} = \sum_{n = 1}^{\infty} \frac{2 α}{α^{2} + n^{2}} .

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

w e h a v e \int_{0}^{1} (π c o t h (π α) - \frac{1}{α}) d α = \sum_{n = 1}^{\infty} \int_{0}^{1} \frac{2 α}{α^{2} + n^{2}} d α

= \sum_{n = 1}^{\infty} {[l n (α^{2} + n^{2})]}_{0}^{1} = \sum_{n = 1}^{\infty} l n (1 + n^{2}) - l n (n^{2}) = \sum_{n = 1}^{\infty} l n (1 + \frac{1}{n^{2}})

= l n (\prod_{n = 1}^{\infty} (1 + \frac{1}{n^{2}})) \Rightarrow

\prod_{n = 1}^{\infty} (1 + \frac{1}{n^{2}}) = e^{\int_{0}^{1} (π c o t h (π α) - \frac{1}{α}) d α} b u t c h . α π = t g i v e

\int_{0}^{1} \frac{α π c o t h (α π) - 1}{α} d α = \int_{0}^{π} \frac{t c o t h t - 1}{\frac{t}{π}} \frac{d t}{π}

= \int_{0}^{π} \frac{t c o t h (t) - 1}{t} d t = \int_{0}^{π} (\frac{e^{t} - e^{- t}}{e^{t} + e^{- t}} - \frac{1}{t}) d t \dots . b e c o n t i n u e d \dots

π